Question

In Figure 3, the measure of angle $QPR$ is $90$ degrees, and the area of triangle $PQR$ is $6$. If triangle $PQR$ is rotated ${360}^o$ about side $PR$, calculate the total surface area of the resulting solid.

• A. $24.00$
• B. $50.27$
• C. $75.40$
• D. $97.39$

Answer 1

Answer: (C)

$75.40$

Area of triangle is $6=\frac{1}{2}\times PQ\times 4$

$\therefore PQ=3$

From pythagoras theorem, we get

$QR=5$
Surface area of the resulting solid after rotating is equal to

$\pi \times {PQ}^2+\pi \times PQ\times QR$

$=9\pi +15\pi$

$=24\pi$

$=75.4$ (the region is cone, so we are effectively finding surface area of cone with radius $PQ$ and height $PR$ )