Pythagoras Theorem ICSE Class-7th Concise Selina Maths

Pythagoras Theorem ICSE Class-7th Concise Selina mathematics Solutions Chapter-16. We provide step by step Solutions of Exercise / lesson-16 Pythagoras Theorem for ICSE Class-7 Concise Selina Mathematics. Our Solutions contain all type Questions of Exe-16 to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7.

## Pythagoras Theorem ICSE Class-7th Concise Selina mathematics Solutions Chapter-16

### Exercise – 16

#### Question 1.

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

Answer

Given: ∆ ABC right angled at A and AB = 18 cm,

AC = 24 cm.

To find: Length of BC.

According to Pythagoras Theorem,

BC2 = AB2 + AC2

= 182 + 242 = 324 + 576

= 900

∴ BC =√900=√30×30

= 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.

Answer

Given: ∆XYZ right angled at Z and XY = 13 cm,

XZ = 12 cm.

To find: Length of YZ.

According to Pythagoras Theorem,

XY2 = XZ2 + YZ2

132 = 122 + YZ2

169= 144 +YZ2

169 − 144 = YZ2

25 = YZ2

∴ YZ = √25cm

=√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.

Answer

Given: ∆PQR right angled at R and PQ = 34 cm,

QR = 33.6 cm.

To find: Length of PR.

According to Pythagoras Theorem,

PR2 + QR2 = PQ2

PR2 + 33.62 = 342

PR2+ 1128.96=

1156

PR2 = 1156 − 1128.96

∴ PR = √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

Answer

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.

ex, If (20)2 = (16)2

(20)2 = (16)2 + (12)2

400 = 256 + 144

400 = 400

6 m, 9 m, and 13 m

The given triangle will be a right-angled triangle if the 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.

Answer

AC = 400 m

AB = 300 m

BC =?

According to the Pythagoras Theorem,

BC2 = AB2 + AC2

BC2 = (300)2 + (400)2

BC2 = 90000 + 160000

BC2 = 250000

BC = √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

Answer

Given :

AC = 12 m

BD = 9 m

PA = PB

= 15 m

(i) In right angle triangle ACP

(AP)2 = (AC)2 + (CP)2

152 = 122 + CP2

225 = 144 + CP2

225 – 144

= CP2

81 = CP

81 = CP

∴ CP = 9 m

(ii) In right angle triangle BPD

(PB)2 = (BD)2 + (PD)2

(15)2 = (9)2 + PD2

225 = 81 + PD2

225 – 81

= PD2

144 = PD2

144 = PD2

∴ PD = 12 m

(iii) CP = 9 m

PD = 12 m

so 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

Answer

(i)

Given:

PQ = 8 cm

QR = 6 cm

PR =?

∠PQR = 90°

According to the Pythagoras Theorem,

(PR)2 = (PQ)2 + (QR)2

PR2 = 82 + 62

PR2 = 64 + 36

PR2 = 100

∴ PR = 100

= 10 cm

(ii)

Given:

PR = 34 cm

QR = 30 cm

PQ =?

∠PQR = 90°

According to the Pythagoras Theorem,

(PR)2 = (PQ)2 + (QR)2

(34)2 = PQ2 + (30)2

1156 = PQ2 + 900

1156 − 900 = PQ2

256 = PQ2

∴ PQ = 256 = 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

Answer

CB = 40 cm

AC = 41 cm

AB = 9 cm

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

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

Answer

∆ ABD

∠ACB = ∠ACD = 90°

and AB = 10 cm,

BC = 6 cm,

AD = 17 cm

To find:

(i) Length of CD

(ii) Length of AC

Proof:

(i) 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 = CD2

225 = CD2

CD =15×15

= 15 cm

(ii) In right-angled triangle ABC

BC = 6 cm, AB = 110 cm

According to Pythagoras Theorem,

AB2 = AC2 + BC2

(10)2 = (AC)2 + (6)2

100 = (AC)2 + 36

AC2 = 100 − 36 = 64 cm

AC2 = 64 cm

∴ AC = 8×8

= 8 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.

Answer

Given:

∆ ABC

∠ADB = 90°,

AC = AB = 26 cm

AD = 24 cm

To find : Length of BC In right angled ∆ ADC

AB = 26 cm,

AD = 24 cm

According to the Pythagoras Theorem,

(AC)2 = (AD)2 + (DC)2

(26)2 = (24)2 + (DC)2

676 = 576 + (DC)2

⇒ (DC)2 = 100

⇒ DC =√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.

Answer

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) =√9+144

=√153 cm

(ii) In right angled ∆ ACD

AD = 13 cm, AC =153

According to Pythagoras Theorem,

DC2 = AB2 − AC2

DC2= 169 − 153

DC = √16

= 4 cm

#### Question 12.

A ladder, 6.5 m long, rests against a vertical wall. If the foot of the ladder is 2.5 m from the foot of the wall, find up to how much height does the ladder reach?

Answer

Given :

Length of the foot of the wall = 2.5 m

Length of ladder = 6.5 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 + AC2

AC2 = 42.25 – 6.25

= 36 m

AC = √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.

Answer

Given: Direction of north = 5 m

ex. AC Direction of east = 12 m

ex. AB

To find: BC

According to Pythagoras Theorem,

In right angled Δ ABC

(BC)2 = (AC)2 + (AB)2

(BC)2 = (5)2 + (12)2

(BC)2 = 25 + 144

(BC)2= 169

∴ BC = √169

=√13×13

= 13 m

#### Question 14.

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

Answer

Given:

AB = 20 cm

so AO = AB

=202 = 10cm

BC = OD

= 24 cm

Find the :

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 = √26×26

AD = 26 cm

— End of Pythagoras Theorem ICSE Class-7th Solutions :–

Return to – Concise Selina Maths Solutions for ICSE Class -7

Thanks

Share with your friends