## Aadarsh Group Title PQRS is a kite. PQ = 7, QR = 11, QS = 10. M is the mid-pint of PR. Find QM^2. 2 years ago 2 years ago

1. Ishaan94 Group Title

Can you draw a figure?

|dw:1334216371667:dw|

4. apoorvk Group Title

|dw:1334216398007:dw| M is mid pt of PR or QS??

5. lgbasallote Group Title

if QS = 10 then QM = 5 5^2 is 25 right?

@lgbasallote , u r wrong.

M is mid-point of PR.

I have done the answer, shall I post it? Will u correct it?

9. lgbasallote Group Title

ahhh found it now >:))

10. thushananth01 Group Title

$\frac{\sqrt{24}+\sqrt{96}}{2}$

Wrong again.

12. thushananth01 Group Title

that is equal to PM

13. apoorvk Group Title

Okay let me think good one. lol. Done.|dw:1334216749935:dw| wait let me type

14. thushananth01 Group Title

Cos rule i guess...what is QRM=

15. thushananth01 Group Title

sorry QPM?

QPM is not given.

17. lgbasallote Group Title

let midpoint of QS = A sqrt (11^2 - 5^2) = AR AR = sqrt (121-25) AR = sqrt (96) AR = 4 sqrt (6) PM + AR = PR + MA PM = PR/2 PR/2 + 4 sqrt 6 = PR + MA lol =)))

18. thushananth01 Group Title

so v can find it using sin x = 5/7

Shall I say how I did? Please. then u will check and rectify me.

20. thushananth01 Group Title

ok?

21. apoorvk Group Title

yeah aadarsh post i am not in the patient mood to type such long replies right now.

Just a minute.

23. thushananth01 Group Title

answer is 31 = QM^2= 32

24. thushananth01 Group Title

31*

How u did @thushananth01 ? Can u say the reason please? U r absolutely right.

26. thushananth01 Group Title

ok first find the full length of PR u can do this first by using pythagoras theorom then u have to add both and divide by 2 to get the midpoint as i gve previous with the radical..that was the length from P to M now we know PM and PQ plus the angle QPM v can find using the sin rule then i got the angle as 45.6 then finally the cose rule$p^{2}= q^{2} + m^{2} - 2qm (cos P)$ $7^{2}+7.35^{2} - 2 \times 7 \times 7.35 \cos 45.6 = p^{2}\rightarrow QM^2$

27. thushananth01 Group Title

$\frac{\sqrt{24} +\sqrt{96}}{2}\rightarrow PM$