Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem

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,
BC² = AB² + AC²
= 18² + 24² = 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,
XY² = XZ² + YZ²
13² = 12² + YZ²


169= 144 +YZ²
169- 144 = YZ²
25 = YZ²
∴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,
PR² + QR² = PQ²
PR² + 33.6² = 34²
PR²+ 1128.96= 1156
PR² = 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,
BC² = AB² + AC²
BC² = (300)² + (400)²
BC² = 90000 + 160000
BC² = 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
15² = 12² + CP²
225 = 144 + CP²
225 – 144 = CP²
81 =CP
\(\sqrt { 81 }\) =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
\(\sqrt { 144 }\)=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
PR² = 8² + 6²
PR² = 64 + 36
PR² = 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 = PQ² + (30)²
1156 = PQ² + 900
1156-900 = PQ²
256 = PQ²
∴ 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

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,
AB² = AC² + BC²
(10)² = (AC)² + (6)²
100 = (AC)² + 36
AC² = 100-36 = 64 cm
AC² = 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 = CD²
225 = CD²
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,
DC² = AB²-AC²
DC²= 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 + AC²
AC² = 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