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.

4. The lateral edge of a square prism is 20 cm long and it is inclined at an angle of 60 degrees with the base. If the volume of the prism is 1,200 cm3, determine the base edge.

Answer 1

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.

Each lateral face is a trapezoid with base lengths 5 and 8.5 m. The 6 of them will have a total area of 160 m², so one of them has an area of $\frac{160}{6}$ = 26.66 m². From the equation of the area of a trapezoid, we know

$\frac{1}{2}(b_1 + b_2)h = area$

$h = \frac{2 \times area}{b_1+b_2} = \frac{2 \times 26.66}{5 + 8.5} = \frac{53.33}{13.5} = 3.95 \;m$

This is the slant height of the face of the frustum.

Answer: 3.95 m

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

Given a regular square pyramid

Altitude (h) $=12 m$

Volume (v) $= 196 m^3$

Slant Height $=?$

Volume of regular square pyramid $=\frac13 a^2h$

$\implies \frac13a^2h=196\\ \implies a^2=\frac{196 \times 3}{h}=\frac{196 \times 3}{12}=49\\ \implies a^2=49 \implies a=\sqrt{49}=7m$

Slant Height

$S^2=r^2+h^2;\ where\ r=\frac a2=\frac 72=3.5m\\ \implies S^2=(3.5)^2+(12)^2\\ \implies S^2=12.25+144 =156.25m\\ \implies S=\sqrt{156.25} \implies S=12.5m\\$

Therefore slant height of pyramid is $12.5m$

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.

Given a regular square pyramid

$Height=8m\\ base\ angle=45^0\\ Lateral\ area\ of\ the\ pyramid=?$

We know

$tan\ \theta=\frac{2h}{a}\\ \implies tan\ 45^0=\frac{2 \times 8}{a} \implies a=16\\$

Lateral area of pyramid

$=a \sqrt{4h^2+a^2}\\ =16 \sqrt{4 \times (8)^2+(16)^2}\\ =16 \sqrt{4 \times 64+256}\\ =16 \sqrt{256+256}\\ =16 \times 16\sqrt{2}\\ =362.0386m^2$

Therefore, the lateral area of the pyramid is $362m^2$

4. The lateral edge of a square prism is 20 cm long and it is inclined at an angle of 60 degrees with the base. If the volume of the prism is 1,200 cm3, determine the base edge.

Given a square prism

Lateral edge $= 20cm$

Base angle $= 60^0$

Value of prism $= 1200cm^3$

Base edge $=?$

Height = Lateral edge x sin $\theta$

$=20 \times sin\ 60^0=20 \times \frac{\sqrt3}{2}\\ =10\sqrt3=17.32m$

We know that

$Volume\ of\ the\ prism=13^2h\\ \implies 1200=13^2 \times 17.32\\ \implies 13^2=\frac{1200}{17.32}=69.284\\ \implies 13=\sqrt{69.284}=8.32cm$

Therefore, base height $=8.12cm$