PYTHAGOREAN pronunciation • How to pronounce PYTHAGOREAN
PYTHAGOREAN pronunciation • How to pronounce PYTHAGOREAN

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.

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.

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.

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

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.

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

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

(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

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

∆ ABD
∠ACB = ∠ACD = 90°
and AB = 10 cm,

BC = 6 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,
(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.

Given:
∆ ABC

AC = AB = 26 cm

To find : Length of BC In right angled ∆ ADC
AB = 26 cm,

According to the Pythagoras Theorem,
(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.

Given :
∆ ACD = ∠ABC = 90°

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?

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.

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.

Given:
AB = 20 cm
so AO = AB

=202 = 10cm
BC = OD

= 24 cm

Find the :

In right angled triangle
AOD (AD)2 = (AO)2 + (OD)2

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

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