SELINA Solution Class 9 Chapter 13 Pythagoras Theorem [Proof and simple applications with converse] Exercise 13B

Question 1

In the figure, given below, AD ⊥ BC.
Prove that: c² = a² + b² - 2ax.

Sol:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔABD and applying Pythagoras theorem we get,
AB² = AD²   + BD² 
c²  = h²  + ( a - x )²  
h²  = c²  - ( a - x )²                      ......(i)
First, we consider the ΔACD and applying Pythagoras theorem we get,
AC² = AD² + CD² 
b²  = h² + x²  
h²  = b² - x²                              ......(ii)

From (i) and (ii) we get,
c²  - ( a - x )² = b² - x²  
c²  - a² - x²  + 2ax = b² - x² 
c² = a² + b²  - 2ax
Hence Proved.

Question 2

In equilateral Δ ABC, AD ⊥ BC and BC = x cm. Find, in terms of x, the length of AD.

Sol:

In equilateral Δ ABC, AD ⊥ BC.
Therefore, BC = x cm.

Area of equilateral ΔABC = ${\displaystyle \frac{\sqrt{3}}{4}\times {\text{side}}^2 =\frac{1}{2}\times \text{base}\times \text{height}}$

= ${\displaystyle \frac{\sqrt{3}}{4}\times x^2=\frac{1}{2}\times x\times \text{AD}}$

AD = ${\displaystyle \frac{\sqrt{3}}{2}x}$

Question 3

ABC is a triangle, right-angled at B. M is a point on BC.
Prove that: AM² + BC² = AC² + BM².

Sol:

The pictorial form of the given problem is as follows,

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔABM and applying Pythagoras theorem we get,
AM² = AB  + BM² 
AB² = AM² - BM²               ......(i)
Now, we consider the ΔABC and applying Pythagoras theorem we get,
AC² = AB²  + BC² 
AB² = AC² - BC²                ......(ii)

From (i) and (ii) we get,
AM²  - BM²  = AC²  - BC² 
AM²  + BC² = AC² + BM²  
Hence Proved.

Question 4

M and N are the mid-points of the sides QR and PQ respectively of a PQR, right-angled at Q.
Prove that:
(i) PM² + RN² = 5 MN²
(ii) 4 PM² = 4 PQ² + QR²
(iii) 4 RN² = PQ² + 4 QR²(iv) 4 (PM² + RN²) = 5 PR²

Sol:

We draw, PM, MN, NR

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Since M and N are the mid-points of the sides QR and PQ respectively, therefore, PN = NQ, QM = RM

(i) First, we consider the ΔPQM, and applying Pythagoras theorem we get,
PM²= PQ² + MQ²
= ( PN + NQ )² + MQ²
= PN + NQ² + 2PN . NQ + MQ²   
= MN²+ PN² + 2PN.NQ  ...[From, ΔMNQ, MN² = NQ² + MQ²]  ......(i)

Now, we consider the ΔRNQ, and applying Pythagoras theorem we get,
RN² = NQ² + RQ² 
= NQ² + ( QM + RM )²
= NQ²  + QM²  + RM²  + 2QM .RM
= MN²  + RM²  + 2QM . RM   .......(ii)

Adding (i) and (ii) we get, 
PM²  + RN²  = MN²  + PN²  + 2PN.NQ + MN²  + RM²  + 2QM. RM
PM²  + RN²  = 2MN²  + PN²  + RM²  + 2PN.NQ + 2QM.RM 
PM²  + RN²  = 2MN²  + NQ²  + QM²  + 2(QN² ) + 2(QM² )
PM²  + RN²  = 2MN²  + MN²  + 2MN² 
PM ² + RN²  = 5MN²                 Hence Proved.

(ii) We consider the ΔPQM, and applying Pythagoras theorem we get,
PM² = PQ² + MQ²
4PM² = 4PQ² + 4MQ²             ...[ Multiply both sides by 4]
4PM² = 4PQ² + 4.( 1/2 QR )² ...[ MQ = ${\displaystyle \frac{1}{2}}$ QR ]
4PM² = 4PQ² + 4PQ + 4 . ${\displaystyle \frac{1}{2}}$ QR²
4PM² = 4PQ²  + QR² 
Hence Proved.

(iii) We consider the ΔRQN, and applying Pythagoras theorem we get,
RN² = NQ² + RQ²
4RN² = 4NQ² + 4QR²  ...[ Multiplying both sides by 4]
4RN² = 4QR² + 4 .(1/2 PQ)² ...[ NQ = ${\displaystyle \frac{1}{2}}$ PQ ]
4RN² = 4QR² + 4 .${\displaystyle \frac{1}{4}}$ PQ²
4RN² = PQ² + 4QR²
Hence Proved.

(iv) First, we consider the ΔPQM, and applying Pythagoras theorem we get,
PM² = PQ² + MQ²
= ( PN + NQ )² + MQ²
= PN² + NQ² + 2PN.NQ + MQ²   
= MN² + PN² + 2PN.NQ  ...[ From, ΔMNQ, = MN² = NQ² + MQ² ]  ......(i)

Now, we consider the ΔRNQ, and applying Pythagoras theorem we get,
RN² + NQ² + RQ² 
= NQ² + ( QM + RM )²
= NQ²  + QM²  + RM²  + 2QM .RM
=MN²  + RM²  + 2QM . RM .......(ii)

Adding (i) and (ii) we get,
PM²  + RN²  = MN²  + PN²  + 2PN . NQ + MN²  + RM²  + 2QM. RM
PM²  + RN²  = 2MN²  + PN²  + RM²  + 2PN . NQ + 2QM . RM 
PM²  + RN²  = 2MN²  + NQ²  + QM²  + 2(QN² ) + 2(QM² )
PM²  + RN²  = 2MN²  + MN²  + 2MN² 
PM ² + RN²  = 5MN² 
4( PM² + RN² ) = 4.5. (NQ² + MQ²)
4( PM² + RN² ) = 4.5. ${\displaystyle [{\left(\frac{1}{2}PQ\right)}^2+{\left(\frac{1}{2}RQ\right)}^2] ....[\because NQ=\frac{1}{2}PQ,MQ=\frac{1}{2}QR]}$
4 ( PM² + RN² ) = 5PR²
Hence Proved.

 

Question 5

In triangle ABC, ∠B = 90^o and D is the mid-point of BC.
Prove that: AC² = AD² + 3CD².

Sol:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

In triangle ABC, ∠B = 90^o and D is the mid-point of BC. Join AD. Therefore, BD = DC

First, we consider the ΔADB, and applying Pythagoras theorem we get,
AD² = AB² + BD²
AB² = AD² - BD²                 ....(i)

Similarly, we get from rt. angle triangles ABC we get,
AC² = AB² + BC²
AB² = AC² - BC²                .....(ii)
From (i) and (ii)
AC² - BC² = AD² - BD² 
AC² = AD² - BD² + BC²
AC² = AD² - CD² + 4CD²   ....[ BD = CD = ${\displaystyle \frac{1}{2}}$ BC ]
AC² = AD² + 3CD² 
Hence proved.

Question 6

In a rectangle ABCD,
prove that: AC² + BD² = AB² + BC² + CD² + DA².

Sol:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Since, ABCD is a rectangle angles A, B, C and D are rt. angles.

First, we consider the ΔACD, and applying Pythagoras theorem we get,
AC² = DA² + CD²                  ....(i)

Similarly, we get from rt. angle triangle BDC we get,
BD² = BC² + CD²
= BC² + AB²          ....[ In a rectangle, opposite sides are equal, ∴ CD = AB ] ...(ii)

Adding (i) and (ii)
AC² + BD² = AB² + BC² + CD² + DA²
Hence proved.

Question 7

In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
Prove that: 2AC² - AB² = BC² + CD² + DA²

Sol:

In quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
So, ΔABC and ΔADC are right-angled triangles.

In ΔABC, using Pythagoras theorem,
AC² = AB² + BC²
⇒ AB² = AC² - BC²                  ....(i)

In ΔADC, using Pythagoras theorem,
AC² = AD² + DC²                   ....(ii)

LHS = 2AC² - AB²
= 2AC² - ( AC² - BC² )           .....[ From(i) ]
= 2AC² - AC² + BC²
= AC² + BC²
= AD² + DC² + BC²            ....[ From(ii) ]
= RHS

Question 8

O is any point inside a rectangle ABCD.
Prove that: OB² + OD² = OC² + OA².

Sol:

Draw rectangle ABCD with arbitrary point O within it, and then draw lines OA, OB, OC, OD. Then draw lines from point O perpendicular to the sides: OE, OF, OG, OH.

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Using Pythagorean theorem we have from the above diagram:
OA² = AH² + OH² = AH² + AE²
OC² = CG² + OG² = EB² + HD²
OB² = EO² + BE² = AH² + BE²
OD² = HD² + OH² = HD² + AE²

Adding these equalities we get:
OA² + OC² = AH² + HD² + AE² + EB²
OB² + OD² = AH² + HD² + AE² + EB²

From which we prove that for any point within the rectangle there is the relation
OA² + OC² = OB² + OD²
Hence Proved.

Question 9

In the following figure, OP, OQ, and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC.

Prove that: AR² + BP² + CQ² = AQ² + CP² + BR²

Sol:

Here, we first need to join OA, OB, and OC after which the figure becomes as follows,

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. First, we consider the ΔARO and applying Pythagoras theorem we get,

AQ² = AR² + OR²
AR² = AO² - OR²            ....(i)
Similarly, from triangles, BPO, COQ, AOQ, CPO, and BRO we get the following results,
BP² = BO² - OP²          ....(ii)
CQ² = OC² - OQ²        ....(iii)
AQ² = AO² - OQ²        ....(iv)
CP² = OC² - OP²          ....(v)
BR² = OB² - OR²          ....(vi)

Adding (i), (ii) and (iii), we get 
AR² + BP² + CQ² = AO² - OR² + BO² - OP² + OC² - OQ²  ....(vii)

Adding (iv), (v) and (vi), we get,
AQ² + CP² + BR² = AO² - OR² + BO² - OP² + OC² - OQ²  ....(viii)

From (vii) and (viii), we get,
AR² + BP² + CQ² = AQ² + CP² + BR²
Hence proved.

