**Chapter 13 Pythagoras Theorem Set 13.1**

Question 1.

In the figures below, find the value of â€˜xâ€™.

Solution:

i. In âˆ†LMN, âˆ M = 90Â°.

Hence, side LN is the hypotenuse.

According to Pythagorasâ€™ theorem,

l(LN)Â² = l(LM)Â² + l(MN)Â²

âˆ´ xÂ² = 72 + 24Â²

âˆ´ xÂ² = 49 + 576

âˆ´ xÂ² = 625

âˆ´ xÂ² = 25Â²

âˆ´ x = 25 units

ii. In âˆ†PQR, âˆ Q = 90Â°.

Hence, side PR is the hypotenuse.

According to Pythagorasâ€™ theorem,

l(PR)Â² = l(PQ)Â² + l(QR)Â²

âˆ´ 412 = 92 + xÂ²

âˆ´ 1681 = 81 + xÂ²

âˆ´ 1681 â€“ 81 = xÂ²

âˆ´ 1600 = xÂ²

âˆ´ xÂ² = 1600

âˆ´ xÂ² = 40Â²

âˆ´ x = 40 units

iii. In AEDF, âˆ D = 90Â°.

Hence, side EF is the hypotenuse.

According to Pythagorasâ€™ theorem,

l(EF)Â² = l(ED)Â² + l(DF)Â²

âˆ´ 17Â² = xÂ² + 8Â²

âˆ´ 289 = xÂ² + 64

âˆ´ 289 â€“ 64 = xÂ²

âˆ´ 225 = xÂ²

âˆ´ xÂ² = 225

âˆ´ xÂ² = 15Â²

âˆ´ x = 15 units

Question 2.

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

Solution:

In âˆ†PQR, âˆ P = 90Â°.

Hence, side QR is the hypotenuse.

According to Pythagorasâ€™ theorem,

l(QR)Â² = l(PR)Â² + l(PQ)Â²

âˆ´ l(QR)Â² = 10Â² + 24Â²

âˆ´ l(QR)Â² = 100 + 576

âˆ´ l(QR)Â² =676

âˆ´ l(QR)Â² = 26Â²

âˆ´ l(QR) = 26 cm

âˆ´ The length of seg QR is 26 cm.

Question 3.

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

Solution:

In âˆ†LMN, âˆ M = 90Â°.

Hence, side LN is the hypotenuse.

According to Pythagorasâ€™ theorem,

l(LN)Â² = l(LM)Â² + l(MN)Â²

âˆ´ 20Â² = 12Â² + l(MN)Â²

âˆ´ l(MN)Â² = 20Â² â€“ 12Â²

âˆ´ l(MN)Â² = 400 â€“ 144

âˆ´ l(MN)Â² = 256

âˆ´ l(MN)Â² = 16Â²

âˆ´ l(MN)= 16 cm

âˆ´ 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 âˆ†ABC, âˆ B = 90Â°

According to Pythagorasâ€™ theorem,

l(AC)Â² = l(AB)Â² + l(BC)Â²

âˆ´ 15Â² = l(BC)Â² + 9Â²

âˆ´ 225 = l(BC)Â² + 81

âˆ´ 225 â€“ 81 = l(BC)Â²

âˆ´ 144 = l(BC)Â²

âˆ´ 12Â² = l(BC)Â²

âˆ´ l(BC) = 12

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

**Intext Questions and Activities**

Question 1.

Write the name of the hypotenuse of each of the rightangled triangles shown below.

i.

The hypotenuse of âˆ†ABC is__

ii.

The hypotenuse of âˆ†LMN is__

iii.

The hypotenuse of âˆ†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 Pythagorasâ€™ theorem. (Textbook pg. no. 87)

Solution:

i. From the figure, by measurement,

l(AB) = 4 cm

Now, in right-angled triangle ABC,

l(AB)Â² + l(BC)Â² = (4)Â² + (3)Â²

= 16 + 9

âˆ´ l(AB)Â² + l(BC)Â² = 25 â€¦. (i)

l(AC)Â² = (5)Â² = 25 â€¦.(ii)

âˆ´ From (i) and (ii),

l(AC)Â² = l(AB)Â² + l(BC)Â²

âˆ´ Pythagorasâ€™ theorem is verified.

(Students should draw the triangles PQR and XYZ and verify the Pythagoras â€™ 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 âˆ†LBM,

mâˆ BLM + mâˆ BML + mâˆ LBM = 180Â° â€¦. (Sum of the measures of the angles of a triangles is 180Â° )

âˆ´ mâˆ BLM + mâˆ BML + 90Â° = 180Â°

âˆ´ mâˆ BLM + mâˆ BML = 90Â° â€¦. (i)

Now, âˆ†LBM and âˆ†LAP are same.

âˆ´ mâˆ BML = mâˆ ALP â€¦. (ii)

âˆ´ mâˆ BLM + mâˆ ALP = 90Â° â€¦. IFrom (i) and (ii)l

Now, mâˆ ALP + mâˆ PLM + mâˆ BLM = 180Â° â€¦. (The measure of a straight angle is 180Â°)

âˆ´ mâˆ ALP + mâˆ BLM + mâˆ PLM = 180Â°

âˆ´ 90Â° + mâˆ PLM = 180Â°

âˆ´ mâˆ PLM = 180Â°- 90Â° = 90Â°

âˆ´ mâˆ 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 Pythagorasâ€™ theorem. (Textbook pg. no. 89)

Solution:

Area of square ABLM = l(AB)Â² = 32 = 9 sq.cm

Area of square BCPN = l(BC)Â²= 42 = 16 sq.cm

Area of square ACQR = l(AC)Â² = 52 = 25 sq.cm

Now, 25 = 16 + 9

i.e. 5Â² = 4Â² + 3Â²

âˆ´ l(AC)Â² = l(BC)Â² + l(AB)Â²

âˆ´ (hypotenuse)Â² = (base)Â² + (height)Â²