SELINA Solution Class 9 Chapter 14 Rectilinear Figures (Quadrilaterals: Parallelogram , Rectangle , Rhombus, Square and Trapezium) Exercise 14C

Question 1

E is the mid-point of side AB and F is the mid-point of side DC of parallelogram ABCD. Prove that AEFD is a parallelogram.

Sol:

Let us draw a parallelogram ABCD Where F is the midpoint Of side DC and E is the mid-point of side AB of a parallelogram  ABCD.

To prove:  AEFD is a parallelogram

Proof: 
In parallelogram ABCD
AB || DC
BC || AD
AB = DC
${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}}$DC
AE = DF
Also AD || EF
Therefore, AEFD is a parallelogram.

Question 2

The diagonal BD of a parallelogram ABCD bisects angles B and D. Prove that ABCD is a rhombus.

Sol:

Given: ABCD is a parallelogram where the diagonal BD bisects
parallelogram  ABCD at angle B and D

To Prove:  ABCD is a rhombus

Proof: Let us draw a parallelogram  ABCD where the diagonal BD bisects the parallelogram at an angle B and D.

Construction: Let us join AC as a diagonal of the parallelogram ABCD

Since ABCD  is a parallelogram
Therefore
AB = DC
AD =BC
Diagonal  BD bisects angle B and D
So ∠COD = ∠DOA
Again AC also bisects at A and C   

Therefore ∠AOB =∠ BOC
Thus ABCD is a rhombus.
Hence proved

Question 3

The alongside figure shows a parallelogram ABCD in which AE = EF = FC.
Prove that:
(i) DE is parallel to FB
(ii) DE = FB
(iii) DEBF is a parallelogram.

Sol:

Construction : 
Join DF and EB
Join diagonal BD

Since diagonals of a parallelogram bisect each other.
∴  OA = OC and OB = OD
Also, AE = EF = FC

Now, OA = OC and AE = FC
⇒  OA - AE = OC - FC
⇒  OE = OF

Thus, in quadrilateral DEFB, bisect each other.
OB = OD and OE = OF
⇒  Diagonals of a quadrilateral DEFB bisect each other.
⇒ DEFB is a parallelogram.
⇒  DE is parallel to FB
⇒  DE = FB                      .....( Opposite sides are equal )

Question 4

In the alongside diagram, ABCD is a parallelogram in which AP bisects angle A and BQ bisects angle B. Prove that : 

(i) AQ = BP
(ii) PQ = CD.
(iii) ABPQ is a parallelogram.

Sol:

Let us join PQ. 
Consider the ΔAOQ and ΔBOP
∠AOQ = ∠BOP                  ..[ opposite angles ]
∠OAQ = ∠BPO                 ...[ alternate angles ]
⇒ ΔAOQ ≅ ΔBOP            ...[ AA test ]
Hence AQ = BP

Consider the ΔQOP and ΔAOB
∠AOB = ∠QOP                 ...[ opposite angles ]
∠OAB = ∠APQ                 ...[ alternate angles ]
⇒ ΔQOP ≅ ΔAOB              ...[ AA test ]
Hence PQ = AB = CD

Consider the quadrilateral QPCD
DQ = CP and DQ || CP ||       ...[ Since AAD = BC and AD || BC ]
Also QP = DC and AB || QP || DC

Hence Quadrilateral QPCD is a Parallelogram.

Question 5

In the given figure, ABCD is a parallelogram. Prove that: AB = 2 BC.

Sol:

Given ABCD is a parallelogram
To prove: AB = 2BC

Proof:  ABCD is a parallelogram
A + D + B + C = 180°

From the AEB we have
⇒ ${\displaystyle \frac{\text{∠A}}{2}+\frac{\text{∠B}}{2}}$ + E = 180°
⇒ ∠A - ${\displaystyle \frac{\text{∠A}}{2}}$ + ∠D + ∠E1 = 180° ...[ taking E1 as new angle ]
⇒ ∠A + ∠D + ∠E1 = 180° + ${\displaystyle \frac{\text{∠A}}{2}}$ 
⇒ ∠E1 = ${\displaystyle \frac{\text{∠A}}{2}}$  ...[ Since ∠A + ∠D = 180°

Again,
similarly ,
∠E1 = ${\displaystyle \frac{\text{∠B}}{2}}$ 
Now
AB = DE + EC
= AD + BC
= 2BC                                ...[ since AD = BC]
Hence proved.

