The section of a certain solid consists of a semicircles,a rectangle,and a triangle

Question

The section of a certain solid consists of a semicircles,a rectangle,and a triangle.The altitude of the triangle is three times the radius of semicircles,the altitude of the triangle is twice the same radius ,and the area of the triangle is 20 sq ft.Find the area of this section.

Answer 1

Ronalyn,

I'm having difficulty understanding your description of the planar section through a solid.  What I deduce and visualize is a semicircle of radius r (and diameter 2r) on top of which sits a rectangle of base 2r spanning the diameter and height  also of 2r (note, you labeled it a second triangle height?) - so it's a square. And finally on top of that is a triangle of base 2r spanning the entire semicircle, and height 3r.  If this is not a correct picture of the section, then you will have to make appropriate adjustments in the math below.

 

The area of the triangle is 1/2 * base * height = 1/2*2r*3r = 3r*r = 20 ft squared, or r*r=20/3 to use below

So r = square root of 20/3

Area of the rectangle (really a square) = base * height = 2r*2r = 4r*r = 4(20/3) = 80/3

Area of the semicircle = 1/2*pi*r*r = 1/2*pi*(20/3) = 10/3*pi

 

So the total area of the section is just those three pieces added together:

Section Area = 20 (triangle) + 80/3 (rectangle) + 10/3* pi,  where pi=3.14159

                   = 20 + 26.667 + 10.472 = 57.139 ft squared

 

So if the actual figure of your section is different, you can rework the above by building up your section using any number of those simpler geometric figures as I have done just keeping their dimensions straight.