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Answer to Q136

Find a, b and x

From Pythagorean theorem,
(9+16)² = a² + b² (i)
b² = 16² + x² (ii)
a² = 9² + x² (iii)

from (i)
625 = a² + b² (iv)

put (ii) into (iv)
625 = a² + 16² + x²
625 = a² +256 + x² (v)

put (iii) into (v)
625 = 9² + x² + 256 + x²
625 = 81 + x² + 256 + x²
625 = 2x² + 337
2x² + 337 - 625 = 0
2x² - 288 = 0
2x² = 288
x² = 288 ÷ 2
x² = 144
x = √144
x = 12

put x into (v)
625 = a² +256 + x²
625 = a² + 256 + 144
625 = a² + 400
a² = 625 - 400
a² = 225
a = √225
a = 15

From (ii)
b² = 16² + x²
b² = 256 + 144
b² = 400
b = √400
b = 20

So, x = 12, a = 15, b = 20



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Answer to Q136

Find a, b and x

From Pythagorean theorem,
(9+16)² = a² + b² (i)
b² = 16² + x² (ii)
a² = 9² + x² (iii)

from (i)
625 = a² + b² (iv)

put (ii) into (iv)
625 = a² + 16² + x²
625 = a² +256 + x² (v)

put (iii) into (v)
625 = 9² + x² + 256 + x²
625 = 81 + x² + 256 + x²
625 = 2x² + 337
2x² + 337 - 625 = 0
2x² - 288 = 0
2x² = 288
x² = 288 ÷ 2
x² = 144
x = √144
x = 12

put x into (v)
625 = a² +256 + x²
625 = a² + 256 + 144
625 = a² + 400
a² = 625 - 400
a² = 225
a = √225
a = 15

From (ii)
b² = 16² + x²
b² = 256 + 144
b² = 400
b = √400
b = 20

So, x = 12, a = 15, b = 20

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