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

Explanation:

Step1: Calculate base - area

The base is an equilateral triangle with side length $a = 14$ mm. The area of an equilateral triangle $A_{base}=\frac{\sqrt{3}}{4}a^{2}=\frac{\sqrt{3}}{4}\times14^{2}=\frac{\sqrt{3}}{4}\times196 = 49\sqrt{3}\approx49\times1.732 = 84.868$ $mm^{2}$.

Step2: Calculate area of one lateral - face

The lateral - faces are isosceles triangles with base $b = 14$ mm and height $h = 18$ mm. The area of a triangle is $A_{lateral}=\frac{1}{2}bh=\frac{1}{2}\times14\times18=126$ $mm^{2}$.

Step3: Calculate total surface area

A triangular pyramid has 1 base and 3 lateral - faces. So $A = A_{base}+3A_{lateral}=84.868 + 3\times126=84.868+378 = 462.868\approx462.87$ $mm^{2}$.

Answer:

$462.87$