(c) The perimeter of an isosceles right-angled triangle is 6(V2+1) cm. Find its area.
Answers
Answer:
Let the length of the equal sides of the isoceles right triangle be x
\textsf{Then, its hypotenuse is}Then, its hypotenuse is
\mathsf{Hypotenuse^2=x^2+x^2=2x^2}Hypotenuse
2
=x
2
+x
2
=2x
2
\implies\mathsf{Hypotenuse=\sqrt{2}x}⟹Hypotenuse=
2
x
\textsf{\underline{Given:}}
Given:
\textsf{Perimeter of the triangle =$6(\sqrt{2}+1)$ cm}Perimeter of the triangle =6(
2
+1) cm
\implies\mathsf{\sqrt{2}x+2x=6(\sqrt{2}+1)}⟹
2
x+2x=6(
2
+1)
\implies\mathsf{\sqrt{2}x(1+\sqrt{2})=6(\sqrt{2}+1)}⟹
2
x(1+
2
)=6(
2
+1)
\implies\mathsf{\sqrt{2}x=6}⟹
2
x=6
\implies\mathsf{x=3\sqrt{2}}⟹x=3
2
\textsf{Then, the are of the triangle }Then, the are of the triangle
\textsf{$=\frac{1}{2}$*base*height}=
2
1
*base*height
\mathsf{=\frac{1}{2}*3\sqrt{2}*3\sqrt{2}}=
2
1
∗3
2
∗3
2
\textsf{=9 square cm}=9 square cm
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