Question 6

Prove that the bisectors of opposite angles of a parallelogram are parallel.

Sol:

Given ABCD is a parallelogram. The bisectors of ∠ADC and ∠BCD meet at E. The bisectors of ∠ABC and ∠BCD meet at F

From the parallelogram ABCD we have

∠ADC + ∠BCD = 180° ...[ sum of adjacent angles of a parallalogram ]
⇒  ${\displaystyle \frac{\text{∠ADC}}{2}+\frac{\text{∠BCD}}{2}}$ = 90°
⇒ ∠EDC + ∠EDC = 90°

In triangle ECD sum of angles = 180°
⇒ ∠EDC + ∠ECD + ∠CED = 180°
= ∠CED = 90°

Similarly taking triangle BCF it can prove that ∠BFC = 90°
Now since
∠BFC = ∠CED = 90°

Therefore the lines DE and BF are parallel
Hence proved

Question 7

Prove that the bisectors of interior angles of a parallelogram form a rectangle.

Sol:

Given:
ABCD is a parallelogram
AE bisects ∠BAD
BF bisects ∠ABC
CG bisects ∠BAD
DH bisects ∠ADC

To prove: LKJI is a rectangle

Proof :
∠BAD + ∠ABC = 180° ...[ adjacent angles of a parallelogram are supplementary ] 
∠BAJ = ${\displaystyle \frac{1}{2}}$ ∠BAD ...[AE bisects  BAD ]
∠ABJ = ${\displaystyle \frac{1}{2}}$ ∠ABC ... [DH  bisect ABC ]
∠BAJ + ∠ABJ = 90° ...[ halves of supplementary angles are complementary ]

ΔABJ is a right triangle because its acute interior angles are complementary.
Similarly
∠DLC = 90°
∠AID = 90°

Then ∠JIL = 90° because ∠AID and ∠JIL are vertical angles

since 3 angles of a quadrilateral, LKJI are right angles, si is the 4^th one and so is LKJI a rectangle, since its interior angles are all right angles
Hence proved.

Question 8

Prove that the bisectors of the interior angles of a rectangle form a square.

Sol:

Given: A parallelogram ABCD in which AR, BR, CP, DP 
Are the bisects of ∠A, ∠B, ∠C, ∠D, respectively forming quadrilaterals PQRS.

To prove: PQRS is a rectangle

Proof :
∠DCB + ∠ ABC =180° ...[ co - interior angles of parallelogram are supplementary ]

⇒ ${\displaystyle \frac{1}{2}}$ ∠DCB + ${\displaystyle \frac{1}{2}}$∠ABC = 90° 
⇒ ∠1 + ∠2 = 90° 
ΔCQB, ∠1 + ∠2 + ∠CQB = 180° 

From the above equation we get
∠CQB = 180° - 90° = 90° 
∠ RQP = 90°         ...[ ∠CQB = ∠ RQP , vertically opposite angles ]
∠QRP = ∠RSP = ∠SPQ = 90° 
So, PQRS is a square.
Hence Proved.

Question 9

In parallelogram ABCD, the bisector of angle A meets DC at P and AB = 2 AD.
Prove that:
(i) BP bisects angle B.
(ii) Angle APB = 90^o.

Sol:

(i) Let AD = x
AB = 2AD = 2x
Also AP is the bisector ∠A
∠1 = ∠2
Now,
∠2 = ∠5                   ...[ alternate angles ]
Therefore ∠1 = ∠5
Now
AP = DP = x ...[ sides opposite to equal angles are also equal ]
Therefore
AB = CD ...[ opposite sides of  parallelogram are equal ]
CD = 2x
⇒ DP + PC = 2x
⇒ x + PC = 2x
⇒ PC = x
Also, BC = x
ΔBPC
⇒ ∠6 = ∠4 ...[ angles opposite to equal sides are equal ]
⇒ In ∠6 = ∠3
Therefore ∠3 =∠ 4
Hence BP bisect ∠B

(ii)
Opposite angles are supplementary
Therefore

∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ 2 ∠2 + 2 ∠3 =180°     .....[ ∠1 = ∠2 , ∠3 = ∠4 ]
⇒ ∠2 + ∠3 = 90°
ΔAPB
∠2 + ∠3 ∠APB = 180°
⇒ ∠APB = 180° - 90° ...[ by angle sum property ] 
⇒ ∠APB = 90° 
Hence proved.

