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705-805 (Hard)|   Geometry|      
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A kite – shaped quadrilateral is cut from a circular sheet of paper su 27 Jan 2021, 08:00
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A
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75% (hard)

Question Stats:

50% (02:47) correct 50% (03:10) wrong based on 38 sessions

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A kite – shaped quadrilateral is cut from a circular sheet of paper such that the vertices of the kite lie on the circumference of the circle. If the lengths of the sides of the kite are in the ratio 3: 3: 4 : 4, then approximately what percentage of the area of the circular sheet of paper remains after the kite has been cut out?

(A) 39%
(B) 40%
(C) 45 %
(D) 51%
(E) 53%
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A
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A kite – shaped quadrilateral is cut from a circular sheet of paper su 27 Jan 2021, 08:27
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Assuming sides of kite are 3,3,4,4

The diameter of the circle is 5 (hypotenuse of the right angle triangles formed by vertices of kite)

Area of kite = 2*(Area of triangle with sides 3,4,5) = 12

Area of circle = π*(2.5)^2 = approximately 19.8

Left out area = 7.8

Percentage remaining = 7.8/19.8*100 = 39%

Answer could be 39% or 40% as I haven't calculated exactly

Going with option A

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A kite – shaped quadrilateral is cut from a circular sheet of paper su 27 Jan 2021, 09:34
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Bunuel
A kite – shaped quadrilateral is cut from a circular sheet of paper such that the vertices of the kite lie on the circumference of the circle. If the lengths of the sides of the kite are in the ratio 3: 3: 4 : 4, then approximately what percentage of the area of the circular sheet of paper remains after the kite has been cut out?

(A) 39%
(B) 40%
(C) 45 %
(D) 51%
(E) 53%
Show more

The graph is labeled below, from the inscribed angle theorem the kite must consist of two right triangles. The ratios are obviously 3:4:5 so we can assume the radius is 2.5 and the area of the triangles added up is 3*4 = 12.

The area of the circle is \((\frac{5}{2})^2 * \pi \approx \frac{25}{4} * \pi \). Then the ratio that is cut out is: \(12/(\frac{25}{4} * \pi) = \frac{4*12}{25\pi} \approx \frac{4*12}{25*3*1.05} = \frac{16}{25}/1.05 = 64\%/1.05 \approx 64\% - 5\%*64\% = 64\% - 3.2\% = 60.8\%\). Then there is roughly 39.2% not cut out.
(The answers were quite close which normally isn't the case, so don't bother too much about the estimations. Most of the time 3.14 ->3 is a sufficient approximation).

Ans: A
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Mugdho
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A kite – shaped quadrilateral is cut from a circular sheet of paper su 09 Jan 2022, 04:15
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Bunuel
A kite – shaped quadrilateral is cut from a circular sheet of paper such that the vertices of the kite lie on the circumference of the circle. If the lengths of the sides of the kite are in the ratio 3: 3: 4 : 4, then approximately what percentage of the area of the circular sheet of paper remains after the kite has been cut out?

(A) 39%
(B) 40%
(C) 45 %
(D) 51%
(E) 53%

The graph is labeled below, from the inscribed angle theorem the kite must consist of two right triangles. The ratios are obviously 3:4:5 so we can assume the radius is 2.5 and the area of the triangles added up is 3*4 = 12.

The area of the circle is \((\frac{5}{2})^2 * \pi \approx \frac{25}{4} * \pi \). Then the ratio that is cut out is: \(12/(\frac{25}{4} * \pi) = \frac{4*12}{25\pi} \approx \frac{4*12}{25*3*1.05} = \frac{16}{25}/1.05 = 64\%/1.05 \approx 64\% - 5\%*64\% = 64\% - 3.2\% = 60.8\%\). Then there is roughly 39.2% not cut out.
(The answers were quite close which normally isn't the case, so don't bother too much about the estimations. Most of the time 3.14 ->3 is a sufficient approximation).

Ans: A
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TestPrepUnlimited

How do we know that "kite must consist of two right triangles"?....can you plz elaborate this portion?

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A kite – shaped quadrilateral is cut from a circular sheet of paper su 09 Jan 2022, 12:38
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Mugdho
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Bunuel
A kite – shaped quadrilateral is cut from a circular sheet of paper such that the vertices of the kite lie on the circumference of the circle. If the lengths of the sides of the kite are in the ratio 3: 3: 4 : 4, then approximately what percentage of the area of the circular sheet of paper remains after the kite has been cut out?

(A) 39%
(B) 40%
(C) 45 %
(D) 51%
(E) 53%

The graph is labeled below, from the inscribed angle theorem the kite must consist of two right triangles. The ratios are obviously 3:4:5 so we can assume the radius is 2.5 and the area of the triangles added up is 3*4 = 12.

The area of the circle is \((\frac{5}{2})^2 * \pi \approx \frac{25}{4} * \pi \). Then the ratio that is cut out is: \(12/(\frac{25}{4} * \pi) = \frac{4*12}{25\pi} \approx \frac{4*12}{25*3*1.05} = \frac{16}{25}/1.05 = 64\%/1.05 \approx 64\% - 5\%*64\% = 64\% - 3.2\% = 60.8\%\). Then there is roughly 39.2% not cut out.
(The answers were quite close which normally isn't the case, so don't bother too much about the estimations. Most of the time 3.14 ->3 is a sufficient approximation).

Ans: A


TestPrepUnlimited

How do we know that "kite must consist of two right triangles"?....can you plz elaborate this portion?

Posted from my mobile device
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Say one of the 4 legs was right on top of the other, on the same side of the center. Well then there would only be one side of 4.

For there to be a distinct side it has to then sweep and of course will become longer than 4 until it starts shrinking in length again as it crosses the center and will eventually hit a length of 4 on the other side of the center.

Same holds for the lengths of 3.

This is how you know the center of the circle is inside the quadrilateral.

You can see that the triangles are similar due to their identical sides.

This means that angle B equals angle D.

Because these two angles together span the full circumference of the circle, together they must add to 180 degrees. And because they're equal, both must be 90 degrees.
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