Question

1. A frustum of a regular hexagonal pyramid has an upper base edge of 5m and a lower base edge of 8.5m. Its lateral area is 160m2. Determine the slant height of the frustum.

2. A regular square pyramid has an altitude of 12 m and volume of 196 m3. Determine the slant height of the pyramid.

3. A regular square pyramid has a height of 8m. If the slant height makes an angle of 45º with the base, find the lateral
area of the pyramid

Answer 1

Its lateral are is equal to 160m² and consists 6 trapezoid which has an upper base edge of 5m and a lower base edge of 8.5m. we can see in the picture above the slant height of the frustum is the altitude of the equilateral trapezoidal height at the side surface:

The are of this trapezoid is $\frac{160}{6}$ =$\frac{80}{3}$ . The formula to find are of trapazoid is $S=\frac{x+y}{2}*h$ .Here x=5m, y=8.5 m. So:

h=$\frac {2*S}{x+y}$ =$\frac{2*\frac{80}{3}}{5+8.5}=$ 3.95 (m)

2.

This picture explain our exercise. We have V=196 m³, h = 12 m, b-? .

Firstly, to find b, we find a. We use formula of finding volume: $V=\frac{1}{3}$ * a² * h.

a=$\sqrt{\frac{3*V}{h}}$ = $\sqrt{\frac{3*196}{12}}$ = 7 (m).

Then we have only to find b. We can use pythagor theorem by this picture:

b =$\sqrt{x^2+y^2}$ = $\sqrt{12^2+7^2}=13$ m.

3.

This picture explain our exercise. We have h =8m.

The angle between base and slant height is 45^o. So h=$\frac{a}{2}$ ---> a=2*h=16 m, and b = $h*\sqrt2$ =$8*\sqrt2$

Lateral area consists 4 equilateral triangles like this:

We will compute one of these triangles and we will multiply by 4.

The formula to find the area of triangle which is above: $S=\frac{a \times h}{2}$ ;

Here is a, h=b;

$S_1=\frac{a\times b}{2}=\frac{16\times 8\times \sqrt2}{2}=64\sqrt2;$

S$_{overal}=4\times 64\times\sqrt2=256\sqrt2;$