Question 1.
The sketch of △XYZ is shown in the attachment .
Area of triangle XYZ
[tex] = \frac{1}{2} bh[/tex]
The base is |XY|=6.
We use the Pythagoras Theorem to find the height, h as follows:
[tex] {h}^{2} + {3}^{2} = {7}^{2} [/tex]
[tex]{h}^{2} + 9 = 4 9[/tex]
[tex]{h}^{2} = 49 - 9 \\ {h}^{2} = 40[/tex]
[tex]h = \sqrt{40} [/tex]
[tex]h = 2 \sqrt{10} [/tex]
The area then becomes:
[tex] = \frac{1}{2} \times 6 \times 2 \sqrt{10} \\ = 6 \sqrt{10} [/tex]
The area of the triangle is 6√10 square units.
Question 2
We want to find the radius of an arc that subtends an angle of
[tex] \frac{13\pi}{18} [/tex]
radians and arc length is 18.4 inches.
We substitute these values into the following relation and solve for r.
[tex]s = r \theta[/tex]
[tex]18 .4= \frac{13\pi}{18} r[/tex]
This implies that:
[tex]r = \frac{18}{13\pi} \times 18.4[/tex]
[tex]r = \frac{324}{13\pi} [/tex]
[tex]r = 8.1 \: inches[/tex]
Answer: C) approximately 8.1 inches
Question 3.
The area of a sector in radians is given by:
[tex]A= \frac{1}{2} {r}^{2} \theta[/tex]
From the question the radius is r=7 inches and
[tex] \theta = \frac{4 \pi} { 15} [/tex]
We substitute into the formula to get:
[tex]A= \frac{1}{2} \times {7}^{2} \times \frac{4\pi}{15} [/tex]
[tex]A= 49 \times \frac{2\pi}{15} [/tex]
[tex]A=\frac{98\pi}{15} [/tex]
[tex]A=6\frac{8\pi}{15} [/tex]
square inches
