+37146 I assume since you need help with #61 , you already solved 60 and 59
then use pythagorean theorem to find RQ
RQ ^2 = RS ^2 + 16^2 ( RS=8 BTW)
+2448 Mhm CPhill helped me with 59 and 60
Also, for 61 would it end up being RQ^2=360
So 17.9^2=360??
Also, I really need help with 58. I know the answer but I need help solving it to get the same answer (5/2 cm.^2)
+983 Since the perimeter of rectangle ABCD is 60 and has a length of 20, the width is:
\((60-(20\cdot2))\div2=10\)
Since they are similar, we can set the equation:
\(\frac{20}{32}=\frac{10}{x}\)
Solving for x, we have
x = 16
Since we know the width and the length, we know the area = 16 * 32 = 512
The answer you are looking for is 512.
I hope this helped,
gavin
+2448 Omg this whole time I thought the answer said 5/2 (cuz my teacher gave me answers but I had to solve the problem to get the answer) and it says 512 I was looking at the wrong problem XDD ty
+2448 That too lol but one of the answers to a previous question was 5/2 and the way I wrote 512 looks like 5/2
61)
16/RQ = RQ/20 The cos of RQS = cos of RQP. Cross multiply
RQ^2 =16 x 20
RQ^2 = 320
RQ = Sqrt(320)
RQ =8sqrt(5)
Need help with 58 and 61
