Look at the picture.
We have in a base an equilateral triangle.
The formula of the area of an equilateral triangle:
[tex]A_B=\dfrac{a^2\sqrt3}{4}[/tex]
We have a = 12. Substitute:
[tex]A_B=\dfrac{12^2\sqrt3}{4}=\dfrac{144\sqrt3}{4}=36\sqrt3[/tex]
The lateral side is a isosceles triangle. The formula of the area of a triangle is:
[tex]A_\triangle=\dfrac{1}{2}bh[/tex]
We must calculate the length of h using the Pythagorean theorem:
[tex]h^2+6^2=10^2[/tex]
[tex]h^2+36=100[/tex] subtract 36 from both sides
[tex]h^2=64\to h=\sqrt{64}\to h=8[/tex]
We have b = 13 and h = 8. Substitute:
[tex]A_\triangle=\dfrac{1}{2}(12)(8)=(6)(8)=48[/tex]
The Total Area:
[tex]T.A.=A_B+3A_\triangle[/tex]
Substitute:
[tex]T.A.=36\sqrt3+3(48)=\boxed{144+36\sqrt3}[/tex]
