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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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Updated on: 18 Feb 2017, 01:06
4
00:00

Difficulty:

55% (hard)

Question Stats:

66% (02:48) correct 34% (03:47) wrong based on 72 sessions

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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

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Originally posted by hazelnut on 18 Feb 2017, 00:25.
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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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

Ans C
Manager
Joined: 15 Mar 2014
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Location: India
Concentration: Technology, General Management
GPA: 3.5
WE: Operations (Telecommunications)
For the rectangular shown above, if (BR)=(BQ)=5√2 and (AR)=(PB)=√119  [#permalink]

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

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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!!!
Manager
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Posts: 141
Location: India
Concentration: Technology, General Management
GPA: 3.5
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Re: For the rectangular shown above, if (BR)=(BQ)=5√2 and (AR)=(PB)=√119  [#permalink]

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

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04 Aug 2017, 02:59
2
1
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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