what is the surface area of this triangular pyramid? square inches

Explanation:

Step1: Calculate base - area

The base is an equilateral triangle with side length $a = 9$ inches. The area formula for an equilateral triangle is $A_{base}=\frac{\sqrt{3}}{4}a^{2}$. So, $A_{base}=\frac{\sqrt{3}}{4}\times9^{2}=\frac{81\sqrt{3}}{4}\approx\frac{81\times1.732}{4}=\frac{140.292}{4}=35.073$ square inches.

Step2: Calculate area of one lateral - face

The lateral - faces are isosceles triangles. The base of each lateral - face is $b = 9$ inches and the height is $h = 8$ inches. The area formula for a triangle is $A=\frac{1}{2}bh$. So, the area of one lateral - face $A_{lateral}=\frac{1}{2}\times9\times8 = 36$ square inches.

Step3: Calculate total surface area

A triangular pyramid has 1 base and 3 lateral - faces. The total surface area $A = A_{base}+3A_{lateral}$. Substitute the values: $A=35.073 + 3\times36=35.073+108 = 143.073$ square inches.

Answer:

$143.073$