Question 10

Points M and N are taken on the diagonal AC of a parallelogram ABCD such that AM = CN. Prove that BMDN is a parallelogram.

Sol:

Points M are N taken on the diagonal AC of a parallelogram ABCD such that.
Prove that BMDN is a parallelogram

construction: Join B to D to meet AC in O.

Proof: We know that the diagonals of a parallelogram bisect each other.
Now, AC and BD bisect each other at O.
OC = OA
AM = CN
OA - AM = OC - CN
OM = ON

Thus in a quadrilateral BMDN, diagonal BD and MN are such that OM = ON and OD = OB

Therefore the diagonals AC and PQ bisect each other.
Hence  BMDN is a parallelogram

Question 11

In the following figure, ABCD is a parallelogram.

Prove that:
(i) AP bisects angle A.
(ii) BP bisects angle B
(iii) ∠DAP + ∠BCP = ∠APB

Sol:

Consider ΔADP and ΔBCP,
AB = BC               ....[ Since ABCD is a parallelogram. ]
DC = AB               ....[ Since ABCD is a parallelogram. ]
∠A ≅ ∠C              ....[ Opposite angles ]
ΔADP ≅ ΔBCP     .....[ SAS ]

Therefore, AP = BP
AP bisects ∠A
BP bisects ∠B

In ΔAPB, AP = BP
AP bisects ∠A
BP bisects ∠B

In ΔAPB,
AP = PB
∠APB = ∠DAP + ∠BCP
Hence proved

Question 12

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:

ABCD is a square and AP = PQ.

Consider ΔDAQ and ΔABP,
∠DAQ = ∠ABP = 90°
DQ = AP 
AD = AB
ΔDAQ ≅ ΔABP
⇒  ∠PAB = ∠QDA

Now,
∠PAB + ∠APB = 90°
also ∠QDA + ∠APB = 90°      ....[ ∠PAB = ∠QDA ]

Consider ΔAOQ by ASP
∠QDA + ∠APB + ∠AOD = 180° 
⇒ 90° + ∠AOD = 180° 
⇒ ∠AOD = 90° 
Hence AP and DQ are perpendicular.

Question 13

In a quadrilateral ABCD, AB = AD and CB = CD.
Prove that :
(i) AC bisects angle BAD.
(ii) AC is the perpendicular bisector of BD.

Sol:

Given: ABCD is quadrilateral,
AB = AD
CB = CD

To prove:
(i) AC bisects angle BAD.
(ii) AC is the perpendicular bisector of BD.

Proof:
In ΔABC and ΔADC,
AB = AD                    ....(given)
CB = CD                   .....(given)
AC = AC                   ......(Common side)
ΔABC ≅ ΔADC         .......(SSS)

∠BAD = ∠DAO       .......(AC bisects A)

Therefore AC bisects ∠BAD
OD = OB
OA = OC                 ......( diagonals bisect each other at O )
Thus AC is perpendicular bisector of BD.
Hence proved.

Question 14

The following figure shows a trapezium ABCD in which AB is parallel to DC and AD = BC.

Prove that:
(i) ∠DAB = ∠CBA
(ii) ∠ADC = ∠BCD
(iii) AC = BD
(iv) OA = OB and OC = OD.

Sol:

Given ABCD is a trapezium, AB || DC and AD = BC.

Prove that:
(i) ∠DAB = ∠CBA
(ii) ∠ADC = ∠BCD
(iii) AC = BD
(iv) OA = OB and OC = OD.