Question 10

Diagonals of rhombus ABCD intersect each other at point O.

Prove that: OA² + OC² = 2AD² - ${\displaystyle \frac{{\text{BD}}^2}{2}}$

SOl:

Diagonals of the rhombus are perpendicular to each other.

In quadrilateral ABCD, ∠AOD = ∠COD = 90°.
So, ΔAOD and ΔCOD are right-angled triangles.

In ΔAOD using Pythagoras theorem,
AD² = OA² + OD²
⇒ OA² = AD² - OD²                 ....(i)

In ΔCOD using Pythagoras theorem,
CD² = OC² + OD²
⇒ OC² = CD² - OD²               ....(ii)

LHS = OA² + OC²
= AD² - OD² + CD² - OD²      ...[ From(i) and (ii) ]
= AD² + CD²  - 2OD²

= AD² + AD² - 2${\displaystyle {\left(\frac{\text{BD}}{2}\right)}^2}$ ...[ AD = CD and OD = ${\displaystyle \frac{\text{BD}}{2}}$]

= 2AD² - ${\displaystyle \frac{{\left(\text{BD}\right)}^2}{2}}$

= RHS.

Question 11

In figure AB = BC and AD is perpendicular to CD.
Prove that: AC² = 2BC. DC.

Sol:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

We consider the ΔACD and applying Pythagoras theorem we get,
AC² = AD² + DC²
= ( AB² - DB² ) + ( DB + BC )²
= BC² - DB² + DB² + BC² + 2DB.BC    ...( Given, AB = BC )
= 2BC² + 2DB.BC
= 2BC( BC + DB )
= 2BC . DC
Hence proved.

Question 12

In an isosceles triangle ABC; AB = AC and D is the point on BC produced.
Prove that: AD² = AC² + BD.CD.

Sol:

In an isosceles triangle ABC; AB = AC and
D is the point on BC produced.
Construct AE perpendicular BC.

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

We consider the rt. angled ΔAED and applying Pythagoras theorem we get,
AD² = AE² + ED²
AD² = AE² + ( EC + CD )²             ....(i)[ ∵ ED = EC + CD ]

Similarly, in ΔAEC,
AC² = AE² + EC²
AE² = AC² - EC²                       ....(ii)
Putting AE² = AC² - EC² in (i), We get,
AD² = AC² - EC² + ( EC + CD )²
        = AC² + CD( CD + 2EC )
AD² = AC² + BD.CD                 .....[ ∵ 2EC + CD = BD ]     
Hence proved.

Question 13

In triangle ABC, angle A = 90^o, CA = AB and D is the point on AB produced.
Prove that DC² - BD² = 2AB.AD.

Sol:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

We consider the rt. angled  ΔACD and applying Pythagoras theorem we get,
CD² = AC² + AD²
CD² = AC² + ( AB + BD )²                    ....[ ∵ AD = AB + BD ]
CD² = AC² + AB² + BD² + 2AB.BD      ...(i)

Similarly, in ΔABC,
BC² = AC² + AB²
BC² = 2AB²                                        ...[ AB = AC ]
AB² = ${\displaystyle \frac{1}{2}}$BC²                                   ...(ii)

Putting, AB2 from (ii) in (i), We get,
CD² = AC² + ${\displaystyle \frac{1}{2}}$BC² + BD² + 2AB . BD

CD² - BD² = AB² + AB² + 2AB . ( AD - AB )

CD² - BD² = AB² + AB² + 2AB . AD - 2AB² 

CD² - BD² = 2AB . AD

DC² - BD² = 2AB . AD                       

Hence Proved.

Question 14

In triangle ABC, AB = AC and BD is perpendicular to AC.
Prove that: BD² - CD² = 2CD × AD.

Sol:

In right-angled ΔADB,
AB² = AD² + BD²
⇒ AD² = AB² - BD²                                      .....(i)

AC = AD + DC
⇒ AC² = ( AD + DC )²
⇒  AC² = AD² + DC² + 2AD x DC
⇒ AC² = AB² - BD² + DC² + 2AD x DC      ...[ From(i) ]
⇒ AC² = AC² - BD² + DC² + 2AD x DC     ...[ AB = AC ]
⇒ BD² - DC² = 2AD x DC.

Question 15

In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1: 3.

Prove that : 2AC² = 2AB² + BC²

Sol:

Here,
BD : DC = 1 : 3.
⇒ BD = ${\displaystyle \frac{1}{4}\text{BC}\text{and}CD=\frac{3}{4}}$BC

AC² = AD² + CD² and AB² = AD² + BD² 
Therefore,
AC² - AB² = CD² - BD²

= ${\displaystyle {\left(\frac{3}{4}\text{BC}\right)}^2-{\left(\frac{1}{4}\text{BC}\right)}^2}$

= ${\displaystyle \frac{9}{16}{\text{BC}}^2- \frac{1}{16}{\text{BC}}^2}$

= ${\displaystyle \frac{1}{2}{\text{BC}}^2}$

∴ 2AC² - 2AB² = BC²
2AC² = 2AB² + BC²
Hence proved.

SELINA Solution Class 9 Chapter 13 Pythagoras Theorem [Proof and simple applications with converse]Exercise 13A

Question 1

A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.

Sol:

The pictorial representation of the given problem is given below,

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Here, AB is the hypotenuse.
Therefore applying the Pythagoras theorem we get,
AB²  = BC² + CA² 
13² = 5² + CA² 
CA² = 13² -  5² 
CA²  = 144
CA = 12 m
Therefore, the distance of the other end of the ladder from the ground is 12m.

Question 2

A man goes 40 m due north and then 50 m due west. Find his distance from the starting point.

Sol:

Here, we need to measure the distance AB as shown in the figure below,

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Therefore, in this case
AB² = BC² + CA² 
AB² = 50² + 40² 
AB² =  2500 + 1600
AB² = 4100
AB = 64.03
Therefore the required distance is 64.03 m.

Question 3

In the figure: ∠PSQ = 90^o, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

Sol:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔPQS and applying Pythagoras theorem we get,
PQ²  = PS² + QS² 
10²  = PS² + 6² 
PS² = 100 - 36
PR  = 8
Now, we consider the ΔPRS and applying Pythagoras theorem we get,
PR²  = RS² + PS² 
PR²  = 15² + 8² 
PR = 17
The length of PR 17 cm.

Question 4

The given figure shows a quadrilateral ABCD in which AD = 13 cm, DC = 12 cm, BC = 3 cm and ∠ABD = ∠BCD = 90^o. Calculate the length of AB.

Sol:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔBDC and applying Pythagoras theorem we get,
DB²  = DC²  + BC² 
DB²  = 12²  + 3² 
DB²  = 144  + 9 
DB²  = 153
Now, we consider the ΔABD and applying Pythagoras theorem we get,
DA²  = DB²  + BA² 
13² = 153  + BA²  
BA²  = 169 - 153 
BA = 4
The length of AB is 4 cm.

Question 5

AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.

Sol:

Since ABC is an equilateral triangle therefore, all the sides of the triangle are of the same measure and the perpendicular AD will divide BC into two equal parts.

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Here, we consider the ΔABD and applying Pythagoras theorem we get,
AB²  = AD²  + BD²  
AD²  = 100²  - 5²    ......[ Given, BC = 10 cm = AB, BC = ${\displaystyle \frac{1}{2}}$ BC ]
AD²  = 100 - 25
AD²  = 75
AD = 8.7
Therefore, the length of AD is 8.7 cm

Question 6

In triangle ABC, given below, AB = 8 cm, BC = 6 cm and AC= 3 cm. Calculate the length of OC.

Sol:

We have Pythagoras theorem which states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔABD, and applying Pythagoras theorem we get,
AB²  = AO²  + OB²  
AO²  = AB²  -  OB²
AO²  = AB²  - ( BD+ OC )²          .....[ Let, OC = x ]
AO²   = AB²   - ( BC+ x )²           ......(i)
First, we consider the ΔACO, and applying Pythagoras theorem we get,
AC²  = AO²  -  x ²
AO² = AC²  -  x ²                        ......(ii)

Now, from (i) and (ii),
AB²  - ( BC+ x )² = AC²  - x ² 
8² - ( 6+ x )² = 3²  - x ²    ...[ Given, AB = 8 cm, BC = 8 cm and AC = 3 cm ]
x = ${\displaystyle 1\frac{7}{12}}$ cm
Therefore , the length of OC will be ${\displaystyle 1\frac{7}{12}}$ cm.

Question 7

In triangle ABC, AB = AC = x, BC = 10 cm and the area of the triangle is 60 cm².
Find x.

Sol:

Here, the diagram will be,

We have Pythagoras theorem which states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Since ABC is an isosceles triangle, therefore perpendicular from vertex will cut the base in two equal segments.

First, we consider the ΔABD, and applying Pythagoras theorem we get,
AB² = AD² + BD²
AD² = x² - 5²
AD² = x² - 25
AD = ${\displaystyle \sqrt{x^2-25}}$                .....(i)
Now,
Area = 60
${\displaystyle \frac{1}{2}\times 10\times \text{AD}}$ = 60
${\displaystyle \frac{1}{2}\times 10\times \sqrt{x^2-25}}$ = 60
x = 13.
Therefore, x is 13 cm.

Question 8

If the sides of the triangle are in the ratio 1: ${\displaystyle \sqrt{2}}$: 1, show that is a right-angled triangle.

Sol:

Let, the sides of the triangle be, x: √2x and x.
Now,
x² + x² = 2x² = ${\displaystyle {\left(\sqrt{2}x\right)}^2}$              ....(i)

Here, in (i) it is shown that a square of one side of the given triangle is equal to the addition of square of the other two sides. This is nothing but Pythagoras theorem which states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Therefore, the given triangle is a right-angled triangle.

Question 9

Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m;
find the distance between their tips.

Sol:

The diagram of the given problem is given below,

We have Pythagoras theorem which states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Here, 11 - 6 = 5m            ...( Since DC is perpendicular to BC )
base = 12 cm

Applying Pythagoras theorem we get,
hypotenuse² = 5² + 12²
h² = 25 + 144
h² = 169
h = 13

Therefore, the distance between the tips will be 13m.

Question 10

In the given figure, AB//CD, AB = 7 cm, BD = 25 cm and CD = 17 cm;
find the length of side BC.

