## Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem

**Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem**

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### Pythagoras Theorem Exercise 16 – Selina Concise Mathematics Class 7 ICSE Solutions

**Question 1. **

**Triangle ABC is right-angled at vertex A. Calculate the length of BC, if AB = 18 cm and AC = 24 cm.**

**Solution: **

**Given**: ∆ABC right angled at A and AB = 18 cm, AC = 24 cm.

** To find :** Length of BC.

According to Pythagoras Theorem,

BC^{2} = AB^{2} + AC^{2 }= 18^{2} + 24^{2 }= 324 + 576 = 900

∴BC = \(\sqrt { 900 }\) = \(\sqrt { 30 x 30 }\)= 30 cm

**Question 2. **

**Triangle XYZ is right-angled at vertex Z. Calculate the length of YZ, if XY = 13 cm and XZ = 12 cm.**

**Solution: **

**Given**: ∆XYZ right angled at Z and XY = 13 cm, XZ = 12 cm.

**To find :**Length of YZ.

According to Pythagoras Theorem,

XY

^{2}= XZ

^{2}+ YZ

^{2 }13

^{2}= 12

^{2}+ YZ

^{2}

^{ }169= 144 +YZ

^{2 }169- 144 = YZ

^{2 }25 = YZ

^{2 }∴YZ = \(\sqrt { 25 }\)cm \(\sqrt { 5×5 }\) = 5 cm

**Question 3. **

**Triangle PQR is right-angled at vertex R. Calculate the length of PR, if:**

**PQ = 34 cm and QR = 33.6 cm.**

**Solution: **

**Given**: ∆PQR right angled at R and PQ = 34 cm, QR = 33.6 cm.

**To find :** Length of PR.

According to Pythagoras Theorem,

PR^{2} + QR^{2} = PQ^{2 }PR^{2} + 33.6^{2} = 34^{2 }PR^{2}+ 1128.96= 1156

PR^{2} = 1156- 1128.96

∴ PR = \(\sqrt { 27.04 }\) = 5.2 cm

**Question 4. **

**The sides of a certain triangle are given below. Find, which of them is right-triangle**

**(i) 16 cm, 20 cm and 12 cm**

**(ii) 6 m, 9 m and 13 m**

**Solution: **

**(i)**16 cm, 20 cm and 12 cm

The given triangle will be a right-angled triangle if square of its largest side is equal to the sum of the squares on the other two sides.

i.e., If (20)

^{2}= (16)

^{2}= (12)

^{2 }(20)

^{2}= (16)

^{2}+ (12)

^{2 }400 = 256 + 144

400 = 400

So, the given triangle is right angled.

**(ii)**6 m, 9 m and 13 m

The given triangle will be a right-angled triangle if square of its largest side is equal to the sum of the squares on the other two sides.

i.e., If (13)

^{2}= (9)

^{2}+ (6)

^{2 }169 = 81+36 169 ≠ 117

So, the given triangle is not right angled.

**Question 5. **

**In the given figure, angle BAC = 90°, AC = 400 m and AB = 300 m. Find the length of BC.**

**Solution:**

AC = 400 m

AB = 300 m

BC = ?

According to Pythagoras Theorem,

BC

^{2}= AB

^{2}+ AC

^{2 }BC

^{2}= (300)

^{2}+ (400)

^{2 }BC

^{2}= 90000 + 160000

BC

^{2}= 250000

BC = \(\sqrt { 250000 }\)= 500 m

**Question 6. **

**In the given figure, angle ACP = ∠BDP = 90°, AC = 12 m, BD = 9 m and PA= PB = 15 m. Find:**

**(i) CP**

** (ii) PD**

** (iii) CD**

**Solution: **

**Given :**AC = 12 m

BD = 9 m

PA = PB= 15 m

**(i)**In right angle triangle ACP

(AP)

^{2}= (AC)

^{2}+ (CP)

^{2 }15

^{2}= 12

^{2}+ CP

^{2 }225 = 144 + CP

^{2 }225 – 144 = CP

^{2 }81 =CP

\(\sqrt { 81 }\) =CP

∴ CP = 9 m

**(ii)**In right angle triangle BPD

(PB)

^{2}= (BD)

^{2}+ (PD)

^{2 }(15)

^{2}= (9)

^{2}+ PD

^{2 }225 = 81 + PD

^{2 }225-81 = PD

^{2 }144 = PD

^{2 }\(\sqrt { 144 }\)=PD ‘

∴ PD = 12 m

**(iii)**CP = 9 m

PD = 12 m

∴ CD = CP + PD

= 9+ 12 = 21 m

**Question 7. **

**In triangle PQR, angle Q = 90°, find :**

**(i) PR, if PQ = 8 cm and QR = 6 cm**

**(ii) PQ, if PR = 34 cm and QR = 30 cm**

**Solution: **

**(i) Given:**

PQ = 8 cm

QR = 6 cm

PR = ?

∠PQR = 90°

According to Pythagoras Theorem,

(PR)

^{2}= (PQ)

^{2}+ (QR)

^{2 }PR

^{2}= 8

^{2}+ 6

^{2 }PR

^{2}= 64 + 36

PR

^{2}= 100

∴ PR

^{=}\(\sqrt { 100 }\)= 10 cm

**(ii)**Given :

PR = 34 cm

QR = 30 cm

PQ = ?

∠PQR = 90°

According to Pythagoras Theorem,

(PR)

^{2}= (PQ)

^{2}+ (QR)

^{2 }(34)

^{2}= PQ

^{2}+ (30)

^{2 }1156 = PQ

^{2}+ 900

1156-900 = PQ

^{2 }256 = PQ

^{2 }∴ PQ = 16 cm

**Question 8. **

**Show that the triangle ABC is a right-angled triangle; if:**

AB = 9 cm, BC = 40 cm and AC = 41 cm

AB = 9 cm, BC = 40 cm and AC = 41 cm

**Solution: **AB = 9 cm

CB = 40 cm

AC = 41 cm

The given triangle will be a right angled triangle if square of its largest side is equal to the sum of the squares on the other two sides.