Proof: (i) Since AD || CE and transversal AE cuts them at A and E respectively.
Therefore, ∠A + ∠B = 180°
Since, AB || CD and AD || BC
Therefore, ABCD is a parallelogram.
∠A = ∠C
∠B = ∠D              ....[ Since ABCD is a parallelogram ]
Therefore,
∠DAB = ∠CBA
∠ADC = ∠BCD

In ΔABC and ΔBAD, we have
BC = AD                 ....( given )
AB = BA                 ....( Common )
∠A = ∠B                 ....( proved )
ΔABC ≅ ΔBAD        ....( SAS )  
ΔABC ≅ ΔBAD
Since, Therefore AC = BD....( Corresponding parts of congruent triangles are equal. )
OA = OB
Again OC = OD      ....( Since diagonals bisect each other at O )
Hence proved.

Question 15

In the given figure, AP is the bisector of ∠A and CQ is the bisector of ∠C of parallelogram ABCD.

Prove that APCQ is a parallelogram.

Sol:

Construction: Join AC

Proof:
∠BAP = ${\displaystyle \frac{1}{2}}$∠A             ...( AP is the bisector  of ∠A )

∠DCQ = ${\displaystyle \frac{1}{2}}$∠C            ...( CQ is the bisector of ∠C ) 
⇒ ∠BAP = ∠DCQ           ....(i)....[ ∠A = ∠R ( Opposite angles of a parallelogram.) ]
Now,
∠BAC = ∠DCA                ....(ii)....[ Alternate angles since AB || DC ]

Subtracting (ii) from (i), We get
∠BAP - ∠BAC = ∠DCQ - ∠DCA 
⇒ ∠CAP = ∠ACQ
⇒ AP || QC                    .....( Alternate angles are equal )
Similarly, PC || AQ.
Hence, APCQ is a parallelogram.

Question 16

In case of a parallelogram
prove that:
(i) The bisectors of any two adjacent angles intersect at 90^o.
(ii) The bisectors of the opposite angles are parallel to each other.

Sol:

ABCD is a parallelogram, the bisectors of ∠ADC and ∠BCD meet at a point E and the bisectors of ∠BCD and ∠ABC meet at F.

We have to prove that the ∠CED = 90° and ∠CFG = 90°

Proof: In the parallelogram ABCD
∠ADC + ∠BCD = 180°       ....[ sum of adjacent angles of a parallelogram ]

⇒ ${\displaystyle \frac{\text{∠ADC}}{2}+\frac{\text{∠BCD}}{2}}$ = 90°

⇒ ∠EDC + ∠ECD + ∠CED = 180°
⇒ ∠CED = 90°

Similarly taking triangle BCF it can be proved that ∠BFC = 90°
∠BFC + ∠CFG = 180°               ....[ adjacent angles on a line ]
Also ⇒ ∠CFG = 90°
Now since ∠CFG = ∠CED = 90° ....[ It means that the lines DE and BG are parallel ]
Hence proved.

Question 17

The diagonals of a rectangle intersect each other at right angles. Prove that the rectangle is a square.

Sol:

To prove: ABCD is a square,
that is, to prove that sides of the quadrilateral are equal
and each angle of the quadrilateral is 90°,
ABCD is a rectangle,
⇒ ∠A = ∠B = ∠c = ∠D = 90° and diagonals bisect each other.

that is, MD = BM                     ..(i)
Consider ΔAMD and ΔAMB,
MD =  BM                               ....( from(i) )
∠AMD = ∠AMB = 90°             .....(given)
AM = AM                                ......( common side )
ΔAMD ≅ ΔAMB                      ....(SAS congruence criterion)
⇒ AD = AB                                 ...( c.p.c.t.c. )
Since ABCD is a rectangle, AD = BC and AB = CD
Thus, AB = BC = CD = AD and ∠A = ∠B = ∠C = ∠D = 90°
⇒ ABCD is a square.

Question 18

In the following figure, ABCD and PQRS are two parallelograms such that ∠D = 120° and ∠Q = 70°.
Find the value of x.

