## Maharashtra State Board Class 7 Maths Solutions Chapter 13 Pythagoras’ Theorem Practice Set 48

Question 1.

In the figures below, find the value of ‘x’.

Solution:

i. In ∆LMN, ∠M = 90°.

Hence, side LN is the hypotenuse.

According to Pythagoras’ theorem,

l(LN)² = l(LM)² + l(MN)²

∴ x² = 72 + 24²

∴ x² = 49 + 576

∴ x² = 625

∴ x² = 25²

∴ x = 25 units

ii. In ∆PQR, ∠Q = 90°.

Hence, side PR is the hypotenuse.

According to Pythagoras’ theorem,

l(PR)² = l(PQ)² + l(QR)²

∴ 412 = 92 + x²

∴ 1681 = 81 + x²

∴ 1681 – 81 = x²

∴ 1600 = x²

∴ x² = 1600

∴ x² = 40²

∴ x = 40 units

iii. In AEDF, ∠D = 90°.

Hence, side EF is the hypotenuse.

According to Pythagoras’ theorem,

l(EF)² = l(ED)² + l(DF)²

∴ 17² = x² + 8²

∴ 289 = x² + 64

∴ 289 – 64 = x²

∴ 225 = x²

∴ x² = 225

∴ x² = 15²

∴ x = 15 units

Question 2.

In the right-angled ∆PQR, ∠P = 90°. If l(PQ) = 24 cm and l(PR) = 10 cm, find the length of seg QR.

Solution:

In ∆PQR, ∠P = 90°.

Hence, side QR is the hypotenuse.

According to Pythagoras’ theorem,

l(QR)² = l(PR)² + l(PQ)²

∴ l(QR)² = 10² + 24²

∴ l(QR)² = 100 + 576

∴ l(QR)² =676

∴ l(QR)² = 26²

∴ l(QR) = 26 cm

∴ The length of seg QR is 26 cm.

Question 3.

In the right-angled ∆LMN, ∠M = 90°. If l(LM) = 12 cm and l(LN) = 20 cm, find the length of seg MN.

Solution:

In ∆LMN, ∠M = 90°.

Hence, side LN is the hypotenuse.

According to Pythagoras’ theorem,

l(LN)² = l(LM)² + l(MN)²

∴ 20² = 12² + l(MN)²

∴ l(MN)² = 20² – 12²

∴ l(MN)² = 400 – 144

∴ l(MN)² = 256

∴ l(MN)² = 16²

∴ l(MN)= 16 cm

∴ The length of seg MN is 16 cm.

Question 4.

The top of a ladder of length 15 m reaches a window 9 m above the ground. What is the distance between the base of the wall and that of the ladder?

Solution:

The wall and the ground are perpendicular to each other. Hence, the ladder leaning against the wall forms a right-angled triangle.

In ∆ABC, ∠B = 90°

According to Pythagoras’ theorem,

l(AC)² = l(AB)² + l(BC)²

∴ 15² = l(BC)² + 9²

∴ 225 = l(BC)² + 81

∴ 225 – 81 = l(BC)²

∴ 144 = l(BC)²

∴ 12² = l(BC)²

∴ l(BC) = 12

∴ The distance between the base of the wall and that of the ladder is 12 m.

**Maharashtra Board Class 7 Maths Chapter 13 Pythagoras’ Theorem Practice Set 48 Intext Questions and Activities**

Question 1.

Write the name of the hypotenuse of each of the right angled triangles shown below.

i.

The hypotenuse of ∆ABC is__

ii.

The hypotenuse of ∆LMN is__

iii.

The hypotenuse of ∆XYZ is__

Solution:

i. AC

ii. MN

iii. XZ

Question 2.

Draw right-angled triangles with the lengths of hypotenuse and one side as shown in the rough figures below. Measure the third side. Verify the Pythagoras’ theorem. (Textbook pg. no. 87)

Solution:

i. From the figure, by measurement,

l(AB) = 4 cm

Now, in right-angled triangle ABC,

l(AB)² + l(BC)² = (4)² + (3)²

= 16 + 9

∴ l(AB)² + l(BC)² = 25 …. (i)

l(AC)² = (5)² = 25 ….(ii)

∴ From (i) and (ii),

l(AC)² = l(AB)² + l(BC)²

∴ Pythagoras’ theorem is verified.

(Students should draw the triangles PQR and XYZ and verify the Pythagoras ’ theorem)

Question 3.

Without using a protractor, can you verify that every angle of the vacant quadrilateral in the adjacent figure is a right angle? (Textbook pg. no. 89)

Solution:

In the square ABCD the shaded triangles are right-angled and are the same.

In ∆LBM,

m∠BLM + m∠BML + m∠LBM = 180° …. (Sum of the measures of the angles of a triangles is 180° )

∴ m∠BLM + m∠BML + 90° = 180°

∴ m∠BLM + m∠BML = 90° …. (i)

Now, ∆LBM and ∆LAP are same.

∴ m∠BML = m∠ALP …. (ii)

∴ m∠BLM + m∠ALP = 90° …. IFrom (i) and (ii)l

Now, m∠ALP + m∠PLM + m∠BLM = 180° …. (The measure of a straight angle is 180°)

∴ m∠ALP + m∠BLM + m∠PLM = 180°

∴ 90° + m∠PLM = 180°

∴ m∠PLM = 180°- 90° = 90°

∴ m∠PLM is a right angle.

Similarly, we can prove that the other angles of the vacant quadrilateral are right angles.

Question 4.

On a sheet of card paper, draw a right-angled triangle of sides 3 cm, 4 cm and 5 cm. Construct a square on each of the sides. Find the area of each of the squares and verify Pythagoras’ theorem. (Textbook pg. no. 89)

Solution:

Area of square ABLM = l(AB)² = 32 = 9 sq.cm

Area of square BCPN = l(BC)²= 42 = 16 sq.cm

Area of square ACQR = l(AC)² = 52 = 25 sq.cm

Now, 25 = 16 + 9

i.e. 5² = 4² + 3²

∴ l(AC)² = l(BC)² + l(AB)²

∴ (hypotenuse)² = (base)² + (height)²