Question

Find the sides of a right angled triangle, if the hypotenuse is 4 units more than the base of the triangle and the third side is 8 units

Answer 1

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let's take,

• base,b=b
• hieght,h=8
• hypotenuse,hy=(b+4)

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according to Pythagorean theorem,

${hy}^{2} = {b}^{2} + {h}^{2}$

substituting the values,we get,

${(b + 4)}^{2} = {b}^{2} + {8}^{2}$

on expanding terms,we get,

${b}^{2} + 2 \times 4 \times b + {4}^{2} = {b}^{2} + {8}^{2} \\ {b}^{2} - {b}^{2} + 2 \times 4 \times b = {8}^{2}-{4}^{2} \\ 2 \times 4 \times b = {8}^{2} -{4}^{2} \\ 8 \times b = 64 -16\\ b = 6$

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therefore,

b=6 units

h=8 units

hypt=6+4=10 units

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