Question

The perimeter of an isosceles right angled triangle is 2p.Find out the area of the same triangle

A. [ 3 - 2√2]p²
B. [ 2 - √2] p²
C. [3 - √2]p²
D. [4 - 2√2]p²

Show necessary calculations.

Answer 1

using basic property of iso. ∆ we can solve easily...

see attachment....

correct option is (A)

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hope it will help u

Answer 2

Answer:

The area of the triangle having the perimeter 2p = $\bold{[3-\sqrt{2}] p^{2}}$

Step-by-step explanation:

The diagram shows a Right-Angled Isosceles Triangle XYZ, where, XY = YZ and Angle XYZ = 90°.

Let, XY = YZ = a  

According to Pythagoras Theorem,

$\begin{array}{l}{X Y^{2}+Y Z^{2}=X Z^{2}} \\ {=>a^{2}+a^{2}=X Z^{2}} \\ {=>X Z^{2}=2 a^{2}} \\ {=>X Z=\sqrt{2 a^{2}}=\sqrt{2} a}\end{array}$

∴ Perimeter = a + a + √2a = 2a + √2a = a (2 + √2)  

Given: Perimeter = 2p

$\begin{array}{l}{\Rightarrow \mathrm{a}(2+\sqrt{2})=2 \mathrm{p}} \\ {\Rightarrow \mathrm{a}=\frac{2 p}{2+\sqrt{2}}=\frac{2 p(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}}\end{array}$ [Rationalizing The denominator]

$\begin{aligned} &=\frac{2 p(2-\sqrt{2})}{(2)^{2}-(\sqrt{2})^{2}}=\frac{2 p(2-\sqrt{2})}{4-2}=\frac{2 p(2-\sqrt{2})}{2} \\ &=p(2-\sqrt{2}) \end{aligned}$

∴ Area of the triangle XYZ

$\begin{array}{l}{=\frac{1}{2} \times X Y \times Y Z} \\ {=\frac{1}{2} \times a \times a} \\ {=\frac{1}{2} \times p(2-\sqrt{2}) \times p(2-\sqrt{2})} \\ {=\frac{1}{2} p^{2}(2-\sqrt{2})^{2}} \\ {=\frac{p^{2}}{2}(4-4 \sqrt{2}+2)=\frac{p^{2}}{2}(6-4 \sqrt{2})} \\ {=\frac{p^{2}}{2} \times 2(3-2 \sqrt{2})} \\ {=p^{2}(3-2 \sqrt{2})}\end{array}$

Hence, the correct option is (c) $\bold{[3-\sqrt{2}] p^{2}}$