Sol:

ABCD is a parallelogram.
⇒ Opposite angles of a parallelogram are congruent.
⇒ ∠DAB = ∠BCD and ∠ABC = ∠ADC = 120°
In ABCD,
∠DAB + ∠BCD + ∠ABC + ∠ADC = 360°   ....( sum of the measures of angles of a quadrilateral )
⇒ ∠BCD + ∠BCD + 120° + 120° = 360°
⇒  2∠BCD = 360° - 240°
⇒  2∠BCD = 120°
⇒  ∠BCD = 60°
PQRS is parallelogram.
⇒ ∠PQR = ∠PSR = 70°
In ΔCMS,
∠CMS + ∠CSM + ∠MCS = 180°      ....( angle sum property )
⇒ x + 70° + 60°  = 180° 
⇒ x  = 50°

Question 19

In the following figure, ABCD is a rhombus and DCFE is a square.

If ∠ABC =56°, find:
(i) ∠DAE
(ii) ∠FEA
(iii) ∠EAC
(iv) ∠AEC

Sol:

ABCD is a rhombus.
⇒ AD = CD and ∠ADC = ∠ABC = 56°
DCFE is a square.
⇒ ED = CD and ∠FED = ∠EDC = ∠DCF = ∠CFE = 90°
⇒ AD = CD = ED
In ΔADE,
AD = ED
⇒ ∠DAE = ∠AED                ...(i)
∠DAE + ∠AED + ∠ADE = 180°
⇒ 2∠DAE + 146° = 180°            ....( Since ∠ADE = ∠EDC + ∠ADC = 90° + 56° = 146° )
⇒ 2∠DAE = 34°
⇒ ∠DAE = 17°
⇒ ∠DEA = 17°                   ....(ii)

In ABCD,
∠ABC + ∠BCD + ∠ADC + ∠DAB = 360°
⇒ 56° + 56° + 2 ∠DAB = 360°   ....( ∵ Opposite angles of a rhombus are equal.)
⇒ 2∠DAB = 248°
⇒ ∠DAB = 124°
We know that diagonals of a rhombus, bisect its angles.

⇒ ∠DAC = ${\displaystyle \frac{124°}{2}}$ = 62°

⇒ ∠EAC = ∠DAC - ∠DAE = 62° - 17° = 45°
Now,
∠FEA = ∠FED - ∠DEA  
          = 90° - 17°             ....( From(ii) and each angle of a square is 90° )
        = 73°       
We know that diagonals of a square bisect its angles.
⇒ ∠CED = ${\displaystyle \frac{90°}{2}}$ = 45°
So,
∠AEC = ∠CED - ∠DEA
          = 45° - 17°
          = 28°
Hence, ∠DAE = 17°, ∠FEA = 73°, ∠EAC = 45° and ∠AEC = 28°.           

SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14C

  Exercise 14 C

Question 1 

Find the area of each parallelogram.

$\begin{array}{|l|l|l|l|l|l|}\hline & \text { (i) } & \text { (ii) } & \text { (iii) } & \text { (iv) } & \text { (v) } \\\hline \text { Base } & 8 \mathrm{~cm} & 12 \mathrm{~mm} & 6.5 \mathrm{~m} & 1 \mathrm{~m}5 \cdot \mathrm{cm} & 4.2 \mathrm{dm} \\\hline \text { Height } & 3 \mathrm{~cm} & 8.7 \mathrm{~cm} & 4.8 \mathrm{~m} & 45\mathrm{~cm} & 25 \mathrm{~cm} \\\hline\end{array}$

 (i) Area = $=B \times H$

$=8 \times 3$

$=24 \mathrm{~cm}^{2}$

(ii) Base $212 \mathrm{~mm}$

$1 c m=10 \mathrm{~mm}$

B= 1.2cm 

Area = $=B \times H$

$=1.2 \times 3.7=10.44\mathrm{~cm}2$

(iii) 

$\begin{aligned} A &=B \times H \\ &=6.5 \times 4.8 \\ &=31.20 \\ & m^{2} \end{aligned}$

(iv) $1 m=100 \mathrm{~cm}$

$B=105 \mathrm{~cm}$

Area $=105 \times 45^{-}$

$=4725 \mathrm{~cm}^{2}$ $=0.4725 \mathrm{~m}^{2}$

Question 2

(i) Area of ||gm ABCD = $48 \mathrm{~cm}^{2}$ DE = 6cm AB =? 