Sol:

Take M to be the point on CD such that AB = DM.
So DM = 7cm and MC = 10 cm

Join points B and M to form the line segment BM.
So BM || AD also BM = AD.

In right-angled ΔBAD,
BD² = AD² + BA²
(25)² = AD² + (7)²
AD² = (25)² - (7)²
AD² = 576
AD = 24

In right-angled ΔCMB,
CB² = CM² + MB²
CB² = (10)² + (24)²            ...[ MB = AD ]
CB² = 100 + 576
CB² = 676
CB = 26 cm

Question 11

In the given figure, ∠B = 90^°, XY || BC, AB = 12cm, AY = 8cm and AX : XB = 1 : 2 = AY : YC.
Find the lengths of AC and BC.

Sol:

Given that AX : XB = 1 : 2 = AY : YC.
Let x be the common multiple for which this proportion gets satisfied.
So, AX = 1x and XB = 2x
AX + XB = 1x + 2x = 3x
⇒ AB = 3x                                        .….(A - X - B)
⇒ 12 = 3x
⇒ x = 4

AX = 1x = 4 and  XB = 2x = 2 × 4 = 8
Similarly,
AY = 1y and YC = 2y
AY = 8                                               …(given)
⇒ 8 = y

∴ YC = 2y = 2 × 8 = 16
∴ AC = AY + YC = 8 + 16 = 24 cm
∆ABC is a right angled triangle.        ….(Given)

∴ By Pythagoras Theorem, we get
⇒ AB² + BC² = AC²
⇒ BC² = AC² - AB²
⇒ BC² = (24)² - (12)²
⇒ BC² = 576 - 144
⇒ BC² = 432
⇒ BC = 12√3 cm
∴ AC = 24 cm and BC = 12√3 cm. 

Question 12

In ΔABC,  Find the sides of the triangle, if:
(i) AB =  ( x - 3 ) cm, BC = ( x + 4 ) cm and AC = ( x + 6 ) cm

(ii) AB = x cm, BC = ( 4x + 4 ) cm and AC = ( 4x + 5) cm

Sol:

(i) In right-angled ΔABC,
AC² = AB² + BC²
⇒ ( x + 6 )² = ( x - 3 )² + ( x + 4 )²
⇒ ( x² + 12x + 36 ) = ( x² - 6x + 9 ) + ( x² + 8x + 16 )
⇒ x² - 10x - 11 = 0
⇒ ( x - 11 )( x + 1 ) = 0
⇒ x = 0             or         x = - 1
But length of the side of a triangle can not be negative.
⇒ x = 11 cm
∴ AB = ( x - 3 ) = ( 11 - 3 ) = 8 cm
BC = ( x + 4 ) = ( 11 + 4 ) = 15 cm
AC = ( x + 6 ) = ( 11 + 6 ) = 17 cm.

(ii) In right-angled ΔABC,
AC² = AB² + BC²
⇒ ( 4x + 5 )² = ( x )² + ( 4x + 4 )²
⇒ ( 16x² + 40x + 25 ) = ( x² ) + ( 16x² + 32x + 16 )
⇒ x² - 8x - 9 = 0
⇒ ( x - 9 )( x + 1 ) = 0
⇒ x = 9             or      x = - 1
But length of the side of a triangle can not be negative.
⇒ x = 9 cm
∴ AB = x = 9 cm
BC = ( 4x + 4 ) = ( 36 + 4 ) = 40 cm
AC = ( 4x + 5 ) = ( 36 + 5 ) = 41 cm.

S.chand Class 8 Maths Solution Chapter 13 Parallelograms Exercise 13 B

 Exercise 13 B

Question 1

1. For each of the statements given below, indicate if it is true (T) or false $(F)$ :

(a) A rectangle is a parallelogram.

(b) A square is a rectangle.

(c) A parallelogram is a rhombus.

(d) A square is a rhombus.

(e) A rectangle is a square.

(f) A square is a parallelogram.

2. What kind of quadrilateral is formed when the mid points of the sides of the following are joined?

(a) a rectangle,

(b) a rhombus,

(c) a kite,

(d) an isosceles trapezium.

3. $A B C D$ is a rectangle, $E F G H$ is a rhombus and $P Q R S$ is a square.

Complete the followings :

(i) $F H \perp $ ___

(ii) $G E$ bisects ___

(iii) If $\angle F E H=130^{\circ}, \angle E H G=$ ___

(iv) If $A C=4, B D=$ ___

(v) If PQ=4, PS=___

(vi) If $\angle E H G=48^{\circ}, \angle H G O=$ ___

(vii) If $H E=12, G H=$ ___

(viii) $\angle E O F=$ ___

4. The diagonals of a rectangle $A B C D$ intersect at $O$. If $\angle B O C=44^{\circ}$, find $\angle O A D$.

5. $A B C D$ is a rhombus with $\angle A B C=56^{\circ}$. Determine $\angle C A D$.

<fig>

6. $A B C D$ is a rhombus. If $\angle D A C=50^{\circ}$, find 

(a) $\angle A C D$

(b) $\angle C A B$

(c) $\angle A B C$.

7. $A B C D$ is a trapezium in which $A B \| D C$. If $\angle A=\angle B=40^{\circ}$, what are the measures of the other two angles?

8. Calculate the angles marked with small letters in the following diagrams.

<fig>

9. $A B C D$ is a kite. If $\angle B C D=40^{\circ}$, find (a) $\angle B D C$ (b) $\angle A B C$.

<fig>

10. $A B C D$ is a trapezium and $A B E D$ is a square. If $B E=E C$, find: (a) $\angle B A E$

(b) $\angle A B C$

(c) What shape is the figure $A B C E$ ?

11. $A B C D$ is a square and $A B R S$ is a rhombus. If $\angle S A D=120^{\circ}$, find: (a) $\angle A S D \quad$ (b) $\angle S R B$.

12. $A B C D$ is a kite and $\angle A=\angle C$. If $\angle C A D=70^{\circ}, \angle C B D=65^{\circ}$, find : (a) $\angle B C D$ (b) $\angle A D C$.

13. If the diagonals of a rhombus are $12 \mathrm{~cm}$ and $16 \mathrm{~cm}$, find the length of each side.

Multiple Choice Questions (MCQs)

14. What value of $x$ makes the given parallelogram a square ?

(a) $7.5^{\circ}$

(b) $6.5^{\circ}$

(c) $4.5^{\circ}$

(d) $8.5^{\circ}$

15. Quadrilateral $A B C D$ is a kite. What is the length of the line segment $A E$ ?

(a) $4 \mathrm{~cm}$

(b) $6 \mathrm{~cm}$

(c) $5 \mathrm{~cm}$

(d) $13 \mathrm{~cm}$

16.$C D E F$ is a square. Find $x$.

(a) $142^{\circ}$

(b) $92^{\circ}$

(c) $113^{\circ}$

(d) $83^{\circ}$

S.chand Class 8 Maths Solution Chapter 13 Parallelograms Exercise 13 A

 Exercise 13 A

Question 1

1. The measure of one angle of a parallelogram is $80^{\circ}$. What are the measures of the remaining angles ?

2. Two adjacent angles of a parallelogram are congruent. What is the measure of each?

3. Two adjacent sides of a parallelogram are $5 \mathrm{~cm}$ and $6 \mathrm{~cm}$ respectively. Find its perimeter.

4. Find each angle of a parallelogram if two consecutive angles are in the ratio $1: 3$.

5. The perimeter of a parallelogram is $180 \mathrm{~cm}$. One of its sides is greater than the other by $30 \mathrm{~cm}$. Find the length of the sides of the parallelogram.

6. Find the sizes of the angles of a parallelogram if one angle is $20^{\circ}$ less than twice the smallest angle.

7. $A B C D$ is a parallelogram. Find $x, y$ and $z$.

8. In the figure, find the four angles $A, B, C$ and $D$ of the parallelogram $A B C D$.

9. $A B C D$ is a parallelogram. $C E$ bisects $\angle C$ and $\mathrm{AF}$ bisects $\angle A$. In each of the following, if the statement is true, give a reason for the same.

(i) $\angle A=\angle C$

(ii) $\angle F A B=\frac{1}{2} \angle A$

(iii) $\angle D C E=\frac{1}{2} \angle C$.

(iv) $\angle F A B=\angle D C E$

(v) $\angle D C E=\angle C E B$

(vi) $\angle C E B=\angle F A B$

(vii) $C E \| A F$

(viii) $A E \| F C$.

Multiple Choice Questions (MCQs)

10. If the quadrilateral $\mathrm{ABCD}$ is a parallelogram. What is the value of $x$ ?

(a) $45^{\circ}$

(b) $30^{\circ}$

(c) $36^{\circ}$

(d) $37^{\circ}$

11. What values of a would make the given quadrilateral a parallelogram.

(a) $5 \frac{1}{4}$

(b) $5 \frac{1}{2}$

(c) $5 \frac{3}{4}$

(d) 5

High Order Thinking Skills (HOTS)

12. The diagonal of a rectangle is thrice its smaller side. Find the ratio of its sides.

[Hint. Use Pythagoras theorem]

(a) $\sqrt{2}: 1$

(b) $2 \sqrt{2}: 1$

(c) $3: 2$

(d) $\sqrt{3}: 1$

SChand Composite Mathematics Class 7 Chapter 13 Congruence of Triangle Exercise 13C

 Exercise 13 C 

Question 1 

(a) What will be the other angles of a right angles isosceles triangle? 

Sol: $x+x+90=180^{\circ} \Rightarrow 2x=90 \Rightarrow x=45^{\circ} \quad 45^{\circ}$

(b) Can you draw an obtuse angles isosceles triangle ? 

Ans: Yes

Question 2

(a) The vertical angle of an isosceles triangle is 110 degree. What must be the size of its base angles? 

(IMAGE TO BE ADDED)

Sol: $\Rightarrow \quad x+x+110^{\circ}=180^{\circ}$

$\Rightarrow \quad 2 x=180^{\circ}-110^{\circ} \Rightarrow \quad 2 x=70^{\circ}$

x = $35^{\circ}$  Ans

(b) What is the size of each exterior angle of an equilateral triangle ? 

