Given:
Length of one side of square = 4
Length of one side of triangle = 7
Let's find the area of the triangle.
Let's find the diagonal which is the base of the triangle.
Apply Pythagorean Theorem to find the diagonal.
We have:
[tex]\begin{gathered} c=\sqrt{4^2+4^2} \\ \\ c=\sqrt{16+16} \\ \\ c=\sqrt{32} \\ \\ c=\sqrt{2*16} \\ \\ c=4\sqrt{2} \end{gathered}[/tex]Now, to find the area, apply the formula:
[tex]A=\frac{1}{2}*b*h[/tex]Where:
is the area
b is the base = 4√2
h = 7
Thus, we have:
[tex]\begin{gathered} A=\frac{1}{2}*4\sqrt{2}*7 \\ \\ A=14\sqrt{2} \\ \\ A=19.8 \end{gathered}[/tex]The area of the triangl is 19.8 square units.
Now, to find the primetrer, let's find the hypotenuse using the Pythagorean Theorem:
[tex]\begin{gathered} c=\sqrt{7^2+(4\sqrt{2})^2} \\ \\ c^=\sqrt{49+32} \\ \\ c=\sqrt{81} \\ \\ c=9 \end{gathered}[/tex]Now, to find the perimeter of the triangle, we have:
[tex]\begin{gathered} P=4\sqrt{2}+7+9 \\ \\ P=5.7+7+9 \\ \\ P=21.7 \end{gathered}[/tex]Therefore, the perimeter of triangle 21.7 units.
ANSWER:
• Area = 19.8 square units
,• Perimeter = 21.7 units