Question 4
The sketch of the isosceles right triangle is given below
For an isosceles right triangle, the two legs are equal
So we will get the value x as follow
[tex]\begin{gathered} x^2+x^2=8^2 \\ 2x^2=8^2 \\ 2x^2=64 \\ x^2=32 \\ x=4\sqrt[]{2} \end{gathered}[/tex]
The perimeter of the triangle can be obtained as follow
The perimeter is simply the sum of all the sides of the triangle
[tex]\begin{gathered} \text{Perimeter}=x+x+8 \\ \text{Perimeter}=4\sqrt[]{2}+4\sqrt[]{2}+8=8\sqrt[]{2}+8 \\ \text{Perimeter}=8\sqrt[]{2}+8 \\ \text{Perimeter}=8(\sqrt[]{2}+1) \\ \text{Perimeter}=19.31\text{ units} \end{gathered}[/tex]
To get the area of the triangle
we will use the formula
[tex]\begin{gathered} \text{Area}=\frac{1}{2}\times base\times\text{height} \\ \text{Area}=\frac{1}{2}\times4\sqrt[]{2}\times4\sqrt[]{2} \\ \text{Area}=2\sqrt[]{2}\times4\sqrt[]{2} \\ \text{Area}=2\times4\times2 \\ \text{Area}=16 \end{gathered}[/tex]
The area of the triangle is 16 square units
