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# A painting at an art gallery is framed such that the area

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A painting at an art gallery is framed such that the area [#permalink]

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12 Mar 2013, 21:29
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A painting at an art gallery is framed such that the area of the square frame is 1/4 the area of the painting itself. If the diagonal line connecting corners of the frame has a length of 10, what is the area of the painting inside the frame?

A. 10
B. 20
C. 30
D. 40
E. 50
[Reveal] Spoiler: OA

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Re: A painting at an art gallery is framed such that the area [#permalink]

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12 Mar 2013, 23:26
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emmak wrote:
A painting at an art gallery is framed such that the area of the square frame is 1/4 the area of the painting itself. If the diagonal line connecting corners of the frame has a length of 10, what is the area of the painting inside the frame?
a) 10
b) 20
c) 30
d) 40
e) 50

Figure attached
Suppose side of the painting(which is a square) is "a" and side of the outer square(paiting+frame) is "b"

Area of paiting = a^2 and we know that the area of the frame is (1/4) of that of the paiting so
Area of frame = (1/4) * a^2

Area of Frame + Paiting = a^2 + (1/4)*a^2 = (5/4)*a^2 which is equal to b^2

Line connecting the corners of the frame is the diagonal of the outer square which is equal to b*sqrt2

so, b * sqrt2 = 10
so, b = 5 * sqrt2
b^2 = 50

we know that b^2 = (5/4)*a^2
so, a^2 = (4/5)*b^2 = (4/5)*50 = 40

And area of paiting = a^2 = 40

Hope it helps!
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Re: A painting at an art gallery is framed such that the area [#permalink]

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12 Mar 2013, 23:34
It has been established that both the painting and the frame are square.

So let the side of the frame be 'A' and the side of the painting be 'a'

It is stated that $$A^2-a^2 = \frac{1}{4}*a^2$$

It has also stated that the diagonal of the frame is 10, so $$\sqrt{2}*A = 10$$

So $$A=\frac{10}{\sqrt{2}}$$ which implies $$A^2=50$$

$$50-a^2 = \frac{1}{4}*a^2$$
$$50 = \frac{5}{4}*a^2$$
So, $$a^2 = 40$$

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Re: A painting at an art gallery is framed such that the area [#permalink]

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13 Mar 2013, 20:03
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Expert's post
emmak wrote:
A painting at an art gallery is framed such that the area of the square frame is 1/4 the area of the painting itself. If the diagonal line connecting corners of the frame has a length of 10, what is the area of the painting inside the frame?

A. 10
B. 20
C. 30
D. 40
E. 50

Start the solution with what is given to you in the question. You may not need to use any variables in that case.

Diagonal = 10
Then, side of the frame = $$5\sqrt{2}$$
Area of the entire painting + frame = $$(5\sqrt{2})^2 = 50$$

Since area of frame is 1/4th of the area of painting,
Area of frame = 10 and Area of painting = 40
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Re: A painting at an art gallery is framed such that the area [#permalink]

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09 Apr 2013, 10:56
VeritasPrepKarishma wrote:
emmak wrote:
A painting at an art gallery is framed such that the area of the square frame is 1/4 the area of the painting itself. If the diagonal line connecting corners of the frame has a length of 10, what is the area of the painting inside the frame?

A. 10
B. 20
C. 30
D. 40
E. 50

Start the solution with what is given to you in the question. You may not need to use any variables in that case.

Diagonal = 10
Then, side of the frame = $$5\sqrt{2}$$
Area of the entire painting + frame = $$(5\sqrt{2})^2 = 50$$

Since area of frame is 1/4th of the area of painting,
Area of frame = 10 and Area of painting = 40

Oh , that's such a neat way !
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Re: A painting at an art gallery is framed such that the area [#permalink]

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09 Apr 2013, 16:29
emmak wrote:
A painting at an art gallery is framed such that the area of the square frame is 1/4 the area of the painting itself. If the diagonal line connecting corners of the frame has a length of 10, what is the area of the painting inside the frame?

A. 10
B. 20
C. 30
D. 40
E. 50

Let d=diagonal length
Area of a square = $$\frac{d^2}{2}$$

So the entire area of the Frame + Painting would be.
$$A= \frac{10^2}{2}$$
$$A= \frac{100}{2}$$
$$A= 50$$

We know that the area of the frame is 1/4 that of the painting. So it is clear that the frame is 10 and the painting is 40.
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Re: A painting at an art gallery is framed such that the area [#permalink]

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02 Jul 2014, 15:11
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