Selina Class 7 ICSE Solutions Mathematics : Chapter 16- Pythagoras Theorem

Class 7: Maths Chapter 16 solutions. Complete Class 7 Maths Chapter 16 Notes.

Selina Class 7 ICSE Solutions Mathematics : Chapter 16- Pythagoras Theorem

Selina 7th Maths Chapter 16, Class 7 Maths Chapter 16 solutions

Exercise 16 page: 189

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

Solution:

It is given that

Triangle ABC is right-angled at vertex A

AB = 18 cm and AC = 24 cm

Using Pythagoras Theorem

BC² = AB² + AC²

Substituting the values

BC² = 18² + 24²

By further calculation

BC² = 324 + 576 = 900

BC = √900 = √ (30 × 30)

So we get

BC = 30 cm

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

Solution:

It is given that

Triangle XYZ is right-angled at vertex Z

XY = 13 cm and XZ = 12 cm

Using Pythagoras Theorem

XY² = XZ² + YZ²

Substituting the values

13² = 12² + YZ²

By further calculation

YZ² = 13² – 12²

YZ² = 169 – 144 = 25

YZ = √25 = √ (5 × 5)

So we get

YZ = 5 cm

3. Triangle PQR is right-angled at vertex R. Calculate the length of PR, if:
PQ = 34 cm and QR = 33.6 cm.

Solution:

It is given that

Triangle PQR is right-angled at vertex R

PQ = 34 cm and QR = 33.6 cm

Using Pythagoras Theorem

PQ² = PR² + QR²

Substituting the values

34² = PR² + 33.6²

By further calculation

1156 = PR² + 1128.96

PR² = 1156 – 1128.96

PR = √27.04

So we get

PR = 5.2 cm

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

Here we have 20², 16² and 12²

We can write it as

20² = 16² + 12²

By further calculation

400 = 256 + 144

So we get

400 = 400

Hence, the given triangle is right angled.

(ii) 6 m, 9 m and 13 m

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

Here 13² = 9² + 6²

By further calculation

169 = 81 + 36

So we get

169 ≠ 117

Hence, the given triangle is not right angled.

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

Solution:

It is given that

BAC = 90°, AC = 400 m and AB = 300 m

Using Pythagoras Theorem

BC² = AB² + AC²

Substituting the values

BC² = 300² + 400²

By further calculation

BC² = 90000 + 160000 = 250000

BC = √250000

So we get

BC = 500 m

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:

It is given that

ACP = ∠BDP = 90°

AC = 12 m

BD = 9 m

PA= PB = 15 m

(i) In the right angled triangle ACP

AP² = AC² + CP²

Substituting the values

15² = 12² + CP²

By further calculation

225 = 144 + CP²

CP² = 225 – 144 = 81

So we get

CP = √81 = 9 m

(ii) In the right angled triangle BPD

PB² = BD² + PD²

Substituting the values

15² = 9² + PD²

By further calculation

225 = 81 + PD²

PD² = 225 – 81 = 144

So we get

PD = √144 = 12 m

(iii) We know that

CP = 9 m

PD = 12 m

So we get

CD = CP + PD

Substituting the values

CD = 9 + 12 = 21 m

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) It is given that

PQ = 8 cm and QR = 6 cm
∠PQR = 90⁰

Using Pythagoras Theorem

PR² = PQ² + QR²

Substituting the values

PR² = 8² + 6²

By further calculation

PR² = 64 + 36 = 100

PR = √100

So we get

PR = 10 cm

(ii) It is given that

PR = 34 cm and QR = 30 cm

∠PQR = 90⁰

Using Pythagoras Theorem

PR² = PQ² + QR²

Substituting the values

34² = PQ² + 30²

By further calculation

1156 = PQ² + 900

PQ² = 1156 – 900 = 256

PQ = √256

So we get

PQ = 16 cm

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

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

Solution:

It is given that

AB = 9 cm

BC = 40 cm

AC = 41 cm

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

Using Pythagoras Theorem

AC² = BC² + AB²

Substituting the values

41² = 40² + 9²

By further calculation

1681 = 1600 + 81

So we get

1681 = 1681

Therefore, ABC is a right-angled triangle.

