# Area of segment.

• May 8th 2010, 05:30 AM
Kane535
Area of segment.
A circle has the center of A and the radius of 5. |BC| = 8; |AC|=|AB|=5
Find the area of the gray bit of the circle.

http://i42.tinypic.com/9h6r85.jpg
• May 8th 2010, 05:58 AM
sa-ri-ga-ma
Quote:

Originally Posted by Kane535
A circle has the center of A and the radius of 5. |BC| = 8; |AC|=|AB|=5
Find the area of the gray bit of the circle.

http://i42.tinypic.com/9h6r85.jpg

Hi Kane535, welcome to MHF.
Altitude from A on BC is 3.
So area of ABC = 1/2*8*3 = 12 sq. units
Angle ABC is given by
θ = 2*arctan(4/3)
Area of the sector ABC = 1/2*r^2*θ, where θ is in radians.
Now find the area of the shaded portion.
• May 8th 2010, 06:05 AM
e^(i*pi)
Quote:

Originally Posted by Kane535
A circle has the center of A and the radius of 5. |BC| = 8; |AC|=|AB|=5
Find the area of the gray bit of the circle.

http://i42.tinypic.com/9h6r85.jpg

Use the cos rule to find angle A (I will use radians because it's easier)

$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{5^2+5^2-8^2}{2 \cdot 5 \cdot 5} = -\frac{7}{25}$

$A = \arccos \left(-\frac{7}{25}\right) \approx 1.855$

The area of a sector is given by $A = \frac{1}{2}r^2 \theta$. In this case $\theta$ is angle A and $r=5$

$A_s = \frac{1}{2} \cdot 5^2 \cdot \arccos \left(-\frac{7}{25}\right)$

Since this includes the white triangle we must remove it's area which is given by $A_t = \frac{1}{2} bc \sin A = \frac{1}{2} \cdot 5^2 = \frac{25}{2}$

$A_g = A_s - A_t = \frac{1}{2} \cdot 5^2 \cdot \arccos \left(-\frac{7}{25}\right) - \frac{25}{2}$

Evaluate for the answer, I get an answer of $10.7 \text{ 3sf }$