1. Area - composite shapes

For this question (see image) we are supposed to find the area of the shaded (blue) area of this shape. I know that I can work out the entire area of this shape using the area formula for a kite. How can I work out the area of the white part, though?

2. Hello, starship!

If we don't know exactly what kind of curve that is,
. . there is no way to calculate the area of the blue region.

However, I do have a suspicion . . .

Is that a right circular cone (a three-dimensional solid)?
. . And we are asked to find its lateral area?

3. I am assuming that it is just a two-dimensional shape as all of the other shapes I had to do were two-dimensional.

I checked the answer sheet, and the area of the blue part is 163cm². The only problem is finding a way to end up with that answer.

4. Originally Posted by starship
I am assuming that it is just a two-dimensional shape as all of the other shapes I had to do were two-dimensional.

I checked the answer sheet, and the area of the blue part is 163cm². The only problem is finding a way to end up with that answer.
There is no way without knowing what the curve at the base of the shape is,
and even then it will be a nasty calculation if it is what I think it is.

RonL

5. Originally Posted by starship
For this question (see image) we are supposed to find the area of the shaded (blue) area of this shape. I know that I can work out the entire area of this shape using the area formula for a kite. How can I work out the area of the white part, though?

Let us say the arc is circular. So the unshaded area is that of a sector of a cirlce whose central angle is 50 degrees, and whose circle's radius is 17.75 cm.
The area of the sector is the sum of the areas of the isosceles triangle and the secant of the circle. So, the area of the secant is area of sector minus area of isosceles triangle.

A1 = (1/2)(17.75)[(50(pi/180))(17.75)]
A1 = 137.47 sq.cm

Area of isosceles triangle, A2 = "(1/2)(ab)sinC" = (1/2)(17.75)(17.75)sin(50)
A2 = 120.66 sq.cm.

So, Area of secant = A1 -A2 = 16.81 sq.cm.

--------------
The isosceles triangle containing the shaded area.

Height of apex based from the chord of the sector is (40 minus the height of the apex of the unshaded isosceles triangle based on the same chord).
H = 40 -17.75cos(25degrees) = 23.913 cm.

Area of isosceles triangle containing the shaded area, A3 = "(1/2)(base)(height)"
A3 = (1/2)(15)(23.913) = 179.35 sq.cm.

-------------
Another way of finding the A2 is by the A = (1/2)(base)(altitude) also.

Altitude of unshaded isosceles triangle, h = 17.75cos(25deg) = 16.087cm
So, A2 = (1/2)(15)(16.087) = 120.65 sq.cm.

Another way of finding h is by using the Pythagorean theorem,
(17.75)^2 = (15/2)^2 +h^2
h = sqrt[(17.75)^2 -(7.5)^2) = 16.088 cm.

6. Thanks, ticbol! It's making sense now.

7. I actually found an easier way of doing it, but you sent me down the right track.

I worked out the entire area of the shape using the area formula for a kite and then subtracted the area of the sector:

Area of kite = 15*40/2
= 300cm²

Area of sector = (50/360)*pi*17.75
²
= 137.47
cm² (rounded to two dp)

Area of shaded area = 300 - 137.47
= 162.53
cm²
= 163
cm²