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

(i) AC

(ii) CD

Solution:

It is given that

angle ACB = 90° = angle ACD

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

(i) In the right angled triangle ABC

BC = 6 cm and AB = 10cm

Using Pythagoras Theorem

AB² = AC² + BC²

Substituting the values

10² = AC² + 6²

By further calculation

100 = AC² + 36

AC² = 100 – 36 = 64

AC = √64 = √ (8 × 8)

So we get

AC = 8 cm

(ii) In the right angled triangle ACD

AD = 17 cm and AC = 8cm

Using Pythagoras Theorem

AD² = AC² + CD²

Substituting the values

17² = 8² + CD²

By further calculation

289 = 64 + CD²

CD² = 289 – 64 = 225

CD = √225 = √ (15 × 15)

So we get

CD = 15 cm

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:

It is given that

angle ADB = 90°

AC = AB = 26 cm

BD = DC

Using Pythagoras Theorem

AC² = AD² + DC²

Substituting the values

26² = 24² + DC²

By further calculation

676 = 576 + DC²

DC² = 676 – 576 = 100

DC = √100

So we get

DC = 10 cm

Here the length of BC = BD + DC

Substituting the values

Length of BC = 10 + 10 = 20 cm

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:

It is given that

AD = 13 cm

BC = 12 cm

AB = 3 cm

angle ACD = angle ABC = 90°

(i) In a right angled triangle ABC

AB = 3 cm and BC = 12 cm

Using Pythagoras Theorem

AC² = AB² + BC²

Substituting the values

AC² = 3² + 12²

By further calculation

AC² = 9 + 144 = 153

So we get

AC = √153 cm

(ii) In a right angled triangle ACD

AD = 13 cm and AC = √153 cm

Using Pythagoras Theorem

DC² = AD² – AC²

Substituting the values

DC² = 13² + √153²

By further calculation

DC² = 169 – 153 = 16

So we get

DC = √16 = 4 cm

Hence, the length of DC is 4 cm.

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 upto how much height does the ladder reach?

Solution:

It is given that

Length of ladder = 6.5 m

Length of foot of the wall = 2.5m

Using Pythagoras Theorem

BC² = AB² + AC²

Substituting the values

6.5² = 2.5² + AC²

By further calculation

AC² = 42.25 – 6.25 = 36

So we get

AC = √36 = 6 m

Hence, the ladder reaches upto 6 m.

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:

It is given that

Direction of north AC = 5 m

Direction of east AB = 12 m

Using Pythagoras Theorem

BC² = AC² + AB²

Substituting the values

BC² = 5² + 12²

By further calculation

BC² = 25 + 144 = 169

BC = √169 = √ (13 × 13)

So we get

BC = 13 m

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

Solution:

It is given that

AB = 20 cm

AO = AB/2 = 20/2 = 10 cm

BC = OD = 24 cm

Using Pythagoras Theorem

AD² = AO² + OD²

Substituting the values

AD² = 10² + 24²

By further calculation

AD² = 100 + 576 = 676

AD = √676 = √ (26 × 26)

So we get

AD = 26 cm

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Selina Class 7 ICSE Solutions Mathematics : Chapter 16- Pythagoras Theorem

Download PDF: Selina Class 7 ICSE Solutions Mathematics : Chapter 16- Pythagoras Theorem PDF

Chapterwise Selina Publishers ICSE Solutions for Class 7 Mathematics :

• Chapter 1- Integers
• Chapter 2- Rational Numbers
• Chapter 3- Fraction (Including Problems)
• Chapter 4- Decimal Fractions (Decimals)
• Chapter 5- Exponents (Including Laws of Exponents)
• Chapter 6- Ratio and Proportion (Including Sharing in a Ratio)
• Chapter 7- Unitary Method (Including Time and Work)
• Chapter 8- Percent and Percentage
• Chapter 9- Profit, Loss and Discount
• Chapter 10- Simple Interest
• Chapter 11- Fundamental Concepts (Including Fundamental Operations)
• Chapter 12- Simple Linear Equations (Including Word Problems)
• Chapter 13- Set Concepts
• Chapter 14- Lines and Angles (Including Construction of Angles)
• Chapter 15- Triangles
• Chapter 16- Pythagoras Theorem
• Chapter 17- Symmetry (Including Reflection and Rotation)
• Chapter 18- Recognition of Solids (Representing 3-D in 2-D)
• Chapter 19- Congruency: Congruent Triangles
• Chapter 20- Mensuration (Perimeter and Area of Plane Figures)
• Chapter 21- Data Handling
• Chapter 22- Probability

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