Question

Felicia wants to draw triangle PQR such that the conditions shown are true. I already did something as you can see , but Im not sure where to go from bere and how to explain question a either. Please show how u got ur work. Please help!! quickly IF POSSIBLE!!

Answer 1

The triangle
`$\triangle P Q R$` with
`$\overline{P R}=10$` units,
`$\overline{P Q}=(50)/(3)$` units, and area
`$(250)/(3)$` square units is one possible solution.

Choose the length of
`$\overline{P R}$`

We know that the area of
`$\triangle \mathrm{PQR}$` is not 6 square units, so we can choose any convenient base length that is not divisible by 6 . In this example, we will choose
`$\overline{P R}=10$` units.

Calculate the height of the triangle

Since
`$\cos P=(b)/(h)=\frac{\overline{P R}}{\overline{P Q}}$`, we can solve for the height
`$\overline{P Q}$` as follows:

`$$\overline{P Q}=\frac{\overline{P R}}{\cos P}=(10)/(0.6)=(50)/(3) \approx 16.67 \text { units }$$`

Draw the triangle

Using the Connect Line tool, draw
`$\overline{P R}$` with a length of 10 units. Then, choose a point
`$\overline{P Q}$` above
`$\overline{P R}$` such that the height of the triangle is approximately 16.67 units. Label the points as P, Q, and R.

Check the area

The area of
`$\triangle \mathrm{PQR}$` is calculated as:

`$K=(1)/(2) \cdot \overline{P R} \cdot \overline{P Q}=(1)/(2) * 10 * (50)/(3)=(250)/(3) \approx 83.33$` square units

Since the area is not 6 square units, our triangle satisfies the given conditions.

Answer 2

Cos p = 3/5=0.6

Therefore, Hypotenuses, PR = 5 and height, h = 3

If this was a right angle triangle such that q=90°, then QR= sqrt (5^2-3^) = 4
and the area, A = 1/2*3*4 = 6 (this is against the conditions)

Lets assume that PQ=PR = 5, and h changes to Adjacent side to angle p= 3.
Then half QR = 3,

Now, h= Sqrt (5^2-3^2) = 4
The new triangle has an area of,

A= 1/2 (3*2)*4 = 3*4 =12 (different from 6 as required).
The triangle will look as the attached photo.