We have to calculate the area and perimeter of ABC.
Area:
We can calculate the area by substracting from the area of the big triangle ABD the area of the little triangle BCD. Both are right triangles.
The area of ABD is:
[tex]A_{\text{ABD}}=\frac{b\cdot h}{2}=\frac{(15+5)\cdot12}{2}=\frac{20\cdot12}{2}=\frac{240}{2}=120[/tex]The area of BCD is:
[tex]A_{\text{BCD}}=\frac{b\cdot h}{2}=\frac{5\cdot12}{2}=\frac{60}{2}=30[/tex]Then, the area of ABC is:
[tex]A_{\text{ABC}}=A_{\text{ABD}}-A_{\text{BCD}}=120-30=90[/tex]The area of ABC is 90 cm^2.
Perimeter:
We calculate the perimeter by adding the length of the three sides. We know only 2 of the sides, so we have to calculate the other one (BC).
The length of BC can be calculated using Pythagorean theorem for the triangle BCD, so we can write:
[tex]\begin{gathered} BC^2=CD^2+BD^2=5^2+12^2=25+144=169 \\ BC=\sqrt[]{169}=13 \end{gathered}[/tex]Now, we can calculate the perimeter as:
[tex]P_{\text{ABC}}=AB+BC+AC=25+13+15=53[/tex]The perimeter is 53 cm.