(DIAGRAM TO BE ADDED)

Sol: Area $=B \times 4$

$\begin{aligned}&48=D E \times A B \\&A B=\frac{43}{6} \\&A B=8 \mathrm{~cm}\end{aligned}$

(ii) 

Area of $\| \mathrm{gm}$ PQRS $=252 \mathrm{~cm}^{2}$ PQ = 9CM RT= ? 

Sol: 

$\begin{aligned} \text { Area } &=P Q \times RT \\ \Rightarrow 252 &=9 \times R T \\ \Rightarrow R T &=\frac{252}{9} \Rightarrow R T=28 \mathrm{~cm} \end{aligned}$

Question 3

The side of a rhombs is 7.2cm and its altitude is 5cm. Find its area. 

[Hint. Since a rhombus is a parallelogram with all its sides equal, the formula for area of a ||gm is applicable to it also]

Sol: Area = Base $ \times $ Height (altitude)

$=7.2 \times 5$

$=36 \mathrm{~cm}^{2}$

Question 4

The adjacent sides of a parallelogram are 36 cm and 27cm in length . if the perpendicular distance between the shorter sides is 12 cm, find the distacne between the longer sides. 

(IMAGE TO BE ADDED)

Sol: Given AE = 12CM 

Area of parallelogram ABCD = BASE $\times$ height 

$=D C \times A E$

$=27 \times 12 \mathrm{~cm}^{2}$

As we know Area of ||gm ABCD = $ BC \times AF $ = $DC \times AE $

$\Rightarrow A F=\frac{27 \times 12}{36} \Rightarrow A F=9 \mathrm{~cm}$,

Question 5

The area of a parallelogram and a square are the same. If the perimeter of the square is $160 \mathrm{~m}$ and the height of the parallelogram is $20 \mathrm{~m}$, find the length of the corresponding base of the parallelogram.

Sol: Perimeter of square = 160 m 

$4 \times$ side $=160 \Rightarrow$ side $=\frac{160}{4}$ =  Side = 40 m

Area of square = side $^{2}$ $=40^{2}=1600 \mathrm{~m}^{2}$

Area of parallelogram = Base $ \times $ Height $=1600$

$\Rightarrow$ Base $=\frac{1600}{20}$

 Base $=80 \mathrm{~m}$

Question 6

The area of a rhombus is $42 \mathrm{~m}^{2}$. If its perimeter is $24 \mathrm{~m}$, find its altitude.

Sol: Perimeter = 24 m 

$4 \times $ side $=24 \Rightarrow$ side $=6 \mathrm{~m}$

Area = Base $ \times $ Height= 42

Height = $\frac{42}{6}$

H= 7m answer

S Chand Class 10 CHAPTER 14 Circle Exercise 14C

 Exercise 14C 

Question 1

Ans: (a) In the figure APB is tangent to the circle with center O 

$\angle Q P D=50^{\circ}$

if $O P$ is the radius and $A P B$ is tangent

So $O P \perp A P B$

So $\angle O P B=90^{\circ} \Rightarrow \angle O P Q+\angle Q P B=90^{\circ}$

$\begin{aligned}&\Rightarrow \angle O P Q+30^{\circ}=90^{\circ} \\&\Rightarrow \angle O P Q=90^{\circ}-50^{\circ} \\&\Rightarrow O P Q=40^{\circ}\end{aligned}$

But in $\triangle O P Q, O P=O Q$

So

$\begin{aligned}&\angle O P Q=\angle O Q P=40^{\circ} \\&\angle P O Q+\angle O P Q+\angle O Q P=180^{\circ} \\&\Rightarrow \angle P O Q+40^{\circ}+40^{\circ}=180^{\circ} \\&\Rightarrow \angle P O Q+80^{\circ}=180^{\circ} \\&\text { So } \angle P OQ=180^{\circ}-80=100^{\circ}\end{aligned}$

(b) In circle, two tangent $A B$ and $A C$ are drawn, a point A outside the circle

so $A C=A B=4 \mathrm{~cm}$

(c) In the figure a circle with center O from a point P out side the circle Two tangents PQ and PR are 