(IMAGE TO BE ADDED)

Sol: $180^{\circ}-60^{\circ}=x$

$x=120^{\circ}$

Question 3

$\triangle A B C$ is isosceles with AB= AC, if $\angle A=80^{\circ}$ what is the measure of angle b?

(IMAGE TO BE ADDED)

Sol: Let $\angle B=\angle C=x$

$\Rightarrow x+x+80^{\circ}=180^{\circ}$

$\Rightarrow 2 x=180^{\circ}-80^{\circ}$

$\Rightarrow 2 x=100^{\circ}$

$\Rightarrow  x=50^{\circ}$

$\angle B=\angle L=50^{\circ}$

Question 4

In $\triangle B B C, \angle A=\angle B=50^{\circ}$. Name the pair of sides which are equal . 

(IMAGE TO BE ADDED)

Side AC = BC 

Question 5

In the figure given below AN =AC, $\angle B A C=52^{\circ}$ $\angle A C K=84^{\circ}$ and BCK is  a straight line . prove that NB =NC

(IMAGE TO BE ADDED)

Sol: $\triangle A B C$

$\angle A+\angle B=84^{\circ}$

$52^{\circ}+\angle B=84^{\circ}$

$\begin{aligned} \angle B &=84^{\circ}-52^{\circ} \\ \Rightarrow \angle B &=32^{\circ} \end{aligned}$

$\triangle A N C$

$\Rightarrow \quad 52^{\circ}+x+x=180^{\circ}$

$\Rightarrow \quad 2 x=180^{\circ}-52^{\circ}$ $2x=123^{\circ}$

$\Rightarrow x=\frac{128}{2} \quad \Rightarrow$

x = $64^{\circ}$

∵ BCK Straight line 

$\angle B C N+\angle A C N+\angle A C K=180^{\circ}$

$\angle B C N+64^{\circ}+84^{\circ}=180^{\circ}$

$\angle B C N=180^{\circ}-148$

$\angle B C N=32^{\circ}$

$\because \angle B C N=\angle B$

BN = NC Hence proved 

Question 6

In the figure AB = AC. Prove that BD = BC

Sol: $\Rightarrow x+x+40^{\circ}=180$

$\Rightarrow 2 x=180^{\circ}-40^{\circ}$

$\Rightarrow 2 x=140^{\circ}$

$\Rightarrow x=\frac{140^{\circ}}{2}$

$\Rightarrow x=70^{\circ}$

By ext. angle prop

$\Rightarrow 30^{\circ}+40^{\circ}=y$

$\Rightarrow y=70^{\circ}$

$\because x=y=70^{\circ}$

$\therefore B D=B C$ Hence proved 

SChand Composite Mathematics Class 7 Chapter 13 Congruence of Triangle Exercise 13B

 Exercise 13 B

Question 1

AB and CD bisect each other at K. Prove that AC = BD .

(IMAGE TO BE ADDED)

Sol: AK= BK

$\angle A K C=\angle B K D$ (Vertically opposite angle )

CK = DK 

By SAS $\triangle A K C, \cong \triangle B K D$

C.P.C.T $\quad A C=B D$ Hence proved 

Question 2

The sides BA and CA have been produced such that BA = AD and CA = AE prove that DE||BC 

(IMAGE TO BE ADDED)

Sol: $A B=A D$

$\angle B A C=\angle E A D$  (Vertically opposite angle )

by $S A S \quad \triangle A B C \cong \triangle A E D$

by C.P.CT $\angle B=\angle D \quad \because B C \| D E$  Hence proved .

Question 3

$O A=O B, O C=O D$, $\angle A O B=\angle C O D$ prove that AC = BD 

(IMAGE TO BE ADDED)

Sol: 

$\begin{aligned} O A &=O B \\ \angle A O B &=\angle C O D \\ \angle A O C+\angle C O B=\angle C O B+\angle B O D \\ \angle A O C &=\angle B O D \\ O C &=O D \end{aligned}$

By SAS $\triangle A O C \cong \triangle B O D$

by C.P.CT $A C=B D$ hence proved 

Question 4

If AB and MN bisect each other at O and AM and BN are $\perp$ on xy. Prove that $\triangle S$, OAM and OBN are congruent and hence prove that AM= BN 

(IMAGE TO BE ADDED)

Sol: $\angle A M O=\angle B N O=90^{\circ}$

$O B=O A$

$O M=O N$

by RHS $\triangle A M O \cong \triangle B N O$ 

by $C \cdot P \cdot c \cdot T \quad A M=B N$

Question 5

$\angle XYZ$ is bisected by YP . L is any point on YP and MLN is Perpendicular to YP. Prove that LM = LN 

(image to be added)

Sol: $\angle YLM=\angle YL N=90^{\circ}$

YL = YL (Common)

$\angle M YL=\angle N YL$ (Bisect by YP )

by ASA $\triangle MY L \cong \triangle N Y L$

By C.P.C.T LM = LN Hence proved 

Question 6

In the fig, PM =PN , PM $\perp A B$ AND $P N \perp A C$ Prove $\triangle A M P \cong \triangle A N P$

Sol: (image to be added)

$\angle A N P=\angle A M P=90^{\circ}$ 

$A P=A P \quad($ Comimon $)$

$M P=N P \quad$ (given)

By R.H.S $\triangle A M P \cong \triangle A N P$ Hence proved 

Question 7

In the figure , AD = BC and AD ||BC. Prove that AB =DC 

Sol: (IMAGE TO BE ADDED)

AD=BC 

$\angle C A D=\angle A C B$ [Alt. angles]

AC = AC 

By ASA $\triangle A C D \cong \triangle A B C$

by C.P.CT. $A B=D C$ Hence proved 

Question 8

In the given figure, triangles ABC and DCB are right angles at A and D respectively and AC = DB . Prove that $\triangle A B C \cong \triangle D C B$

 (IMAGE TO BE ADDED)

Sol: $\angle B A C=\angle B D C=90^{\circ}$

$B C=B C=$ Common

$A C=D B$

By RHS $\triangle A B C \cong \triangle D C B$ Hence proved

SChand Composite Mathematics Class 7 Chapter 13 Congruence of Triangle Exercise 13A

 Exercise 13 A

Question 1 

State whether or not the following pairs of triangles are congruent. If they are, give reasons.

(i) (Image to be added)

Sol: $5 \frac{3}{8} \mathrm{~cm}$ = $\frac{43}{8}$ = 5.375 cm 

1 cm = 10mm

$5.375 \times 10$ = 53.75mm

 $4 \frac{5}{8} \mathrm{~cm}$ = 46.25mm

$6 \frac{1}{8} \mathrm{~cm}$ = 61.25mm

(BY SSS Triangle are congruent each other)

(ii) (Image to be added)

Sol: 62mm converted into Cm = 6.2cm

Triangle are not congruent

(iii)   (Image to be added)

Sol: 40 mm = 4cm 

70 mm = 7cm 

Acc SAS triangle are congruent 

(iv)   (Image to be added)

Sol: not congruent 

(v) (Image to be added)

Sol: (Acc ASA Triangle are equal)

(v) (Image to be added)

Sol: By RHS Triangle are Congruent 

Question 2

Show by comparing angles and sides, which of the triangles given here are congruent to each other .

(Image to be added)

Question 3

In the following figure, state the condition you would use to show that $\triangle A B C$ and $\triangle C D E$ are congruent . 

(IMAGE TO BE ADDED)

Sol: A C=C E

B C=C D

$\angle A C B=\angle DCE$  (Vertically opposite angle)

 By SAS  $\triangle A B C \cong \triangle CDE$

Question 4

 In the figure, $A B C D$ is a rhombus. Are $\triangle A D C$ and $\triangle A B C$ congruent? What can you say about $\triangle A B D$ and $\triangle B C D$ ?

(IMAGE TO BE ADDED)

Sol: $A D=A B$

$C D=B C$

$A C=A C$

Hence By SSS $\triangle A C D \cong \triangle A B C$

$A D=C D$

$A B=B C$

$B D=B D$

$\triangle A B D \cong \triangle  C B D$ (BY SSS)

Question 5

In this figure, $T S \| P Q$ and $T S=P Q$. Prove that the triangles $P Q R$ and $S T R$ are congruent.

(IMAGE TO BE ADDED)

Sol: $\angle R S T=\angle R P Q$

TS= PQ = 5cm

$\angle R T S=\angle R Q P $ 

(By ASA)

$STR\cong \triangle P Q R$

Question 6

$\triangle P R Q \cong \triangle L M N$. If $P Q=6 \mathrm{~cm}, P R=5 \mathrm{~cm}$ and $\angle P=50^{\circ}$, find $N L$ and $\angle L$ if $L M=5 \mathrm{~cm}$ and $Q R=M N$.

(IMAGE TO BE ADDED)

Sol: $\triangle P R Q \cong \triangle L M N$ (Given )

PR = LM 

PQ = LN =NL = 6cm 

Hence corresponding side and angle of congruent angle are equal 

Question 7

In the diagrams of the following figures. Name the triangle which is congruent to $\triangle A B C$ Keeping the letters in the right order. State the congruence condition also. 

(IMAGE TO BE ADDED)

Question 8

In each of the following name the congruent triangles and state the congruence condition.

(IMAGE TO BE ADDED)

ML Aggarwal Solution Class 9 Chapter 13 Rectilinear Figures Exercise 13.1

 Exercise 13.1

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Q1 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

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Question 1

If two angles of a quadrilateral are 40° and 110° and the other two are in the ratio 3 : 4, find these angles.

Sol :

Sum of four angles of a quadrilateral=360°

Sum of two given angles =40°+110°=150°

∴Sum of remaining two angles

=360°-150°=210°

∴Third angle$=\frac{210^{\circ} \times 3}{3+4}$

$=\frac{210^{\circ} \times 3}{7}=90^{\circ}$

and fourth angle $=\frac{210^{\circ} \times 4}{3+4}$

$=\frac{210^{\circ} \times 4}{7}$

=120°

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Q2 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

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Question 2

If the angles of a quadrilateral, taken in order, are in the ratio 1 : 2 : 3 : 4, prove that it is a trapezium.

