Question

The difference between the sides at right angles in a right angled triangle us 14cm. The area if the triangle 120cm^2.Calculate the perimeter of the Triangle.​

Answer 1

Answer:-

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Perimeter of the triangle = $\bf{60 \: cm}$

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• Given:-

Difference between the sides at right angles in a right angled triangle = $\bf{14 \: cm}$

Area of the triangle = $\bf{120 \: cm^2}$

• Solution:-

Let the sides containing the right angle be $\bf{"x" cm}$ and $\bf{(x-14)cm}$.

Then it's Area will be::

$\boxed {{\frac{1}{2}}.x.(x-14)cm^2}$

But, Area = $\bf{120cm^2}$

$\therefore {\frac{1}{2}}x(x-14) = 120$

$\implies{x^2-14x - 240 = 0}$

$\implies{x^2-24x+10x-240=0}$

$\implies{x(x-24)+10(x-24)=0}$

$\implies{(x-24)(x+10)=0}$

$\implies{\bf{x=24}}$ $\: \: \: \: \: [Neglecting \: \bf{x = -10}]$

Therefore,

one side = $\bf{24 \: cm}$

other side = $(24-14)cm$ =$\bf{10 \: cm}$

$\\ Hypotenuse = \sqrt{(24)^{2}+(10)^{2}}cm$

$\longrightarrow{\sqrt{576 + 100 cm}}$

$\longrightarrow{\sqrt{576}cm}$

$\implies \bf{26 \: cm}$

$\therefore$Perimeter of the triangle =

$(24+10+26)cm$

$\implies \bf{60 \: cm}$

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Answer 2

Answer:

$the \: answer \: is \: 60cm$