Drawn and $\angle Q P R=80^{\circ}$

So $\angle Q P R$ and $\angle Q O R$ are supplementry

So $\angle Q P R+\angle Q O R=180^{\circ}$

$\begin{aligned}&\Rightarrow 80^{\circ}+\angle Q O R=180^{\circ} \\&\Rightarrow \angle Q O R=180^{\circ}-80^{\circ}=100^{\circ} \\&\text { SO } \angle Q O R=100^{\circ}\end{aligned}$

Question 2

Ans: In the figure a circle with center 0 from a point $P$ outside of it , tangents $P T$ and $P S$ are drawn to the circle and $\angle T P O=30^{\circ}$

In $\triangle P T O, O T \perp P T$

So $\angle O T P=90^{\circ}$

So $\angle T O P+\angle T P O=90^{\circ}$

$\begin{aligned}&\Rightarrow \angle \text { TOP }+30^{\circ}=90^{\circ} \\&\Rightarrow \angle T O P=90^{\circ}-30^{\circ}=60^{\circ}\end{aligned}$

So OP is the bisects of ㄥTOS

So $\angle T O P=\angle P O S=60^{\circ}$

Question 3

Ans: In the figure $B D$ is the diameter of the circle $P Q$ it tangent to the circle at $A$

$\angle A D B=30^{\circ}, \angle O B C=60^{\circ}$

(i) If QAP is tangent and AB is chord of the circle 

So  $\angle Q A B=\angle A D B=30^{\circ}$

(ii) $\angle P A D+\angle D A B+\angle Q A B=180^{\circ}$

$\Rightarrow \angle P A D+90^{\circ}+30^{\circ}=180^{\circ}$

$\begin{aligned}&\Rightarrow \angle P A D+120^{\circ}=180^{\circ} \\&\Rightarrow \angle P A D=180^{\circ}-120^{\circ}=60^{\circ}\end{aligned}$

(iii) In $\triangle B C D$

$\begin{aligned}& \angle C O B+\angle C B D+\angle B C D=180^{\circ} \\\Rightarrow & \angle C D B+60^{\circ}+90^{\circ}=180^{\circ} \\\Rightarrow & \angle C D B+150^{\circ}=180^{\circ} \\\Rightarrow & \angle C D B=180^{\circ}-150^{\circ}=30^{\circ}\end{aligned}$

Question 4

Ans: In the figure PQ and PR the tangents drawn from P outside the circle such that 

PQ= PR =9cm and $\angle Q P R=60^{\circ}$

$Q R$ is joined

In $\triangle P Q R 1<Q P R=60^{\circ}$

if $P Q=P R$

So $\angle P Q R=\angle P R Q=60^{\circ}$

So $\triangle P Q R$ is an equilateral triangle $P Q=P R=Q R=9 \mathrm{~cm}$

Question 5

Ans: In a circle of radius 3cm, point P is 3cm  away from the center O of the circle 

PQ and PR are the tangents drawn from P to the circle 

(IMAGE TO BE ADDED)

if OQ is radius and PQ is tangent

So $O Q \perp Q P$ or $\angle O Q P=90^{\circ}$

Now in right angled $\triangle O P Q$.

$\begin{aligned}&O P^{2}=O Q^{2}+P Q^{2} \\&=(5)^{2}=(3)^{2}+P Q^{2}\end{aligned}$

$\Rightarrow 25=9+P Q^{2} \Rightarrow P Q^{2}=25-9=16=(4)^{2}$

So $P Q=4 \mathrm{~cm}$

BUT PQ $=P R$

So $P Q=P R=4 \mathrm{~cm}$

Question 6

Ans:  (IMAGE TO BE ADDED)

if from A, A Q and A R the tangents drawn to the circle

So $\quad A Q=A R=5 \mathrm{~cm}$.......(i)

Similarly from B tangent BQ and BP are drawn 

So $B Q=B P$...........(ii)

and from C

C P=C R

 Now perimeter of $\triangle A B C$,

$=A B+A C+B C=A B+A C+B P+C P$

$=A B+A C+B Q+C R$

$=A B+B Q+A C+C R$

$=A Q+A R=5 \mathrm{~cm}+5 \mathrm{~cm}$

$=10 \mathrm{~cm}$

Question 7

Ans:  Two circle, which are concentric and their center is O, are radii 5cm and 3cm 

i.e OA = 5cm and OP = 3cm 

AB is chord of larger circle which touches the smaller circle at P 

Question 8

Ans: $A B C$ is a triangle and with center $A \perp B$ and $C$, three circle are drawn touching each other externally at $P, Q$ and $R$ respectively $A B=4 \mathrm{~cm}, B C=7 \mathrm{~cm}$ and $A C=6 \mathrm{~cm}$