Sol :

In trapezium ABCD

∠A : ∠B : ∠C : ∠D=1 : 2 : 3 : 4

Sum of angles of the quad ABCD=360°

Sum of the ratio's=1+2+3+4=10

∴$ \angle \mathrm{A}=\frac{360^{\circ} \times 1}{10}=36^{\circ}$

$\angle B=\frac{360^{\circ} \times 2}{10}=72^{\circ}$

$\angle C=\frac{360^{\circ} \times 3}{10}=108^{\circ}$

$\angle D=\frac{360^{\circ} \times 4}{10}=144^{\circ}$

Now , ∠A+∠D=36°+114°=180°

∵∠A+∠D=180° and there are co-interior angles

∴AB||DC

Hence ABCD is a trapezium

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Q3 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

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Question 3

If an angle of a parallelogram is two-thirds of its adjacent angle, find the angles of the parallelogram.

Sol :

Here ABCD is a parallelogram 

Let ∠A=x°

then ∠B$=\frac{2}{3} x^{\circ}$

(given condition an angle of a parallelogram is two third of its adjacent angle)

∴∠A+∠B=180°

(∵Sum of adjacent angle in parallelogram is 180°)

⇒$x^{\circ}+\frac{2}{3} x^{\circ}=180^{\circ}$

⇒$\frac{3 x+2 x}{3}=180$

⇒5x=180×3

⇒$x=\frac{180 \times 3}{5}$

⇒x=36×3

⇒x=108

∴∠A=108°

$\angle B=\frac{2}{3} \times 108^{\circ}=2 \times 36^{\circ}=72^{\circ}$

∠B=∠D=72°

(opposite angle in parallelogram is same)

Also,∠A=∠C=180°

(opposite angles in parallelogram is same)

Hence, angles of parallelogram are 108° , 72° , 108° , 72°

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Q4 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

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Question 4

(a) In figure (1) given below, ABCD is a parallelogram in which ∠DAB = 70°, ∠DBC = 80°. Calculate angles CDB and ADB.

(b) In figure (2) given below, ABCD is a parallelogram. Find the angles of the ΔAOD.

(c) In figure (3) given below, ABCD is a rhombus. Find the value of x.

Sol :

(a) ∵ABCD is parallelogram

∴ADB=80°  [∵∠DBC=80° (given)]

In ΔADB,

⇒∠A+∠ADB+∠ABD=180°  (sum of all angles in a triangle is 180°)

⇒70°+80°+∠ABD=180°

⇒150°+∠ABD=180°

⇒∠ABD=180°-150°

⇒∠ABD=30°...(2)

Now ∠CDB=∠ABD ...(3)

[∵AB||CD (alternate angles)]

From (2) and (3)

⇒∠CDB=30°...(4)

From (1) and (4)

⇒∠CDB=30° and ∠ABD=80°

(b) Given : ∠BCO=35° , ∠CBO=77° , 

In ΔBOC

⇒∠BOC+∠BCO+∠CBO=180°

(sum of all angles in a triangle is 180°)

⇒∠BOC=180°-112°=68°

Now in parallelogram ABCD,

We have 

⇒∠AOD=∠BOC

(vertically opposite angles)

∴∠AOD=68°

(c) ABCD is a rhombus ∠A+∠B=180°

(In rhombus sum of adjacent angle is 180°)

⇒72°+∠B=180°

⇒∠B=180°-72°=108°

∴$x=\frac{1}{2} \angle \mathrm{B}=\frac{1}{2} \times 108^{\circ}$

=54°

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Q5 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

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Question 5

(a) In figure (1) given below, ABCD is a parallelogram with perimeter 40. Find the values of x and y.

(b) In figure (2) given below. ABCD is a parallelogram. Find the values of x and y.

(c) In figure (3) given below. ABCD is a rhombus. Find x and y.

Sol :

(a) Since ABCD is a parallelogram 

∴AB=CD and BC=AD

∴3x=2y+2  (AB=CD)

3x-2y=2...(1)

Also, AB+BC+CD+DA=40

⇒3x+2x+2y+2+2x=40

⇒7x+2y=40-2

⇒7x+2y=38...(2)

Adding (1) and (2)

$\begin{array}{l}3 x-2 y=2 \\7 x+2 y=38 \\\hline 10 x \quad=40\end{array}$

⇒$x=\frac{40}{10}=4$

Subtracting the value of x in (1), we get

⇒3×4-2y=2

⇒12-2y=2

⇒-2y=2-12

⇒-2y=-10

⇒$y=\frac{-10}{-2}$

∴y=5

Hence , x=4 and y=5

(b) In parallelogram ABCD

⇒∠A=∠C (opposite angles are same in parallelogram)

⇒3x-20°=x+40°

⇒3x-x=40°+20°

⇒2x=60°

⇒$x=\frac{60^{\circ}}{2}$

⇒x=30°

Also , ∠A+∠B=180°

(sum of adjacent angles in parallelogram is equal to 180°)

⇒3x-20°+y+15°=180°

⇒3x+y-5°=180°

⇒3x+y=180+5°

⇒3x+y=185°

⇒3×30°+y=180°

[putting the value of x from (1)]

⇒90°+y=185°

⇒y=185°-90°

⇒y=95°

Hence, x=30° , y=95°

(c) ABCD is a rhombus

∴AB=AD

⇒3x+2=4x-4

⇒3x-4x=-4-2

⇒-x=-6

⇒x=6...(1)

In ΔABD,

∴∠BAD=60° ,Also, AB=AD

∴∠ADB=∠ABD

∴∠ADB$=\frac{180^{\circ}-\angle \mathrm{BAD}}{2}$

$=\frac{180^{\circ}-60^{\circ}}{2}=\frac{120^{\circ}}{2}$

=60°

ΔABD is equilateral triangle (∵each angles of this triangle are 60°)

∴AB=BD

⇒3x+2=y-1

⇒3×6+2=y-1

⇒3x+2=y-1

⇒20=y-1

⇒y-1=20

⇒y=20+1=21

Hence , x=6 and y=21

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Q6 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

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Question 6

The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, find ∠DPC.

Sol :

ABCD is a rectangle

Since diagonals of rectangle are same and bisects each other

∴AP=BP

∴∠PAB=∠PBA

(equal sides have equal opposite angles)

⇒∠PAB=50°  [∵∠PBA=50° (given)]

In ΔAPB,

⇒∠APB+∠ABP+∠BAP=180°

⇒∠APB+50°+50°=180°

⇒∠APB=180°-100°=80°...(1)

∴∠DPC=∠APB...(2) (vertically opposite angles)

From (1) and (2)

⇒∠DPC=80°

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Q7 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

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Question 7

(a) In figure (1) given below, equilateral triangle EBC surmounts square ABCD. Find angle BED represented by x.

(b) In figure (2) given below, ABCD is a rectangle and diagonals intersect at O. AC is produced to E. If ∠ECD = 146°, find the angles of the ∆ AOB.

(c) In figure (3) given below, ABCD is rhombus and diagonals intersect at O. If ∠OAB : ∠OBA = 3:2, find the angles of the ∆ AOD.

Sol :

(a) Since EBC is an equilateral triangle

⇒EB=BC=EC

∴EB=BC=EC...(1)

Also, ABCD is a square

⇒AB=BC=CD=AD...(2)

From (1) and (2)

⇒EB=EC=AB=BC=CD=AD..(3)

In ΔECD,

⇒∠ECD=∠BCD+∠ECB

(BEC is an equilateral triangle)

⇒∠ECD=90°+60°=150°...(4)

Also, EC=CD [From (3)]

∴∠DEC=∠CDE ...(5)

∴∠ECD+∠DEC+∠CDE=180°

(sum of all angles in a triangle is 180°)

⇒150°+∠DEC+∠DEC=180°

(using (4) and (5))

⇒2∠DEC=180°-150°

⇒2∠DEC=30°

⇒∠DEC$=\frac{30^{\circ}}{2}$

⇒∠DEC=15°...(6)

Now ∠BEC=60° (BEC is an equilateral triangle)

⇒∠BED+∠DEC=60°

⇒∠x+15°=60°

[From (6)]

⇒x=60°-15°=45°

Hence , the value of x=45°

(b) Since ABCD is a rectangle 

⇒∠ECD=146° (given)

∴ACE is straight line

∴146°+∠ACD=180°

(linear pair)

⇒∠ACD=180°-146°=34°...(1)

∴∠CAB=∠ACD (alternate angles)...(2)

[∵AB||CD]

From (1) and (2)

⇒∠CAB=34°

⇒∠OAB=34°...(3)

In ∠AOB

⇒AO=OB

(In rectangle diagonals are same and bisects each other)

⇒∠OAB=∠OBA ...(4)

(equal sides have equal angles opposite to them)

From (3) and (4)

⇒∠OBA=34°...(5)

∴∠AOB+∠OBA+∠OAB=180°

(sum of all ang;es in a triangle is 180°)

⇒∠AOB+34°+34°=180°

⇒∠AOB+68°=180°

⇒∠AOB=180°-68°=112°

Hence ,∠AOB=112° ,∠OAB=34° ,

and ∠OBA=34°

(c) Here ABCD is a rhombus and diagonals intersects at O

and ∠OAB : ∠OBA= 3 : 2

Let ∠OAB=2x

then ∠OBA=2x

We know that diagonals of rhombus intersects at right angle 

∴∠OAB=90 in ΔAOB

∴∠OAB+∠OBA=180°

⇒90°+3x+2x=180°

⇒90°+5x=180°

⇒5x=180°-90°

⇒$x=\frac{90^{\circ}}{5}$

⇒x°=18°

∴∠OAB=3x=3×18=54°

⇒∠OBA=2x=2×18=36°

and ∠AOB=90°

---

Q8 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

OPEN IN YOUTUBE

Question 8

(a) In figure (1) given below, ABCD is a trapezium. Find the values of x and y.

(b) In figure (2) given below, ABCD is an isosceles trapezium. Find the values of x and.y.

(c) In figure (3) given below, ABCD is a kite and diagonals intersect at O. If ∠DAB = 112° and ∠DCB = 64°, find ∠ODC and ∠OBA.

