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In the figure shown above, line segment QR has length 12 and rectangle

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Re: In the figure shown above, line segment QR has length 12 and rectangle  [#permalink]

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New post 24 Mar 2018, 13:16
oss198 wrote:
Image
In the figure shown above, line segment QR has length 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?

(A) 144
(B) 216
(C) 324
(D) 360
(E) 396

Attachment:
Picture 1.png



why isn't E an answer if QR has length 12, and rectangle MPQT is a square -- hence area of square is 144

if area of rectangle TQRS? is 540 so 540-144 = 396

so whats wrong with my reasoning ?
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In the figure shown above, line segment QR has length 12 and rectangle  [#permalink]

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New post 24 Mar 2018, 13:35
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dave13 wrote:
oss198 wrote:
Image
In the figure shown above, line segment QR has length 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?

(A) 144
(B) 216
(C) 324
(D) 360
(E) 396

Attachment:
The attachment Picture 1.png is no longer available



why isn't E an answer if QR has length 12, and rectangle MPQT is a square -- hence area of square is 144


if area of rectangle TQRS? is 540 so 540-144 = 396

so whats wrong with my reasoning ?



Hi dave13

Attachment:
Picture 1.png
Picture 1.png [ 9.84 KiB | Viewed 208 times ]


As you can see, we are given the following details

1. Length of a line segment of QRST which might be a square/rectangle - QR = 12
2. MPQT(the highlighted portion) is a square.

We have been asked to find the area of QRST, and know that MPQT is a square.
That's the reason your reasoning is wrong!

Hope that clears the confusion.
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Re: In the figure shown above, line segment QR has length 12 and rectangle  [#permalink]

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New post 15 Aug 2018, 23:28
How to solve LARGE COEFFICIENT quadratic equations?
ax^2 +bx+c =0 (where a ,b and c are constants/coefficient)

Example
For example, try factoring . (3x^2+10x-1000)
It's relatively simple to factor it to (3x-50)(X+20) but that would take a little while or at least longer than the way that I'm about to discuss.

We begin with the expression
.(3x^2+10x-1000)
Then we divide the second coefficient by 10
and the third by 100,
and we are left with the expression (3x^2+x-10=0)
which we can easily factor to (3x-5)(X+2).
Finally, we multiply the second term in each factor by 10
we have (3x-50)(X+20)=0 . Looks familiar, doesn't it?

Basically, what I have done is that I divided the second coefficient by any one of its factors (in this case 10) and then divided the third coefficient by the square of that factor while leaving the first untouched.

This method applies to irrational and imaginary coefficients as well.

Do the same for
PQ^2+12PQ-540=0

HCF of 12 and 540 = 6
Now divide constant"b" ( in this case 12) with 6
And divide constant"c" with 36( sqayre of 6)
== pq^2+2-15=0
== (PQ+5)(qp-3)=0
Multiply 6 to both the constant of both the factors.
(PQ+30)(pq-18)=0

USE THIS FOR LARGER COEFFICIENTS.

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Re: In the figure shown above, line segment QR has length 12 and rectangle &nbs [#permalink] 15 Aug 2018, 23:28

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In the figure shown above, line segment QR has length 12 and rectangle

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