MATH SOLVE

10 months ago

Q:
# In the diagram, SR = and QR = . What is the perimeter of parallelogram PQRS?

Accepted Solution

A:

Since the figure is a parallelogram, then SP=RQ, and SR=PQ.

To find SR, we draw right triangle SAR, as shown in the picture, where the coordinates of A are clearly (-3, -2).

Thus, we can see that SA=4 units, and AR=4 units. By the Pythagorean theorem,

[tex]SR= \sqrt{4^2+4^2}= \sqrt{2\cdot4^2}=4 \sqrt{2}. [/tex]

Similarly, to find SP, which is equal to QR, we draw the triangle PBS, where the coordinates of B are (-2, 2).

Clearly, PB=3 and SB=1, thus using the Pythagorean theorem in right triangle PBS:

[tex] SP= \sqrt{1^2+3^2}= \sqrt{10} [/tex] units.

The perimeter of the parallelogram is SP+RQ+SR+PQ=

[tex]\sqrt{10} +\sqrt{10} +4 \sqrt{2}+4 \sqrt{2}=2\sqrt{10} +8 \sqrt{2}[/tex].

Answer: second choice

To find SR, we draw right triangle SAR, as shown in the picture, where the coordinates of A are clearly (-3, -2).

Thus, we can see that SA=4 units, and AR=4 units. By the Pythagorean theorem,

[tex]SR= \sqrt{4^2+4^2}= \sqrt{2\cdot4^2}=4 \sqrt{2}. [/tex]

Similarly, to find SP, which is equal to QR, we draw the triangle PBS, where the coordinates of B are (-2, 2).

Clearly, PB=3 and SB=1, thus using the Pythagorean theorem in right triangle PBS:

[tex] SP= \sqrt{1^2+3^2}= \sqrt{10} [/tex] units.

The perimeter of the parallelogram is SP+RQ+SR+PQ=

[tex]\sqrt{10} +\sqrt{10} +4 \sqrt{2}+4 \sqrt{2}=2\sqrt{10} +8 \sqrt{2}[/tex].

Answer: second choice