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.