Question

Two questions please help?

Answer 1

Answer:

The area of triangle PQR is

$A=21.5\ cm^{2}$

Step-by-step explanation:

we know that

The area of a triangle is equal to

$A=\frac{1}{2}bh$

see the attached figure to better understand the problem

we have

$b=14\ cm$

$h=PS\ cm$

Find the length side of segment PS

In the right triangle PSR

$sin(20\°)=\frac{PS}{PR}$

Solve for PS

$PS=PRsin(20\°)$

we have

$PR=9\ cm$

substitute

$PS=(9)sin(20\°)$

Find the area of the triangle PQR

$A=\frac{1}{2}(14)((9)sin(20\°))$

$A=21.5\ cm^{2}$

Answer 2

21.5, just took the test!!!!