# Selina Concise Class 9 Maths Chapter 13 Pythagoras Theorem Exercise 13A Solutions

## Selina Concise Class 9 Maths Chapter 13 Pythagoras Theorem Exercise 13A Solutions EXERCISE – 13A

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

=>

Given : A ladder 13m long rest against a vertical wall.
∴ hypotenuse = AC = 13m
If the foot of the ladder is 5m from the foot of the wall.
∴ Base = BC = 5m

By Pythagoras Theorem,
(hypotenuse)2 = (Base)2 + (height)2
(AC)2 = (BC)2 + (AB)2
(13)2 = (5)2 + (AB)2
169 = 25 + (AB)2
169 – 25 = (AB)2
144 = (AB)2
Taking square root on both sides,
√144 = √(AB)2
12cm = AB
∴ AB = 12cm

∴ The distance of the other end of the ladder from the ground is 12m.

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

=>

Given, A man goes 40m due north. Base = 40m
and 50m due west
height = 50m

By Pythagoras Theorem,

(hypotenuse)2 = (Base)2 + (height)2

H2 = (40)2 + (50)2
H2 = 1600 + 2500
H2 = 4100

Taking square root on both sides,
√H2 = √4100
√H2 = √41 × 100
H = 10√41 m

∴ His distance from the starting point is 10√41 m.

(3) In the figure : ∠PSQ = 90°, PQ = 10cm, QS = 6cm and RQ = 9cm. Calculate the length of PQ.

=>

Given : ∠PSQ = 90°
PQ = 10cm
QS = 6cm
and RQ = 9cm.
In △ PSQ,

By Pythagoras Theorem,

(hypotenuse)2 = (Base)2 + (height)2
(PQ)2 = (QS)2 + (PS)2
(10)2 = (6)2 + (PS)2
100 = 36 + (PS)2
100 – 36 = (PS)2
64 = (PS)2

Taking square root on both sides,

√64 = √(PS)2
8cm = PS
PS = 8cm
In △ PSR,

By Pythagoras theorem,

(PR)2 = (RS)2 + (PS)2
(PR)2 = (15)2 + (8)2
= 225 + 64
(PR)2 = 289

Taking square root on both sides,

√(PR)2 = √289

PR = 17cm

(4) The given figure shows a quadrilateral ABCD in which AD = 13cm, DC = 12cm, BC = 3cm and ABC = BCD = 90°. Calculate the length of AB.

=>

Given : DC = 12cm
BC = 13cm
and ∠ABD = BCD = 90°
Now,
We have to find the length of AB.

In △ BCD

By Pythagoras Theorem,
(hypotenuse)2 = (Base)2 + (height)2
(BD)2 = (BC)2 + (CD)2
= (3)2 + (12)2
= 9 + 144
(BD)2 = 153
Taking Square root on both sides,
√(BD)2 = √153
BD = √153

In △ ABD,
by Pythagoras theorem,
(13)2 = (√153)2 + (AB)2
169 = 153 + (AB)2
169 – 153 = (AB)2
16 = (AB)2

Taking square root on both sides,

√16 = √(AB)2
4cm = AB

∴ AB = 4cm

∴ The length of AB is 4cm.

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

=>

Given: BC = 10cm

AD is draw n perpendicular to base BC of an equilateral triangle.

By Pythagoras theorem,

(hypotenuse)2 = (Base)2 + (height)2

Taking square root on both sides,

Now, we have to find the length of  AD, correct to 1 place of decimal.

∴ The length of AD is 8.7 cm. (6) In triangle ABC, given below, AB = 8cm, BC = 6cm and AC = 3cm. Calculate the length of OC.

=> Given, AB = 8cm
BC = 6cm
AC = 3cm
Now,
we have to find the length of OC.

In △ AOC,

By Pythagoras theorem,

(hypotenuse)2 = (Base)2 + (height)2
(AC)2 = (OC)2 + (AO)2
(3)2 = (AO)2 + x2
9 = (AO)2 + x2
9 – x2 = AO2 ……(i)
In △ AOB,
by Pythagoras theorem,
(AB)2 = (AO)2 + (BO)2
(8)2 = (AO)2 + (6 + x)2
64 = (AO)2 + (6 + x)2
64 – (6 + x)2 = (AO)2 …… (ii)
From equation (i) and (ii),
9 – x2 = 64 – (6 + x)2
9 – x2 = 64 – (36 + x2 + 12x)  (∵ a2 + 2ab + b2)
9 – x2 = 64 – 36 – x2 – 12x
9 = 28 – 12x
9 – 28 = – 12x
– 19 = -12x
19 = 12x
19/12 = x
x = 19/12
x = 1 7/12 cm
∴ The length of OC is 1 7/12 cm.

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

=> Given :
AB = AC = X
BC = 10cm and the area of triangle is 60cm2.
Now, we have to find the value of x.

We know that,

Area of triangle = 1/2 × Base × height
60 = 1/2 × x × x
60 × 2 = x2
120 = x2

Taking square root on both sides,
√120 = √x2
x = √120
Now, we have to draw isosceles triangle

Area of isosceles triangle = base/4 √4a2 – b2
60 = 10/4 √4x2 – (10)2
60 = 5/2 √4x2 – 100
60 × 2 = 5 √4x2 – 100
120 = 5 √4x2 – 100
120/5 = √4x2 – 100
24 = √4x2 – 100

Taking square on both sides,
(24)2 = (√4x2 – 100)2
576 = 4x2 – 100
576 + 100 = 4x2
676 = 4x2
676/4 = x2
169 = x2
Taking square root on both sides,
√169 = √x2
13 = x
∴ x = 13
∴ The value of x = 13

(8) If the sides of triangle are in the ratio 1 : 2 : 1, show that is a right angled triangle. => Given, If the sides of triangle are in the ratio 1 : 2 : 1.

By Pythagoras theorem,

(hypotenuse)2 = (Base)2 + (height)2

(√2)2 = (1)2 + ( )2

2 = 1 + 1

2 = 2

R.H.S. = L.H.S.

Right hand side = Left hand side.

∴ This triangle is a right angle triangle.

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

=> Given, Two poles of heights 6m and 11m stand 6m vertically on a plane ground.
If the distance between their feet is 12m.
Now, we have to find the distance between their tips.

By Pythagoras theorem,
(hypotenuse)2 = (Base)2 + (height)2
(AE)2 = (12)2 + (5)2
= 144 + 25
AE2 = 169

Taking square root on both sides,

√169 = √AE2

13m = AE

∴ The distance between their tips is 13m.

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

=> Given, AB parallel CD (AB || CD)
AB = 7cm
BD = 25cm and
CD = 17cm.

Now, we have to find the length of side BC.

by Pythagoras theorem,

(hypotenuse)2 = (Base)2 + (height)2

Taking square root on both sides,

In △ BMC,
by Pythagoras theorem,

(BC)2 = (BM)2 + (CM)2
(BC)2 = (24)2 + (10)2
= 576 + 100
(BC)2 = 676

Taking square root on both sides.

√BC2 = √676

BC = 26cm

Here is your solution of Selina Concise Class 9 Maths Chapter 13 Pythagoras Theorem Exercise 13A

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Updated: January 3, 2023 — 1:19 pm