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

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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 + x

∴ 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,

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