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

Maharashtra BoardEnglish MediumSTD 10MathsPerimeter and Area of Plane Figures5 Marks
Question
The difference between the sides at right angle in a right-angled triangle is $7\ cm$. The area of the triangle is $60\ cm^2$. Find its perimeter.
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Answer

Given:
Area of triangle $= 24\ cm^2$
Let the sides be a and b, where a is the height and b is the base of triangle.
$a - b = 2cm a - b = 2cm$
$a= 2 + b ...(1)$
Area of triangle $=\frac{1}{2}\times\text{b}\times\text{h}$
$\Rightarrow24=\frac12\times\text{b}\times(2+\text{b})$
$\Rightarrow48=\text{b}+\frac12\text{b}^2$
$\Rightarrow48=2\text{b}+\text{b}^2$
$\Rightarrow\text{b}^2+2\text{b}-48=0$
$\Rightarrow(\text{b}+8)(\text{b}-6)=0$
$\Rightarrow\text{b}=-8$ or $6$
Side of a triangle cannot be negative.
Threfore, $b = 6\ cm.$
Substituting the value of $b = 6\ cm$, in equatin (1) we get:
$a = 2 + 6 = 8cm$
Now, $a = 8cm, b = 6cm$
Now, in the given right triangle, we have to find third side.
Using the relation
$(Hyp)^2 = (Oneside)^2 + (Otherside)^2$
$\Rightarrow Hyp^2 = 8^2 + 6^2$
$\Rightarrow Hyp^2 = 64 + 36$
$\Rightarrow Hyp^2 = 100$
$\Rightarrow Hyp = 10cm$
So, the third side is 10cm.
So, Perimeter of a triangle $= a + b + c.$
$\therefore$ required perimeter of the triangle $= 8 + 6 + 10 = 24\ cm$.

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