#2**+1 **

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)

ElectricPavlov
Apr 24, 2018

#3**0 **

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)

RainbowPanda
Apr 24, 2018

#4**+2 **

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

GYanggg
Apr 24, 2018

#6**0 **

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

RainbowPanda
Apr 24, 2018

#12**0 **

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

RainbowPanda
Apr 24, 2018

#11**+1 **

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)**

Guest Apr 24, 2018