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.

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)

S.chand class 6 Mathematics Chapter 13 Exercise 13A

 Exercise 13A

Question 1

Write the following phrases using symbols.

S.No	Using Words	Using Symbols
(i)	8 more than p
(ii)	The sum of p and q
(iii)	13 Less than x
(iv)	10 times a
(v)	three-fourths of a
(vi)	p divides by q
(vii)	t is greater than 12
(viii)	30° less than current temperature t°C

Question 2

Write the following phrases using symbols.

S.No	Using Words	Using Symbols
(i)	18 Less than 6 times a
(ii)	9 times the sum of a and 7
(iii)	Thrice the number a increased by b
(iv)	15 times the number p plus four
(v)	25 less than 4 times q
(vi)	6 times a less than the quotient of b by 4
(vii)	The square of the sum of a and b

Multiple Choice Question (MCQ)

Tick (✔) the correct option.

3. The perimeter of an isosceles triangle with each of the equal sides as $x \mathrm{~cm}$ and third side as $y \mathrm{~cm}$ is :

(a) $x+2 y$

(b) $2 x+y$

(c) $2 x-y$

(d) $x+y$

RS Aggarwal solution class 8 chapter 13 Time and Work Exercise 13A

Exercise 13A

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Question 1:

Rajan can do a piece of work in 24 days while Amit can do it in 30 days. In how many days can they complete it, if they work together?

Answer 1:

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Question 2:

Ravi can do a piece of work in 15 hours while Raman can do it in 12 hours. How long will both take to do it, working together?

Answer 2:

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Question 3:

A and B, working together can finish a piece of work in 6 days, while A alone can do it in 9 days. How much time will B alone take to finish it?

Answer 3:

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Question 4:

Two motor mechanics, Raju and Siraj, working together can overhaul a scooter in 6 hours. Raju alone can do the job in 15 hours. In how many hours can Siraj alone do it?

Answer 4:

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Question 5:

A, B and C can do a piece of work in 10 days, 12 days and 15 days respectively. How long will they take to finish it if they work together?

Answer 5:

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Question 6:

A can do a piece of work in 24 hours while B alone can do it in 16 hours. If A, B and C working together can finish it in 8 hours, in how many hours can C alone finish the work?

Answer 6:

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Question 7:

A, B and C working together can finish a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone do the same work?

Answer 7:

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Question 8:

A and B can finish a piece of work in 16 days and 12 days respectively. A started the work and worked at it for 2 days. He was then joined by B. Find the total time taken to finish the work.

Answer 8:

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Question 9:

A can do a piece of work in 14 days while B can do it in 21 days. They began together and worked at it for 6 days. Then, A fell ill and B had to complete the remaining work alone. In how many days was the work completed?

Answer 9:

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Question 10:

A can do $\frac{2}{3}$ of a certain work in 16 days and B can do $\frac{1}{4}$ of the same work in 3 days. In how many days can both finish the work, working together?

Answer 10:

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Question 11:

A, B and C can do a piece of work in 15, 12 and 20 days respectively. They started the work together, but C left after 2 days. In how many days will the remaining work be completed by A and B?

Answer 11:

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Question 12:

A and B can do a piece of work in 18 days; B and C can do it in 24 days while C and A can finish it in 36 days. In how many days can A, B, C finish it, if they all work together?

Answer 12:

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Question 13:

A and B can do a piece of work in 12 days, B and C in 15 days, and C and A in 20 days. How much time will A alone take to finish the job?

Answer 13:

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Question 14:

Pipes A and B can fill an empty tank in 10 hours and 15 hours respectively. If both are opened together in the empty tank, how much time will they take to fill it completely?

Answer 14:

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Question 15:

Pipe A can fill an empty tank in 5 hours while pipe B can empty the full tank in 6 hours. If both are opened at the same time in the empty tank, how much time will they take to fill it up completely?

Answer 15:

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Question 16:

Three taps A, B and C can fill an overhead tank in 6 hours, 8 hours and 12 hours respectively. How long would the three taps take to fill the empty tank, if all of them are opened together?

Answer 16:

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Question 17:

A cistern has two inlets A and B which can fill it in 12 minutes and 15 minutes respectively. An outlet C can empty the full cistern in 10 minutes. If all the three pipes are opened together in the empty tank, how much time will they take to fill the tank completely?

Answer 17:

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Question 18:

A pipe can fill a cistern in 9 hours. Due to a leak in its bottom, the cistern fills up in 10 hours. If the cistern is full, in how much time will it be emptied by the leak?

Answer 18:

The leak will empty the filled cistern in 90 hours.

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Question 19:

Pipe A can fill a cistern in 6 hours and pipe B can fill it in 8 hours. Both the pipes are opened and after two hours, pipe A is closed. How much time will B take to fill the remaining part of the tank?

Answer 19: