22)

Find the area of the sector $OAB$ and subtract it from the area of the triangle $OAB$.
__________________________________________________ _______________________________________

23)

a) The arc $AB$ is equal to the circumference of the cone's base. With that information you can calculate the radius.

b) Once you know the slanted height of the cone and its base's radius, you can form a triangle to calculate the perpendicular height.

c) Use the volume formula for a cone.

d) The curved surface area is equal to the area of the sector $OAB$.
__________________________________________________ _______________________________________

24)

a) First calculate the area of the sector $OPQ$ in terms of $r$ and $\theta$. Then express $r$ in terms of $\theta$ by using the perimeter and the arc length formula.

b) Differentiate the area formula and identify its maximum value. The domain should be $0 < \theta < \pi$, but it might not matter either way.

How is the solution 4/3 (4pi -3 root 3) i got 4/3(2 Pi - 3 root 3)

How is the solution 4/3 (4pi -3 root 3) i got 4/3(2 Pi - 3 root 3)

Originally Posted by fan96
Sorry q22

Originally Posted by fan96
22)

Find the area of the sector $OAB$ and subtract it from the area of the triangle $OAB$.
__________________________________________________ _______________________________________

23)

a) The arc $AB$ is equal to the circumference of the cone's base. With that information you can calculate the radius.

b) Once you know the slanted height of the cone and its base's radius, you can form a triangle to calculate the perpendicular height.

c) Use the volume formula for a cone.

d) The curved surface area is equal to the area of the sector $OAB$.
__________________________________________________ _______________________________________

24)

a) First calculate the area of the sector $OPQ$ in terms of $r$ and $\theta$. Then express $r$ in terms of $\theta$ by using the perimeter and the arc length formula.

b) Differentiate the area formula and identify its maximum value. The domain should be $0 < \theta < \pi$, but it might not matter either way.
Q22

OA = OB = 4cm

I just realised I made a mistake with that diagram. The radius should be 4cm, not 2cm.

$\angle OAB = 2 \cos^{-1}(0.5) = \frac{2\pi}{3}$

Area of sector $OAB$

$= \frac{1}{2} (4)^2 \left(\frac{2\pi}{3}\right)$

$= \frac{16\pi}{3}$

Area of triangle $OAB$

$= \frac{1}{2} (4) (4) \sin \frac{2\pi}{3}$

$= 4\sqrt 3$

$= \frac{16\pi}{3} - 4\sqrt 3$

$= \frac{4}{3}\left(4\pi - 3\sqrt 3\right)$

There are currently 1 users browsing this thread. (0 members and 1 guests)

Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•