Chapter 10.6, Problem 10E

To determine

To find the surface area of pyramid.

Answer to Problem 10E

  $133$ in²

Explanation of Solution

Given Information: Pyramid Base b = 9 in., Radius $r=\frac{Baseb}{2}=\frac{9}{2}=4.5$ in. and Area of Triangle B = 35 in².

Formula Used:

Surface area of triangular pyramid = $\frac{1}{2}(a\times b)+\frac{3}{2}(b\times s)$ , Area of Triangle $B=\frac{1}{2}(b\times s)$ , Slant Height $s=\sqrt{a^2+r^2}$

Calculation:

Surface area of triangular pyramid = $\frac{1}{2}(a\times b)+\frac{3}{2}(b\times s)$

Now find Slant Height s.

Area of Triangle $B=\frac{1}{2}(b\times s)$

  $35i{n}^2=\frac{1}{2}(9\times s)$

Hence,

Slant Height s = $\frac{70}{9}$ in.

Now find Height a

Slant Height $s=\sqrt{a^2+r^2}$

s = $\frac{70}{9}$ in. and r = $\frac{9}{2}$ in.

  $\frac{70}{9}=\sqrt{a^2+{\left(\frac{9}{2}\right)}^2}$

So,

  ${\left(\frac{70}{9}\right)}^2-{\left(\frac{9}{2}\right)}^2=a^2$

So,

  $60.49-20.25=a^2$

So,

  $a=\sqrt{40.24}$

Hence,

Height $a=6.34$ in.

Surface area of triangular pyramid = $\frac{1}{2}(a\times b)+\frac{3}{2}(b\times s)$

Base b = 9 in., Slant Height s = $\frac{70}{9}=7.77$ in. and Height a = 6.34 in.

Now put the value of a, b and s to find Surface area of triangular pyramid

Surface area of triangular pyramid = $\frac{1}{2}(6.34\times 9)+\frac{3}{2}(9\times 7.77)$ in²

So,

Surface area of triangular pyramid = $\frac{1}{2}\times (57.06)+\frac{3}{2}\times (69.93)$ in²

So,

Surface area of triangular pyramid = $28.53+104.89$ in²

Hence,

Surface area of triangular pyramid = $133.42$ in²

Nearest Surface area of triangular pyramid = $133$ in²