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For the rectangular shown above, if (BR)=(BQ)=5√2 and (AR)=(PB)=√119

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For the rectangular shown above, if (BR)=(BQ)=5√2 and (AR)=(PB)=√119 [#permalink]

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New post 18 Feb 2017, 00:25
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For the rectangular shown above, if \((BR)=(BQ)=5\sqrt{2}\) and \((AR)=(PB)=(CQ)=\sqrt{119}\), what is the area of the triangle PQR?

A. 30
B. 45
C. 60
D. 75
E. 90
[Reveal] Spoiler: OA

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Last edited by hazelnut on 18 Feb 2017, 01:06, edited 1 time in total.

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Re: For the rectangular shown above, if (BR)=(BQ)=5√2 and (AR)=(PB)=√119 [#permalink]

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New post 18 Feb 2017, 01:03
ziyuenlau wrote:
For the rectangular shown above, if \((BR)=(BQ)=5\sqrt{2}\) and \((AR)=(PB)=(CQ)=\sqrt{119}\), what is the area of the triangle PQR?

A. 30
B. 45
C. 60
D. 75
E. 90


PR^2=PQ^2= √119^2 + 5√2^2 = 119 +50= 169
PR=PQ= 13
RQ^2= 5√2^2 + 5√2^2 = 100
RQ= 10

height = 13^2 - 10/2^2 = 169-25= 144 =12
Area of triangle = 1/2 *12 *10 = 60

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For the rectangular shown above, if (BR)=(BQ)=5√2 and (AR)=(PB)=√119 [#permalink]

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New post 18 Feb 2017, 10:56
I'm trying to find the area by directly without using the Pythagoras theorem but looks like a wrong approach. Could anyone please help me out.

Steps done:

Area of PQR = Area of PRB + Area of PBQ + Area of QDR
= 3 * 1/2 * 5√2 * √119 ?
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Re: For the rectangular shown above, if (BR)=(BQ)=5√2 and (AR)=(PB)=√119 [#permalink]

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New post 18 Feb 2017, 11:11
ashwink wrote:
I'm trying to find the area by directly without using the Pythagoras theorem but looks like a wrong approach. Could anyone please help me out.

Steps done:

Area of PQR = Area of PRB + Area of PBQ + Area of QDR
= 3 * 1/2 * 5√2 * √119 ?



ashwink

well ...

first of all your consideration is wrong
PQR is also the back part of whole volume u are considering
so Area of PQR = Area of PRB + Area of PBQ + Area of QDR is never a possible case

u must try with pytha theorem!!!

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Re: For the rectangular shown above, if (BR)=(BQ)=5√2 and (AR)=(PB)=√119 [#permalink]

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New post 18 Feb 2017, 11:16
rohit8865 wrote:

first of all your consideration is wrong
PQR is also the back part of whole volume u are considering
so Area of PQR = Area of PRB + Area of PBQ + Area of QDR is never a possible case

u must try with pytha theorem!!!


Ah didn't take into consideration the back as well! Thanks for the quick reply! :)
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Re: For the rectangular shown above, if (BR)=(BQ)=5√2 and (AR)=(PB)=√119 [#permalink]

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New post 18 Feb 2017, 11:25
ashwink wrote:
rohit8865 wrote:

first of all your consideration is wrong
PQR is also the back part of whole volume u are considering
so Area of PQR = Area of PRB + Area of PBQ + Area of QDR is never a possible case

u must try with pytha theorem!!!


Ah didn't take into consideration the back as well! Thanks for the quick reply! :)



welcome back...

u can replace the highlighted part
kudos is better way of thanking :wink:

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Re: For the rectangular shown above, if (BR)=(BQ)=5√2 and (AR)=(PB)=√119 [#permalink]

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New post 04 Aug 2017, 02:59
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hazelnut wrote:
For the rectangular shown above, if \((BR)=(BQ)=5\sqrt{2}\) and \((AR)=(PB)=(CQ)=\sqrt{119}\), what is the area of the triangle PQR?

A. 30
B. 45
C. 60
D. 75
E. 90


QR = \(\sqrt{{{5[square_root]2}}^2 + {5\sqrt{2}}^2}[/square_root]\) = 10
PQ = PR = \(\sqrt{{{[square_root]119}}^2 + {5\sqrt{2}}^2}[/square_root][/square_root]\) = 13

So,, sides of triangle are 13,13,10
semi perimeter = (13+13+10)/2= 18
Area of triangle PQR = \(\sqrt{18*(18-13)*(18-13)*(18-10)}\) = 60.
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Re: For the rectangular shown above, if (BR)=(BQ)=5√2 and (AR)=(PB)=√119   [#permalink] 04 Aug 2017, 02:59
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