Answer:
Area of an isosceles right triangle is 24.5
Step-by-step explanation:
Given that triangle is an isosceles right angle which means that two side of the triangle are equal and one of the angle is [tex]90\degree[/tex].
Consider triangle [tex]\bigtriangleup ABC[/tex] (Refer to attachment),
[tex]\angle ABC=90^{\circ},AB=BC, AC=7\sqrt{2}[/tex].
Since both sides are equal angle opposite to both sides are same.
Now applying pythagoras theorem to [tex]\bigtriangleup ABC[/tex],
[tex] \left ( AB \right )^{2}+\left ( BC \right )^{2}=\left ( AC \right )^{2}[/tex]
Assume [tex]AB=BC=x[/tex],
Substituting the value,
[tex] \left ( x \right )^{2}+\left ( x \right )^{2}=\left ( 7\sqrt{2} \right )^{2}[/tex]
Simplifying,
[tex] 2\left ( x \right )^{2}=\left (7^2\left(\sqrt{2}\right)^2 \right )[/tex]
[tex] 2\left(x\right)^{2}=\left (49\left(2\right)\right)[/tex]
[tex] 2\left(x\right)^{2}=98[/tex]
Dividing by 2,
[tex] \left(x\right)^{2}=49[/tex]
Taking square root on both sides,
[tex]\sqrt{\left(x\right)^{2}}=\sqrt{49}[/tex]
[tex]x=7[/tex]
[tex]\therefore AB=BC=7[/tex]
Formula for area of triangle is given as,
[tex]Area=\dfrac{1}{2}\times base\times height[/tex]
Now AB=height,BC=base.
Substituting the value,
[tex]Area=\dfrac{1}{2}\times AB\times BC[/tex]
[tex]Area=\dfrac{1}{2}\times 7\times 7[/tex]
[tex]Area=\dfrac{1}{2}\times 49[/tex]
[tex]Area=\dfrac{49}{2}[/tex]
[tex]Area=24.5[/tex]
So, area of an isosceles right triangle is 24.5