(IMAGE TO BE ADDED)

Let radii of circle with center $A, B$ and $C$ respectively be $x, y$ and $z$

So $A B=x+y, B C=y+2, C A=2+x$

$\begin{gathered}\Rightarrow x+y=4 \mathrm{~cm}, y+z=6 \mathrm{~cm}, z+x=7 \mathrm{~cm} \\\text { So } A B+B C+C A=x+y+y+z+z+x \\\Rightarrow 4+6+7=2(x+y+z) \\\Rightarrow x+y+z=\frac{17}{2}=8.5 \mathrm{~cm}\end{gathered}$

Subtracting From $x+y+z$, we qut

$x=8.5-4=4.5 \mathrm{~cm}$

$y=8.5-6=2.5 \mathrm{~cm}$

$z=8.5-7=1.5 \mathrm{~cm}$

Hence their radii are $2.5 \mathrm{~cm} 1.5 \mathrm{~cm}$ and $4.5 \mathrm{~cm}$

Question 9

Ans: Two equal circles with center O and O' touch each other externally at X. OO' is produce to meet the circle O' at A. Through A, a tangent AC is drawn to the circle with center O. O'D$\perp A C$

Let $r$ be the radius of each circle 

In $\triangle A O^{\prime} D$ and $\triangle A O C_{1}$

$\angle D=\angle C$

$\angle A=\angle A$

(i) So $\triangle A O^{\prime} O \sim \triangle A O C$

So $\frac{A O^{\prime}}{A O}=\frac{r}{A X+X 0}=\frac{r}{2 r+r}=\frac{r}{3 r}=\frac{1}{3}$

(ii) So

$\begin{aligned}\frac{\text { area of } \triangle A D O^{\prime}}{\text { area of } \triangle A C O}=\frac{A O^{\prime2}}{A O^{2}} &=\left(\frac{1}{3}\right)^{2} \\&=\frac{1}{9}\end{aligned}$

Question 10

Ans: Two circles with centers $P$ and $Q$ touch externally at $R$ $\times 4$ is their common tangent

$A$ and $B$ are their points af contact

Join $P A, Q B$ and $P Q$

from $Q$, draw QS|| XY

(IMAGE TO BE ADDED)

if $P A$ and $Q B$ are perpendicular to $x 4$ and $Q S$ || $A B$

So $Q S=A B$

$P A=12 \mathrm{~cm}, Q B=3 \mathrm{~cm}, P Q=12+3=15 \mathrm{~cm}$

So $P S=P A-S A=12-3=9 \mathrm{~cm}$

Now in right $\triangle P S Q$,

$P Q^{2}=P S^{2}+Q S^{2}$

$=(15)^{2}=(9)^{2}+Q S^{2} \Rightarrow 225=81+Q S^{2}$

$\Rightarrow Q S^{2}=225-81=144=(12)^{2}$

So $Q S=12 \mathrm{~cm}=A B=Q S=12 \mathrm{~cm}$

S.chand class 6 Mathematics Chapter 14 Exercise 14C

 Exercise 14C

Question 1

1. Multiply:

(i) $2 x$ by $3 y$

(ii) $-a$ by $2 b$

(iii) $-2 m$ by $-3 n$

(iv) $-y^{2}$ by $-1$

(v) $5 a^{3} b$ by $3 a b^{2}$

(vi) $5 y^{2}$ by $3 y^{5}$

(vii) $-5 x^{4}$ by $18 x^{6}$

(viii) $\left(-5 m^{2} n p\right)$ by $\left(-4 m n^{2} p\right)$

Question 2

2. Find the product.

(i) $7(x+4)$

(ii) $5(2 x-4)$

(iii) $-4(-m-5)$

(iv) $5(3 p-2 q)$

Question 3

Simplify: $x(y-z)+y(z-x)+z(x-y)$