According to Pythagoras Theorem,

(AC)

^{2}= (BC)

^{2}+ (AB)

^{2 }(41)

^{2}= (40)

^{2}+ (9)

^{2 }1681 = 1600 + 81

1681 = 1681

Hence, it is a right-angled triangle ABC.

**Question 9. **

**In the given figure, angle ACB = 90° = angle ACD. If AB = 10 m, BC = 6 cm and AD = 17 cm, find :**

**(i) AC**

**(ii) CD**

**Solution: **

**Given:**

∆ABD

∠ACB = ∠ACD = 90°

and AB = 10 cm, BC = 6 cm and AD = 17 cm

**To find:**

(i) Length of AC

(ii) Length of CD

**Proof:**

(i) In right-angled triangle ABC

BC = 6 cm, AB = 110 cm

According to Pythagoras Theorem,

AB

^{2}= AC

^{2}+ BC

^{2 }(10)

^{2}= (AC)

^{2}+ (6)

^{2 }100 = (AC)

^{2}+ 36

AC

^{2}= 100-36 = 64 cm

AC

^{2}= 64 cm

∴ AC = \(\sqrt { 8×8 }\) = 8 cm

(ii) In right-angle triangle ACD

AD = 17 cm, AC = 8 cm

According to Pythagoras Theorem,

(AD)

^{2}= (AC)

^{2}+ (CD)

^{2}

(17)

^{2}= (8)

^{2}+ (CD)

^{2 }289 – 64 = CD

^{2 }225 = CD

^{2}

CD =\(\sqrt { 15×15 }\) = 15 cm

**Question 10. **

**In the given figure, angle ADB = 90°, AC = AB = 26 cm and BD = DC. If the length of AD = 24 cm; find the length of BC.**

**Solution: **

**Given**:

∆ABC

∠ADB = 90° and AC = AB = 26 cm

AD = 24 cm

**To find :**Length of BC In right angled ∆ADC

AB = 26 cm, AD = 24 cm

According to Pythagoras Theorem,

(AC)

^{2}= (AD)

^{2}+ (DC)

^{2 }(26)

^{2}= (24)

^{2}+ (DC)

^{2 }676 = 576 + (DC)

^{2 }⇒ (DC)

^{2}= 100

⇒ DC = \(\sqrt { 100 }\) = 10 cm

∴ Length of BC = BD + DC

= 10 + 10 = 20 cm

**Question 11. **

**In the given figure, AD = 13 cm, BC = 12 cm, AB = 3 cm and angle ACD = angle ABC = 90°. Find the length of DC.**

**Solution: **

**Given :**

∆ACD = ∠ABC = 90°

and AD = 13 cm, BC = 12 cm, AB = 3 cm

**To find :**Length of DC.

**(i)**In right angled ∆ABC

AB = 3 cm, BC = 12 cm

According to Pythagoras Theorem,

(AC)^{2} = (AB)^{2} + (BC)^{2 }(AC)^{2} = (3)^{2} + (12)^{2 (AC) = \(\sqrt { 9+144 }\) = \(\sqrt { 153 }\) cm}

** (ii)** In right angled triangle ACD

AD = 13 cm, AC =\(\sqrt { 153 }\)

According to Pythagoras Theorem,

DC^{2} = AB^{2}-AC^{2 }DC^{2}= 169-153

DC = \(\sqrt { 16 }\) = 4 cm

∴ Length of DC is 4 cm

**Question 12. **

**A ladder, 6.5 m long, rests against a vertical wall. Ifthe foot of the ladcler is 2.5 m from the foot of the wall, find upto how much height does the ladder reach?**

**Solution: **

**Given :**

Length of ladder = 6.5 m

Length of foot of the wall = 2.5 m

**To find :**Height AC According to Pythagoras Theorem,

(BC)

^{2}= (AB)

^{2}+ (AC)

^{2}

^{ }(6.5)

^{2}= (2.5)

^{2}+ (AC)

^{2 }42.25 = 6.25 + AC

^{2 }AC

^{2}= 42.25 – 6.25 = 36 m

AC = \(\sqrt { 6×6 }\) = 6 m

∴ Height of wall = 6 m

**Question 13. **

**A boy first goes 5 m due north and then 12 m due east. Find the distance between the initial and the final position of the boy.**

**Solution: **

**Given :**Direction of north = 5 m i.e. AC Direction of east = 12 m i.e. AB

**To find:** BC

According to Pythagoras Theorem,

In right angled AABC

(BC)^{2} = (AC)^{2} + (AB)^{2 }(BC)^{2} = (5)^{2} + (12)^{2 }(BC)^{2} = 25 + 144

(BC)^{2} = 25 + 144

(BC)^{2}= 169

∴ BC = \(\sqrt { 169 }\) = \(\sqrt { 13×13 }\) = 13 m

**Question 14. **

**Use the information given in the figure to find the length AD.**

**Solution: **

**Given :**

AB = 20 cm

∴AO =\(\frac { AB }{ 2 }\) = \(\frac { 20 }{ 2 }\) =10cm

BC = OD = 24 cm

**To find :**Length of AD

In right angled triangle

AOD (AD)

^{2}= (AO)

^{2}+ (OD)

^{2 }(AD)

^{2}= (10)

^{2}+ (24)

^{2 }(AD)

^{2}= 100 + 576

(AD)

^{2}= 676

∴ AD = \(\sqrt { 26×26 }\)

AD = 26 cm