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In the 7-inch square above, another square is inscribed. Wha
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06 Oct 2018, 15:15
06 Oct 2018, 15:15
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In the 7-inch square above, another square is inscribed. What fraction of the larger square is shaded?
(A) \(\frac{3}{12}\)
(B) \(\frac{24}{49}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{25}{49}\)
(E) \(\frac{7}{12}\)
Re: In the 7-inch square above, another square is inscribed. Wha
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15 Nov 2018, 00:48
15 Nov 2018, 00:48
1
This problem seems more intuitive to me, so I'll do my best to explain it.
The problem starts off by saying that this is a 7-inch square and we need to find the fraction that is shaded. Since this is implying area without saying it, we can multiply both sides to get a total area of 49 for the big square.
To get the smaller square, it is a little harder for me to explain. In the sides, we see that there is a 3 and 4. Because a square is even on all the sides, you can put a 4 at the top bordering the 3 or a 3 on the bottom bordering the 4. From here we realize that both sides are now 3 and 4, which fit into Pythagorean's Theorem of a 3,4,5 area triangle. The hypotenuse of 5, would be a side of the inner triangle. This is important because we can now find the inner square.
If the inner square is 5-inches, it means it covers a total of 25 square inches.
From the original 7-inch square, we see that it is 49 inches. So, 49-25 = 24 inches left unshaded, out of the 49 possible, hence the fraction \frac{24}{49}.
B is your answer.
The problem starts off by saying that this is a 7-inch square and we need to find the fraction that is shaded. Since this is implying area without saying it, we can multiply both sides to get a total area of 49 for the big square.
To get the smaller square, it is a little harder for me to explain. In the sides, we see that there is a 3 and 4. Because a square is even on all the sides, you can put a 4 at the top bordering the 3 or a 3 on the bottom bordering the 4. From here we realize that both sides are now 3 and 4, which fit into Pythagorean's Theorem of a 3,4,5 area triangle. The hypotenuse of 5, would be a side of the inner triangle. This is important because we can now find the inner square.
If the inner square is 5-inches, it means it covers a total of 25 square inches.
From the original 7-inch square, we see that it is 49 inches. So, 49-25 = 24 inches left unshaded, out of the 49 possible, hence the fraction \frac{24}{49}.
B is your answer.
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Re: In the 7-inch square above, another square is inscribed. Wha
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19 Nov 2018, 17:39
19 Nov 2018, 17:39
1
This can be solved in two ways.
First noticing that the small square will be part of pythagorean theorem where x^2 = 4^2 + 3^2 = 25 so the side of this square is 5 and the area would be 25.
The total area of the larger square is (4+3)^2 = 49
We subtract the smaller square from the larger 49-25 = 24
So it would be 24/49.
The other way is to notice that there are 4 triangles with base and height 4 and 3. So it would be 4*3 *1/2 = 6.
Since we have 4 triangles it then becomes 4 * 6 = 24/49
First noticing that the small square will be part of pythagorean theorem where x^2 = 4^2 + 3^2 = 25 so the side of this square is 5 and the area would be 25.
The total area of the larger square is (4+3)^2 = 49
We subtract the smaller square from the larger 49-25 = 24
So it would be 24/49.
The other way is to notice that there are 4 triangles with base and height 4 and 3. So it would be 4*3 *1/2 = 6.
Since we have 4 triangles it then becomes 4 * 6 = 24/49
