Question

Heya !!




In a right angled triangle, one side is 42m and the difference of hypotenuse and the other side is 14 m. Find the lengths of both unknown sides. Calculate it's area by the formula for finding the area of a right triangle. Verify the result by Heron's formula.

Answer 1

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$\huge\boxed{Answer}$

step - by - step explanation

s1 = 126m

hypotenuse - s2 = 42cm = 0.42m

Let the s2 be x.

hypotenuse - x = 0.42

hypotenuse = 0.42 + x

By Pythagoras theorem.

${hypotenuse}^{2} = {s1}^{2} + {s2}^{2}$

${(0.42 + x)}^{2} = 126 + {x}^{2}$

$\begin{lgathered}17.64 + 0.84x + {x}^{2} = \\ 126 + {x}^{2}\end{lgathered}$

17.64 + 0.84x = 12617.64+0.84x=126

0.84x = 126 - 17.640.84x=126−17.64

0.84x = 108.360.84x=108.36

x = $\frac{108.36}{0.84}$x=

x = 129x=129

so

s2 = 129m

hypotenuse = x + 0.42

hypotenuse = 129 + 0.42

hypotenuse = 129.42m

Area of right angled triangle

=1/2 × product of perpendicular sides

=1/2 × 126 × 129

= 63 × 129

= 8127 metre sq.

Therefore,

The area of right angled triangle is 8127sq.m.

Answer 2

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Hi!

here's your answer!

s1 = 126m

hypotenuse - s2 = 42cm = 0.42m

Let the s2 be x.

hypotenuse - x = 0.42

hypotenuse = 0.42 + x

By Pythagoras theorem.



17.64 + 0.84x = 126

0.84x = 126 - 17.64

0.84x = 108.36

x = \frac{108.36}{0.84}

x = 129

so

s2 = 129m

hypotenuse = x + 0.42

hypotenuse = 129 + 0.42

hypotenuse = 129.42m

Area of right angled triangle

=1/2 × product of perpendicular sides

=1/2 × 126 × 129

= 63 × 129

= 8127 metre sq.

Therefore,

The area of right angled triangle is 8127sq.m.

2.6