**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}**

H^{2} = (40)^{2} + (50)^{2}

H^{2} = 1600 + 2500

H^{2} = 4100

Taking square root on both sides,

√H^{2} = √4100

√H^{2} = √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 :

AD = 13cm

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,

(AD)^{2} = (BD)^{2} + (AB)^{2}

(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.

In △ ADB,

**By Pythagoras theorem,**

**(hypotenuse) ^{2 }= (Base)^{2} + (height)^{2}**

(AB)^{2} = (BD)^{2 }+ (AD)^{2}

(10)^{2} = (5)^{2} + (AD)^{2}

100 = 25 + (AD)^{2}

100 – 25 = (AD)^{2}

75 = (AD)^{2}

Taking square root on both sides,

√75 = √(AD)^{2}

√75 = AD

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

∴ AD = √75

AD = 8.66

AD = 8.7 cm

∴ 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} + x^{2}

9 = (AO)^{2} + x^{2}

9 – x^{2} = AO^{2} ……(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 – x^{2} = 64 – (6 + x)^{2}

9 – x^{2} = 64 – (36 + x^{2} + 12x) (∵ a^{2} + 2ab + b^{2})

9 – x^{2} = 64 – 36 – x^{2} – 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 60cm ^{2}. Find x.**

=>

Given :

AB = AC = X

BC = 10cm and the area of triangle is 60cm^{2}.

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 = x^{2}

120 = x^{2}

Taking square root on both sides,

√120 = √x^{2}

x = √120

Now, we have to draw isosceles triangle

Area of isosceles triangle =

base/4 √4a^{2} – b^{2}

60 = 10/4 √4x^{2} – (10)^{2}

60 = 5/2 √4x^{2} – 100

60 × 2 = 5 √4x^{2} – 100

120 = 5 √4x^{2} – 100

120/5 = √4x^{2} – 100

24 = √4x^{2} – 100

Taking square on both sides,

(24)^{2} = (√4x^{2} – 100)^{2}

576 = 4x^{2} – 100

576 + 100 = 4x^{2}

676 = 4x^{2}

676/4 = x^{2}

169 = x^{2}

Taking square root on both sides,

√169 = √x^{2}

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.

In △ ADE,

By Pythagoras theorem,

(hypotenuse)^{2} = (Base)^{2} + (height)^{2}

(AE)^{2} = (AD)^{2} + (DE)^{2}

(AE)^{2} = (12)^{2} + (5)^{2}

= 144 + 25

AE^{2} = 169

Taking square root on both sides,

√169 = √AE^{2}

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.

**In △ BAD,**

**by Pythagoras theorem,**

**(hypotenuse) ^{2} = (Base)^{2} + (height)^{2}**

(BD)

^{2}= (AB)

^{2}+ (AD)

^{2}

(25)

^{2}= (7)

^{2}+ (AD)

^{2}

625 = 49 + (AD)

^{2}

625 – 49 = (AD)

^{2 }576 = AD

^{2}

Taking square root on both sides,

√576 = √AD^{2}

24 = AD

**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.

√BC^{2 }= √676

BC = 26cm

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

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