Sol :

(a) Given : ABCD is a trapezium 

∠A=x+20° , ∠B=y , ∠C=92° , ∠D=2x+10°

Required :Value of x and y

Since ABCD is a trapezium

⇒∠B+∠C=180°  (∵AB||DC)
⇒y+92°=180°

⇒y=180°-92°=88°

Also , ∠A+∠D=180°
⇒x+20°+2x+10°=180°

⇒3x+30°=180°

⇒3x=180°-30°

⇒3x=150°

⇒$x=\frac{150^{\circ}}{3^{+}}$

⇒x=50°

(b) Given : ABCD is an isosceles trapezium BC=AD

∠A=2x ,∠C=y , ∠D=3x

Required : Value of x and y

Since ABCD is a trapezium and AB||DC

∴∠A+∠D=180°
⇒2x+3x=180°
⇒5x=180°
⇒x$=\frac{180^{\circ}}{5}=36$

∴x=36°...(1)

Also , AB=BC and AB||DC

∴∠A+∠C=180°

⇒2x+y=180°

⇒2×36+y=180°

[substituting the value of x from (1)]

⇒72°+y=180°

⇒y=180°-72°=108°

⇒y=108°

Hence, value of x=72° and y=108°

(c) Given : ABCD is a kite and diagonals intersects at O

⇒∠DAB=112° and 

⇒∠DCB=64°

Required : ∠ODC and ∠OBA

∴AC diagonal of kite ABCD 

∴$\angle D O C=\frac{64}{2}^{\circ}=32^{\circ}$

∴∠DOC=90°

(diagonals of kites bisects at right angles )

In ∠OCD

∴∠ODC=180°-(∠DCO+∠DOC)

=180°-(32°+90°)=180°-122°=58°

In ΔDAB

⇒∠OAB$=\frac{112^{\circ}}{2}=56^{\circ}$

⇒∠OAB=90°

(diagonals of kites bisect at right angles)

In ΔOAB

⇒∠OBA=180°-(∠OAB+∠AOB)

=180°-(56°+90°)=180°-146°=34°

Hence , ODC=58° and OBA=34°

---

Q9 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

OPEN IN YOUTUBE

Question 9

(i) Prove that each angle of a rectangle is 90°.

(ii) If the angle of a quadrilateral are equal, prove that it is a rectangle.

(iii) If the diagonals of a rhombus are equal, prove that it is a square.

(iv) Prove that every diagonal of a rhombus bisects the angles at the vertices.

Sol :

(i) A rectangle ABCD

To prove : Each angle of rectangle=90°

Proof : ∵Opposite angles of a rectangle are equal

∴∠A=∠C and ∠B=∠D

But ∠A+∠B+∠C+∠D=360°

(sum of angles of a quadrilateral)

⇒∠A+∠B+∠A+∠B=360°

⇒2(∠A+∠B)=360°

⇒∠A+∠B$=\frac{360}{2}=180$

But ∠A=∠B and ∠C=∠D (adjacent angles of a rectangle are equal)

∠A+∠A=180°

2∠A=180°

∠A=90°

Hence ∠A=∠B=∠C=∠D=90°

(ii) Given : In quadrilateral ABCD

∠A=∠B=∠C=∠D

To prove : ABCD is a rectangle 

Proof : ∠A=∠B=∠C=∠D

⇒∠A=∠C and ∠B=∠D

But these are opposite angles of the quadrilateral 

Also, ∠A=∠B and ∠C=∠D (these are adjacent angle of rectangle)

∴∠A=∠B=∠C=∠D=90°

Hence ABCD is a rectangle 

(iii) Given : ΔABCD is a rhombus in which AC=BD

To prove : ABCD is a square 

Proof : In ΔABC and ΔDCB ,

AB=DC (ABCD is a rhombus)

BC=BC (common)

and AC=BD (given)
∴ΔABC≅ΔDCB

(by S.S.S axiom of congruency)

∴∠ABC=∠DCB (c.p.c.t)..(i)

But these are angle made by transversal

BC on the same side of parallel 

Lines AB and CD

∠ABC+∠DCB=180°

∠ABC+∠ABC=180° (from i)

2∠ABC=180°

∴∠ABC=90°

∴ABCD is a square  (Q.E.D)

(iv) AC and BD bisects ∠A , ∠C and ∠B , ∠D respectively

Proof : 

Statements : Reasons 

(1) In ΔAOD and ΔCOD

AD=CD (each side or rhombus is same)

OD=OD (common)

AO=OC (diagonals of rhombus bisect each other)

(2) ΔAOD≅ΔCOD [S.S.S]

(3) ∠AOD=∠COD (c.p.c.t)

(4) ∠AOD+∠COD=180° AOC is a straight line

⇒∠AOD+∠COD=180° by (3)

⇒2∠AOD=180° 

⇒∠AOD$=\frac{180^{\circ}}{2}$

(5) ∠COD=90°  By (3) and (4)

∴OD⟂AC 

⇒BD⟂AC

(6) ∠ADO=∠CDO  (c.p.c.t)

⇒OD bisects ∠D

⇒BD bisects ∠D

Similarly we can prove that BD bisects ∠B

and AC bisects the ∠A and ∠C

---

Q10 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

OPEN IN YOUTUBE

Question 10

ABCD is a parallelogram. If the diagonal AC bisects ∠A, then prove that:

(i) AC bisects ∠C

(ii) ABCD is a rhombus

(iii) AC ⊥ BD.

Sol :

Given : In parallelogram ABCD, diagonal AC bisects ∠C

To prove : (i) AC bisects ∠C

(ii) ABCD is a rhombus

(iii) AC⊥ BD

Proof : (i) ∵AB||CD (opposite sides of a parallelogram)

∴∠DCA=∠CAB (alternate angles)

Similarly ∠DAC=∠DCB

But ∠CAB=∠DAC  (∵AC bisects ∠A)

∴∠DCA=∠ACB

∴AC bisects ∠C

(iii) ∵AC bisects ∠A and ∠C and ∠A=∠C

∴ABCD is a rhombus

(iii) ∵AC and BD are the diagonals of a rhombus

∴AC and BD bisects each other at right angles

Hence , AC⊥BD

Hence proved

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Q11 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

OPEN IN YOUTUBE

Question 11

(i) Prove that bisectors of any two adjacent angles of a parallelogram are at right angles.

(ii) Prove that bisectors of any two opposite angles of a parallelogram are parallel.

(iii) If the diagonals of a quadrilateral are equal and bisect each other at right angles, then prove that it is a square.

Sol :

(i) Prove that bisectors of any two adjacent angles of a parallelogram are at right angles.

(i) Given : AM bisects angle A and BM bisects angle B of parallelogram ABCD

To prove : ∠AMB=90°

Proof :

Statement : Reasons

(1) ∠A+∠B=180°  AD||BC and AB is the transversal

(2) $\frac{1}{2}(\angle \mathrm{A}+\angle \mathrm{B})=\frac{180^{\circ}}{2}$  Multiplying both sides by $\frac{1}{2}$

⇒$\frac{1}{2} \angle \mathrm{A}+\frac{1}{2} \angle \mathrm{B}=90^{\circ}$

⇒∠MAB+∠MBA=90° 

(i) AM bisects ∠A

∴$\frac{1}{2} \angle \mathrm{A}=\angle \mathrm{MAB}$

(ii) BM bisects ∠B

∴$\frac{1}{2} \angle \mathrm{B}=\angle \mathrm{MBA}$

(3) In ΔAMB ,

∠AMB+∠MAB+∠MBA=180°  [Sum of angles of a triangle is equal to 180° ]

⇒∠AMB+(∠MAB+∠MAB)=180° 

(4) ∠AMB+90°=180°  [From (2) and (3)]

⇒∠AMB=180°-90°

⇒∠AMB=90°

(ii) Prove that bisectors of any two opposite angles of a parallelogram are parallel.

(ii) Given : A parallelogram ABCD in which bisects AR of ∠A meets DC in R and bisector CQ of ∠C meets AB in Q

To prove : AR||CQ

Proof :

Statements  : Reasons

(1) In parallelogram ABCD  [opposite angles of parallelogram are equal]

∠A=∠C

⇒$\frac{1}{2} \angle \mathrm{A}=\frac{1}{2} \angle \mathrm{C}$  [multiplying both sides by $\frac{1}{2}$]

⇒∠DAR=∠BCQ  (i) AR is bisector of $\frac{1}{2} \angle \mathrm{A}=\angle \mathrm{DAR}$

                              (ii) CQ is bisector of $\frac{1}{2} \angle C=\angle B C Q$

(2) In ΔADR and ΔCBQ

⇒∠DAR=∠BCQ  [Proved in (1) opposite sides of parallelogram ABCD are equal]

AD=BC

⇒∠D=∠B   [opposite sides of parallelogram ABCD are equal]

∴ΔADR≅ΔCBQ  [By A.S.A axiom of congruency]

∴∠DRA=∠BCQ  [c.p.c.t]

(3) ∠DRA=∠RAQ  [Alternate angles ]

                             [DC||AB , ∵ABCD is a parallelogram]

(4) ∠RAQ=∠BCQ   [From (2) and (3)]

But there are corresponding angles

∴AR||CQ  (Q.E.D)

(iii) If the diagonals of a quadrilateral are equal and bisect each other at right angles, then prove that it is a square.

(iii) Given : In quadrilateral ABCD, diagonals AC and BD are equal and bisects each other at right angles.

To prove : ABCD is a square

Proof : In ΔAOB and ΔCOD

AO=OC (given)

BO=OD (given)

∠AOB=∠COD (vertically opposite angles)

∴ΔAOB≅ΔCOD(SAS axiom)

∴AB=CD..(i)

and ∠OAB=∠OCD

But theses are alternate angles 

∴AB||CD

∴ABCD is a parallelogram

∵In a parallelogram , the diagonal bisects each other and are equal , opposite sides are equal.

In ΔAOB and ΔBOC

AO=OC

(Diagonals bisect each other at right angle)

OB=OB (common)

∠AOB=∠COB  (each 90°)

∴ΔBOC≅ΔCOD

∴BC=CD...(ii)

From (i) and (ii)

∴ABCD is square

---

Q12 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

OPEN IN YOUTUBE

Question 12

(i) If ABCD is a rectangle in which the diagonal BD bisect ∠B, then show that ABCD is a square.

(ii) Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Sol :

(i) If ABCD is a rectangle in which the diagonal BD bisect ∠B, then show that ABCD is a square.

(i) ABCD is a rectangle and its diagonals AC bisects ∠A and ∠C

To prove : ABCD is a square

Proof : 

∵Opposite sides of a rectangle are equal and parallel .Each vertex angle is 90° 

If BD bisects ∠B then it bisects ∠D as well.

