Pythagoras Theorem Class 7 Practice Set 48 Answers Solutions Chapter 13
Question 1
Maharashtra Board Solution
In the figures below, find the value of ‘x’.
Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
In the right-angled ∆PQR, ∠P = 90°. If l(PQ) = 24 cm and l(PR) = 10 cm, find the length of seg QR.
Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
In the right-angled ∆LMN, ∠M = 90°. If l(LM) = 12 cm and l(LN) = 20 cm, find the length of seg MN.
Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
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 & Step-by-Step Answer:
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
Maharashtra Board Solution
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 & Step-by-Step Answer:
i. AC ii. MN iii. XZ
Question 2
Maharashtra Board Solution
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 & Step-by-Step Answer:
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
Maharashtra Board Solution
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 & Step-by-Step Answer:
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
Maharashtra Board Solution
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 & Step-by-Step Answer:
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