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Perimeter and Area of Plane Figures — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsPerimeter and Area of Plane Figures4 Marks
Question
The difference between the sides at right angle in a right-angled triangle is $7\ cm$. The area of the triangle is $60cm^2$. Find its perimeter.
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Answer

Given,
Area of the triangle $= 60cm^2$
Let the side of the triangle be a, b and c, where a is the height, b is the base and c is hypotense of the triangle.
$a - b = 7cm$
$a = 7 + b ...(1)$
Area of triangle $=\frac{1}{2}\times\text{b}\times\text{h}$
$\Rightarrow6=\frac12\times\text{b}\times(7+\text{b})$
$\Rightarrow120=7\text{b}+\text{b}^2$
$\Rightarrow\text{b}^2+7\text{b}-12=0$
$\Rightarrow(\text{b}+15)(\text{b}-8)=0$
$\Rightarrow\text{b}=-15$ or $8$
Side of a triangle cannot be negative.
Threfore, $b = 8cm.$
Substituting the value of $b = 8cm,$ in equatin $(1):$
$a = 7 + 8 = 15cm$
Now, $a = 15cm, b = 8cm$
Now, in the given right triangle, we have to find third side.
$ (\text { Hyp })^2=(\text { First side })^2+(\text { Second side })^2 $
$ \Rightarrow \text { Hyp }^2=8^2+15^2 $
$ \Rightarrow \text { Hyp }^2=64+225 $
$ \Rightarrow \text { Hyp }^2=289 $
$ \Rightarrow \text { Hyp }=17 \mathrm{~cm}$
So, the third side is $17cm.$
Perimeter of a triangle $= a + b + c.$
$\therefore$ required perimeter of the triangle $= 15 + 8 + 17 = 40cm.$

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