Geometry: Areas

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1. An equilateral triangle is inscribed in a circle with radius 2. What is the area of the triangle?

(A) 3 · √3
(B) 2 · √3
(C) 3 · √2
(D) π · √6
(E) π · √10

Draw the triangle in a circle, draw a radius to a corner of the triangle, and draw a perpendicular from the center of the circle to one of the sides of the triangle. The hypotenuse of your 30-60-90 triangle is 2 (=radius), the height is 1, and the base is √3. The area of this small triangle is √3 / 2. The full triangle has 6 small triangles so its area is 3√3 (A).

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2. The area of the square in the diagram is equal to s². What fraction of its area is occupied by the four circles?

(A) π · s² / 4
(B) 6π · s² / √3
(C) 2π / 5
(D) √6π / 6
(E) π / 4

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The sides of the square have length s which means the radius of each circle is s/4. Calculate the total area of the four circles: 4πs² / 16 and divide by s² to get π / 4 or (E).

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3. A square is inscribed in a circle. The area of the circle is 6. What is the area of the square?

(A) 4
(B) 16 / π
(C) 12 / π
(D) 3 · √2
(E) 8 / π

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The area of the circle is given so you have to solve for the radius using: 6 = πr². Once you have the radius (√6 / √π) double it to get the diameter (2√6 / √π) which is the hypotenuse of a 45-45-90 triangle. The legs of that triangle equal the hypotenuse divided by √2 which gives you 2√3 / √π for the length of a side of the square. Now square this number to get the area of 12 / π or (C).

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4. A square is inscribed in a large circle. A small circle is inscribed in the square. If the area of the small circle is 1, what is the area of the large circle?

(A) √2
(B) 3/2
(C) √3
(D) 2
(E) 2 · √2

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You need the radius of the big circle. This big radius makes a 45-45-90 triangle with the radius of the small circle and half of a side of the square. Since the area of the small circle is 1, you know the radius must be 1 / √π. This means the radius of the big circle is √2 / √π. Square this and multiply by π to get 2 for the area of the big circle (D).

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5. Each circle in the figure above has radius r = 1. What is the area of the shaded region surrounded by the three circles?

(A) (3 · √3 – π) / 2
(B) (2 · √3 – π) / 3
(C) (3 · √2 – π) / 2
(D) (2 · √3 – π) / 2
(E) (3 · √3 – π) / 3

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Each circle has area π (because the radius is 1) and since the angles of an equilateral triangle are all 60 degrees, the area of each sector inside the triangle is π / 6 (60/360 = 1/6). The area of the whole equilateral triangle is √3 (see below). The area of the shaded region is √3 minus the area of all 3 sectors (π / 2). If you get a common denominator and subtract, you get (D).

The area of the equilateral triangle is one-half base times height. Half the base is equal to 1 and the height is √3 because it is the long leg of a 30-60-90 and the short leg is 1. The area of the equilateral triangle is √3.

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6. In the figure, AX / BX = 2. If the area of quadrilateral AXCD = 30, what is the area of rectangle ABCD?

(A) 32
(B) 33
(C) 35
(D) 36
(E) 39

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You can fit six triangles the size of CXB in the rectangle. Start by dropping a perpendicular from point X to line DC. That gives you two triangles. Since point X is 2/3 of the way from point A to point B (AX/BX = 2) you can fit four more triangles. Since there are 5 triangles in AXCD and its area is 30, the area of each triangle must be 6 and the area of the ABCD must be 36 (D).

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7. Circles L and R have radius 4. The radius of circle M is 2 and line LMR is collinear with the diameter of the largest circle. What is the area of the shaded region?

(A) 64π
(B) 48π
(C) 42π
(D) 32π
(E) 24π

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The radius of the large circle is 4+4+2=10. The area of the shaded region is half the area of the large circle minus half the area of the three smaller circles. Or, you can get the area of the big circle, subtract the area of the three smaller circles and divide by 2. Either way you get: (1/2)(100π-16π-16π-4π) = 32π (D).

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8. The center of circle O is one of vertices of square OPQR. The midpoint of OP is S and the midpoint of OR is T. The shaded region occupies what fraction of the square?

(A) π · √2 / 20
(B) π · √3 / 36
(C) π / (12 · √2)
(D) π / 16
(E) π / 20

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The question is asking for the ratio of the area of the shaded region to the area of the square. Call the side of the square s. The radius of the circle is s/2. The area of the circle is πs² / 4 and a quarter of the area of the circle is πs² / 16. Divide this by the area of the square (s²) to get the ratio. You get π / 16 (D).

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Note: Figure not drawn to scale.

9. The area of circle P is twice that of circle O. The measure of angle MON is equal to x. The measure of angle QPR is equal to 2x. If the clockwise distance along circle O from point M to point N is equal to 3, what is the clockwise distance along circle P from point Q to point R?

(A) 6
(B) 6 · √2
(C) 6 · √3
(D) 12
(E) 24

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If you double the area of a circle, you multiply the radius by √2. The circumference and all arc lengths are also multiplied by √2. So doubling the area of circle O means arc MN goes from 3 to 3√2. But then you increase the angle from x to 2x to make arc QR. Doubling the angle doubles the arc length. So arc QR is 6√2 (B).

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Note: Figure not drawn to scale.

10. An isosceles triangle has vertex angle x < 90° and an area equal to 2. A new isosceles triangle is drawn with vertex angle 2x. If the new triangle has the same height as the original triangle, the area of the new triangle is:

(A) 2 · √2
(B) 2 · √3
(C) 4
(D) 6
(E) cannot be determined

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If the angle is pretty small like 1 degree and you double the angle you'll approximately double the length of the base (if the base were the arc of a circle, you would exactly double it). If the angle is 89.999999 degrees and you double it, your base will be HUGE. In fact, there is no limit to how big the base can get so the answer is "cannot be determined" (E).

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