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Re: In the 7-inch square above, another square is inscribed. Wha [#permalink]
if we take the diagonal of the smaller square=7, thus, one side=7/[square_root][2]. area=24.5, shaded area=49-24.5. So, the proportion 24.5/49. But I do not find it matching. Pls correct me.
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Re: In the 7-inch square above, another square is inscribed. Wha [#permalink]
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No.

You do not have enough information to draw the conclusion that the diagonal of the smallest square is 7

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In the 7-inch square above, another square is inscribed. Wha [#permalink]
1
The vertex of the inner square divides one side of the outer square in the ratio of \(3\) and \(4\). We can expect the other four sides of the outer square also to be divided into similar ratios as there are two squares inscribed inside one another.

We automatically see that each shaded region is a \(3,4,5\) right angled square with an area of \(\frac{1}{2} \times 4 \times 3 = 6.\)

Since there are four shaded regions, their total area will be \(4 \times 6 = 24 \)square inches.

The area of the larger square is \(7 \times 7 = 49\) square inches.

The fraction of the larger square shaded is\( \frac{24}{49}\)

The answer is Choice B
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In the 7-inch square above, another square is inscribed. Wha [#permalink]
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