Tinos
Answered

In the figure O is the center of a semi-circle PQR and r is the radius.

PQ is a diameter of semi-circle PTO. Q^ = x.

Determine:


1.) RQ in terms of r and x , and then simplify the expression. (3 marks)

2.)The ratio Area of triangle ORQ/Area of triangle POT if x = 75 degrees. (leave your answer simplified in surd form) (5 marks)

In the figure O is the center of a semi-circle PQR and r is the radius. PQ is a diameter of semi-circle PTO. Q = x. Determine: 1.) RQ in terms of r and x , and class=

Verified Answer :

Hagrid

1.) We can draw the height of triangle ORQ that intersects the center and we form right triangles. Using the definition of cosine, we get
cos x = (1/2) RQ / r
RQ = 2r cos x

2.) We simply use the expression for RQ we got from 1 to solve for the area of triangle ORQ. Similar to 1), the expression for the height of triangle ORQ is: r sin x.
For triangle POT, if we draw the height of the triangle that hits OP, we get special right triangles with angles 30-60-90. The height is simply twice that of the radius divided by the square root of 3.

So,
ratio of areas = (1/2)bh of triangle ORQ / (1/2)bh of triangle POT

b is the base  and h is the height

ratio of areas = (1/2)(2r cos 75)(r sin 75) / (1/2)(r)(2/sqrt(3))(r)
ratio of areas = sqrt(3) / 8

Related Questions