∠1=∠2 (BD bisects ∠B)

But ∠B=∠D=90°

∴∠1=45° and ∠2=45°

Also,

∵AB||CD ,BD is transversal 

∴∠3=∠2=45°

∵BC||AD ,BD is transversal 

∴∠1=∠4=45°

Also,

∠1=∠2=∠3=∠4=45°

∵∠2=∠4

∴AB=BC (opposite sides of equal angles)

But AB=CD and BC=AD

∴AB=BC=CD=DA

∴ABCD is a square

(ii) Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

(ii) In quadrilateral ABCD diagonals AC and BD are equal and bisects each other at right angle 

To prove : ABCD is a square

Proof : 

In ΔAOB and ΔCOD

AO=OC (given)

BO=OD (given)

∠AOB=∠COD (vertically opposite angles)

∴ΔAOB≅ΔCOD(SAS axiom)

∴AB=CD...(i)

and ∠OAB=∠OCD

But theses are alternate angles 

∴AB||CD

∴ABCD is a parallelogram

∵In a parallelogram , the diagonal bisects each other and are equal , opposite sides are equal.

In ΔAOB and ΔBOC

AO=OC

(Diagonals bisect each other at right angle)

OB=OB (common)

∠AOB=∠COB  (each 90°)

∴ΔAOB≅ΔBOC

∴BC=CD...(ii)

From (i) and (ii)

∴ABCD is a square

---

Q13 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

OPEN IN YOUTUBE

Question 13

P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.

Sol :

ABCD is a parallelogram P and Q are the points on AB and DC. Diagonals AC and BD intersects each other at O

To prove : OP=OQ

Proof : ∴Diagonals of parallelogram ABCD bisects each other at O

∴AO=OC and BO=OD

Now in ΔAOP and ΔCOQ

AO=OC  (proved)

∠OAP=∠OCQ (alternate angles)

∠AOP=∠COQ (vertically opposite angles)

∴ΔAOP≅ΔCOQ  (SAS axiom)

∴OP=OQ

Hence O bisects PQ

---

Q14 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

OPEN IN YOUTUBE

Question 14

(a) In figure (1) given below, ABCD is a parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram ABPQ is completed. Prove that:

(i) the triangles ABX and QCX are congruent;

(ii)DC = CQ = QP

(b) In figure (2) given below, points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.

Sol :

(a) Given : ABCD is a parallelogram and X is mid point of BC. The line AX produced meets DC produced at Q and ABPQ is a parallelogram

To prove : (i) ΔABX≅ΔQCX

(ii) DC=CQ=QP

Proof :

Statements  :  Reasons 

(1) In ΔABX and ΔQCX

BX=XC   (X is the mid point of BC)

∠AXB=∠CXQ  (vertically opposite angles)

∠XCQ=∠XBA  (Alternate angle)

(∵AB||CQ)

∴ΔABX≅ΔQCX  [A.S.A]

(2) ∴CQ=AB  [c.p.c.t]

(3) AB=DC  [ABCD is a parallelogram]

(4) AB=QP  [ABPQ is a parallelogram]

(5) DC=CQ=QP  [From (2),(3) and (4)]

(Q.E.D)

(b) In figure (2) given below, points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.

(b) In parallelogram ABCD,

P and Q are points on AB and CD respectively

PQ and AC intersects each other at O and AP=CQ

To prove : AC and PQ bisects each other

i.e. AO=OC , PO=OQ

Proof : AC and PQ bisects each other

AP=CQ (Given)

∠AOP=∠COQ (vertically opposite angles)

∠OAP=∠OCQ  (Alternate angles)

∴ΔAOP≅ΔCOQ  (AAS axiom)

∴OP=OQ  (c.p.c.t)

and OA=OC  (c.p.c.t)

Hence AC and PQ bisects each other 

---

Q15 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

OPEN IN YOUTUBE

Question 15

ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP=DQ, prove that AP and DQ are perpendicular to each other.

Sol :

Given : ABCD is a square . P is any point on BC and Q is any point on AB and these points are taken such that AP=DQ

To prove : AP⟂DQ

Proof :

Statements   :   Reasons

(1) In ΔABP and ΔADQ

AP=DQ  (given)

AD=AB   (ABCD is a square)

∠DAQ=∠ABP  (ABCD is a square and each 90°)

∴ΔABP≅ΔADQ   [R.H.S axiom of congruency]

∴∠BAP=∠ADQ  

(2) But ∠BAD=90°   (each angle of square is 90°)

(3) ∠BAD=∠BAP+∠PAD

90°=∠BAP+∠PAD  [From (2)]

⇒∠BAP+∠PAD=90°

⇒∠PAD+∠ADQ=90°  [From (1)]

(4) In ΔADM,

⇒∠MAD+∠ADM+∠AMD=180°   [Sum of all angles in a triangle is 180°]

⇒∠90°+∠AMD=180°

⇒∠AMD=180°-90°

⇒∠AMD=90°

∴DM⟂AP

⇒DQ⟂AP

Hence , AP⟂DQ  (Q.E.D)

---

Q16 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

OPEN IN YOUTUBE

Question 16

If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ || AP.

Sol :

Given : ABCD is a parallelogram in which BP=PQ=QD

To prove : CQ||AP

Proof :

Statements   :   Reasons

(1) In parallelogram ABCD   [opposite sides of parallelogram are equal]

AB=CD

(2) In parallelogram ABCD  [From (1)]

AB=CD

and BD is the transversal

∴∠1=∠2   [Alternate angles]

(3) In ΔABP and ΔDCQ,

AB=CD  [opposite side of parallelogram are equal]

∠1=∠2   [From (2)]

BP=QD  [Given]

∴ΔABP≅ΔDCQ   [S.A.S axiom of congruency]

∴AP=QC   [c.p.c.t]

Also, ∠APB=∠DQC   [c.p.c.t]

⇒-∠APB=-∠DQC  [Multiplying both sides by (-1)]

⇒180°-∠APB  [adding 180° both sides]

⇒∠APQ=∠CQP

But there are alternate angles

∴AP||QC

⇒CQ||AP  (Q.E.D)

---

Q17 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

OPEN IN YOUTUBE

Question 17

A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected and so are the two interior angles at B ; the four bisectors form a quadrilateral ABCD. Prove that

(i) ABCD is a rectangle.

(ii) CD is parallel to the original parallel lines.

Sol :

Given : LM||PQ , AB transversal line cut ∠M at A and PQ at B

⇒AC, AD, BC and BD is the bisector of ∠LAB

⇒∠BAM , ∠PAB and ∠ABQ respectively

AC and BC intersects at C and AD and BD intersects at D.  A quadrilateral ABCD is formed

To prove : (i) ABCD is a rectangle 

(ii) CD||LM and PQ

Proof :

Statement   :   Reasons 

(1) ∠LAB+∠BAM=180°   [LAM is a straight line]

⇒$\frac{1}{2}$(∠LAB+∠BAM)   [Multiplying both sides by $\frac{1}{2}$]

=90° 

⇒$\frac{1}{2} \angle \mathrm{LAB}+\frac{1}{2} \angle \mathrm{BAM}$

=90°

⇒∠2+∠3=90°   [AC and AD is bisector of ∠LAB and ∠BAM respectively]

                     [∴$\frac{1}{2}$∠LAB=∠2 and $\frac{1}{2}$∠LAB=∠3]

⇒∠CAD=90°

⇒∠A=90°

(2) Similarly , ∠PBA+PBQ is a straight line ∠QBA=180°

⇒$\frac{1}{2}$∠PBA+$\frac{1}{2}$∠QBA   (Multiplying both sides by $\frac{1}{2}$)

⇒∠6+∠7=90°   [∵BC and BD is bisector of ∠PBA and ∠QBA respectively]

              $\frac{1}{2}$∠PBA=∠6

              $\frac{1}{2}$∠QBA=∠7

⇒∠CBD=90°

⇒∠B=90°

(3) ∴∠LAB+∠ABP=180°   [Sum of co-interior angles is 180° LM||PQ given]

 

$\frac{1}{2}$∠LAB+$\frac{1}{2}$∠ABP=90°    [Multiplying both sides by $\frac{1}{2}$ ]

 

∠2+∠6=90°   [∴AC and BC is bisector of ∠LAB and ∠PBA respectively 

∴$\frac{1}{2}$∠LAB=∠2 and $\frac{1}{2}$∠APB=∠6]

(4) In ΔACB

⇒∠2+∠6+∠C=180° [Sum of all angles in a triangle is 180°]

⇒(∠2+∠6)+∠C=180°

⇒90°+∠C=180°  [using (6)]

⇒∠C=90°

(5) ∴∠MAB+∠ABQ=180°  [sum of co-interior angles is 180°]

                 [(LM||PQ) given]

⇒$\frac{1}{2}$∠MAB+$\frac{1}{2}$∠ABQ$=\frac{180}{2}$   [Multiplying both sides by $\frac{1}{2}$]

⇒∠3+∠7=90°  ∵AD and BD bisect the ∠MAB and ∠ABQ

∴$\frac{1}{2}$∠MAB=∠3

and $\frac{1}{2}$∠ABQ=∠7

(6) In ΔADB ,

∵∠3+∠7+∠D=180°  [sum of all angles in a triangle is 180°]

⇒(∠3+∠7)+∠D=180°

⇒90°+∠D=180° [from (5)]

⇒∠D=180°-90°

⇒∠D=90°

(7) ∠LAB+∠BAM=∠BAM=∠ABP  [From (1) and (3)]

⇒$\frac{1}{2}$∠BAM=$\frac{1}{2}$∠ABP  [Multiplying both sides by $\frac{1}{2}$]

⇒∠3=∠6  ∵AD and BC is bisector of ∠BAM and ∠ABP respectively

∴$\frac{1}{2}$∠BAM=∠3 and $\frac{1}{2}$∠ABP=∠6

Similarly ∠2=∠7

(8) In ΔABC and ΔABD

⇒∠2=∠7  [From (7)]

⇒AB=AB (common)

⇒∠6=∠3 [From (7)]

∴ΔABC≅ΔABD  [By A.S.A axiom of congruency]

∴AC=DB  [c.p.c.t]

Also ,CB=AD  [c.p.c.t]

(9) ∠A=∠B=∠C=∠D=90°  [From (1) ,(2), (4) and (6)]

⇒AC=DB  [proved in (8)]

⇒CB=AD  [proved in (8)]

∴ABCD is a rectangle 

(10) ∵ABCD is a rectangle  [From (9)]

⇒OA=OD  [Diagonals of rectangle bisect each other]

(11) In ΔAOD [From (11)]

⇒OA=OD  Angles opposite to equal sides are equal 

∴∠9=∠3

(12) ∠3=∠4 AD bisects ∠MAB 

(13) ∠9=∠4 [From (11) and (12)]

But these are alternate angles

∴OD||LM

⇒CD||LM

Similarly we can prove that

⇒∠10=∠8

But these are alternate angles

∴OD||PQ

⇒CD||PQ

(14) CD||LM  [proved in (13)]

CD||PQ [proved in (19)]

(Q.E.D)

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Q18 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

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Question 18

In a parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2 AD. Prove that

(i) BE bisects ∠B

(ii) ∠AEB = a right angle.

