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in Pythagoras Theorem by (46.9k points)
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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

∴ x​​​​​​​2= 72 + 24​​​​​​​2

.∴ x​​​​​​​2= 49 + 576 

∴ x​​​​​​2​= 625

∴ x​​​​​​2= 25​​​​​​2

∴ 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 + x​​​​​​2​

∴ 1681 = 81 + x​​​​​​2​

∴ 1681 – 81 = x​​​​​​2​

∴ 1600 = x​​​​​​2​

∴ x​​​​​​2​= 1600 

∴ x​​​​​​2​= 40​​​​​​2​

∴ 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​

∴ 17​​​​​​2​= x​​​​​​2​+ 8​​​​​​2​

∴ 289 = x2+ 64 

∴ 289 – 64 = x2

∴ 225 = x2

∴ x= 225 

∴ x2= 152

∴ x = 15 units

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