Sol :

Given : ABCD is parallelogram in which bisectors of angle A and B meets in E and AB=2AD

To prove : (i) BE bisects ∠B

(ii) ∠AEB= a right angle i.e. ∠AEB=90°

Proof : 

Statements  :  Reasons

(1) In parallelogram ABCD 

∠1=∠2  [AD bisectors of ∠A]

(2) AB||DC and AE is the transversal

∴∠2=∠3  (alternate angles)

(3) ∠1=∠2  [From (1) and (2)]

(4) In ΔADE

∠1=∠3  [prove in (3)]

∴DE=AD  [sides opposite equal angles are equal]

⇒AD=DE

(5) AB=2AD  [given]

⇒$\frac{A B}{2}=A D$

⇒$\frac{\mathrm{AB}}{2}=\mathrm{DE}$

⇒$\frac{\mathrm{DC}}{2}=\mathrm{DE}$ [AD=DC]

(∵opposite sides of parallelogram are equal)

∴E is mid point of D

∴DE=EC

(6) AD=BC  [opposite sides of parallelogram are equal]

(7) DE=BC  [From (4) and (6)]

(8) EC=BC  [From (5) and (7)]

(9) In ΔBCE

⇒EC=BC  [proved in (8)]

∴∠6=∠5  [angles opposite equal sides are equal]

(10) AB||DC and BE is the transversal

∴∠4=∠5 [alternate angles]

(11) ∠4=∠6  [From (9) and (10)]

∴BE is bisector of ∠B 

(12) ∠A+∠B=180°  [sum of co-interior angles is equal to 180° (AD||BC)]

$\frac{1}{2} \angle \mathrm{A}+\frac{1}{2} \angle \mathrm{B}=\frac{180^{\circ}}{2}$  [Multiplying both sides by $\frac{1}{2}$]

∠2+∠4=90°  [AE is bisector of ∠A and BE is bisector of ∠B]

(13) In ΔAPB,

⇒∠AEB+∠2+∠4=180°

⇒∠AEB+90°=180° [From (12)]

⇒∠AEB=180°-90°

⇒∠AEB=90°

(Q.E.D)

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Q19 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

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Question 19

ABCD is a parallelogram, bisectors of angles A and B meet at E which lie on DC. Prove that AB=2AD

Sol :

Given : ABCD is a parallelogram in which bisector of ∠A and ∠B meets DC in E

To prove : AB=2AD

Proof :

Statements  :  Reasons 

(1) In parallelogram ABCD 

AB||DC

∠1=∠5  [Alternate angles (∵AE is transversal)]

(2) ∠1=∠2 [AE is bisector of ∠A (given)]

(3) ∠2=∠5 [From (1) and (2) equal angles have equal sides opposite to them]

In ΔAED,

DE=AD

(4) ∠3=∠6 [alternate angles]

(5) ∠3=∠4 [∵BE is bisector of ∠B (given)]

(6) ∠4=∠6 [From (4) and (5)]

In ΔBCE   [equal angles have equal sides opposite to them]

BC=EC 

(7) AD=BC [opposite sides of parallelogram are equal]

(8) AD=DE=EC  [From (3), (6) and (7)]

(9) AB=DC  [opposite sides of parallelogram are equal]

⇒AB=DE+EC

⇒AB=AD+AD [From (8)]

⇒AB=2AD

(Q.E.D)

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Q20 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

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Question 20

ABCD is a square and the diagonals intersect at O. If P is a point on AB such that AO =AP, prove that 3 ∠POB = ∠AOP.

Sol :

Given : ABCD is a square and the diagonals intersect at O. P is a point on AB such that 

AO=AP

To prove : 3∠POB=∠AOP

Proof : 

Statement  :   Reasons 

(1) In square ABCD, AC In square is diagonal

∴∠CAB=45° make 45° with side

⇒∠OAP=45°

(2) In ΔAOP

⇒∠OAP=45° [From (1) equal side have a equal angles opposite to them]

⇒AO=AP

∴∠AOP+∠APO+∠OAP=180°  [Sum of all angles in a triangle is 180°]

⇒∠AOP+∠AOP+45°=180°

⇒2∠AOP=180°-45°

⇒2∠AOP=135°

⇒$\angle \mathrm{AOP}=\frac{135^{\circ}}{2}$

(3) ∠AOB=90°  [In square ABCD diagonals bisect at right angles]

⇒∠AOP+∠POB=90°

⇒$\frac{135^{\circ}}{2}+\angle \mathrm{POB}=90^{\circ}$  [From (2)]

⇒$\angle \mathrm{POB}=90^{\circ}-\frac{135^{\circ}}{2}$

⇒$\angle \mathrm{POB}=\frac{180^{\circ}-135^{\circ}}{2}$

⇒$\angle \mathrm{POB}=\frac{45^{\circ}}{2}$

⇒$3 \angle \mathrm{POB}=\frac{135^{\circ}}{2}$  [Multiplying both sides by 3]

(4) ∠AOP=3∠POB  [From (2) and (3)]

(Q.E.D)

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Q21 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

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Question 21

ABCD is a square. E, F, G and H are points on the sides AB, BC, CD and DA respectively such that AE = BF = CG = DH. Prove that EFGH is a square.

Sol :

Given : ABCD is a square in which E,F,G and H are points AB, BC, CD and DA

Such that AE=BF=CG=DH

EF,FG,GH and HE are joined

To prove : EFGH is a square

Prove : ∵AE=BF=CG=DH

∴EB=FC=GD=HA

Now in ΔAEH and ΔBFE

⇒AE=BF (given)

⇒AH=EB (proved)

⇒∠A=∠B  (each 90°)

∴ΔAEH≅ΔBFE  (S.A.S. axiom)

∴EH=EF (c.p.c.t)

and ∠4=∠2 (c.p.c.t)

But ∠1+∠4=90°

∴∠1+∠2=90°  (∵∠4=∠2)

∴∠HEF=90°

Hence EFGH is square

Hence proved

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Q22 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

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Question 22

(a) In the Figure (1) given below, ABCD and ABEF are parallelograms. Prove that

(i) CDFE is a parallelogram

(ii) FD = EC

(iii) Δ AFD = ΔBEC.

(b) In the figure (2) given below, ABCD is a parallelogram, ADEF and AGHB are two squares. Prove that FG = AC

Sol :

(a) Given : ABDC and ABEF are parallelogram 

To prove : (i) CDEF is parallelogram

(ii) FD=EC

(iii) ΔAFD≅ΔBEC

Proof :

Statements  :   Reasons

(1) DC||AB and DC=AB  [ABCD is a parallelogram]

(2) FE||AB and FE=AB [ABEF is a parallelogram] 

(3) DC||FE and DC=FE [From (1) and (2)]

∴CDFE is a parallelogram  [If a pair of opposite sides of a quadrilateral are parallel and equal]

It is a parallelogram 

(4) CDFE is a parallelogram  [opposite sides of parallelogram CDFE are equal]

FD=EC

(5) In ΔAFD and ΔBEC  [opposite sides parallelogram ]

⇒AD=BC [ABCD are equal]

⇒AF=BE [opposite sides of parallelogram ABEF are equal]

⇒FD=EC [From (4)]

∴ΔAFD≅ΔBEC  [By S.S.S. axiom of congruency]

(Q.E.D)

(b) Given : ABCD is a parallelogram , ADEF and AGHB are two squares

To prove : FG=AC

Proof :

Statements  :  Reasons 

(1) ∠FAG+90°+90°+∠BAD=36°    [At a point total angle is 360°]

⇒∠FAG=360°-90°-90°-∠BAD

⇒∠FAG=180°-∠BAD  [ABCD is a parallelogram]

(2) ∠B+∠BAD=180°  [Sum of adjacent angle in parallelogram is equal to 180°]

⇒∠B=180°-∠BAD

(3) ∠FAG=∠B  [From (1) and (3)]

(4) In ΔAFG and ΔABC  [FA, DE and ABCD ]

⇒AF=BC  [both are square on the same base DA]

⇒Similarly AG=AB

⇒∠FAG=∠B  [From (3)]

∴ΔAFG≅ΔABC  [By S.A.S axiom of congruency]

∴FG=AC  [c.p.c.t]

(Q.E.D)

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Q23 | Ex-13.1 |Class 9 | ML Aggarwal | Rectilinear Figures | Ch-13 | myhelper

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Question 23

ABCD is a rhombus in which ∠A = 60°. Find the ratio AC : BD.

Sol :

Let each side of the rhombus ABCD=a 

∵∠A=60°

∴ΔABD is an equilateral triangle 

∴BD=AB=a

∵The diagonals of a rhombus bisect each other at right angles

∴In right ΔAOB,

⇒AO²+OB²=AB²

⇒AO²=AB²-OB²

⇒$=a^{2}-\left(\frac{1}{2} a\right)^{2}$

⇒$=a^{2}-\frac{a^{2}}{4}=\frac{3}{4} a^{2}$

∴$\mathrm{AO}=\sqrt{\frac{3}{4}} a^{2}=\frac{\sqrt{3}}{2} a$

But AC=2AO$=2 \times \frac{\sqrt{3}}{2} a=\sqrt{3} a$

Now AC : BD$=\sqrt{3} a: a=\sqrt{3}: 1$