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23 May 2018, 07:05
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In the figure above, O and P are the centers of the two circles. If each circle has radius r, what is the area of the shaded region?
A) \((\frac{\sqrt{2}}{2})(r^2)\)
B) \((\frac{\sqrt{3}}{2})(r^2)\)
C) \((\sqrt{2})(r^2)\)
D) \((\sqrt{3})(r^2)\)
E) \((2\sqrt{3})(r^2)\)
Attachment:
b1pxiJI.jpg [ 12.17 KiB | Viewed 13861 times ]
23 May 2018, 07:32
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OA: B
Area of rhombus =\(\frac{1}{2}*d*\sqrt{4a^2-d^2}\)
where d = length of one of the diagonals of rhombus
a = side of the rhombus
here a=d=r
so Area = \(\frac{1}{2}*r*\sqrt{4r^2-r^2}\)
Area = \(\frac{\sqrt{3}r^2}{2}\)
Area of rhombus =\(\frac{1}{2}*d*\sqrt{4a^2-d^2}\)
where d = length of one of the diagonals of rhombus
a = side of the rhombus
here a=d=r
so Area = \(\frac{1}{2}*r*\sqrt{4r^2-r^2}\)
Area = \(\frac{\sqrt{3}r^2}{2}\)
23 May 2018, 14:34
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Bunuel
Attachment:
b1pxiJI123.png [ 52.1 KiB | Viewed 13020 times ]
I. Area of shaded region = (Area of equilateral triangle * 2)
Area of figure OXPY = area of ∆ OPX + ∆ OPY
∆ OPX and ∆ OPY are equilateral
Diagram, on left, red outline: all side lengths = \(r\)
Area of equilateral triangle = \(\frac{r^2 \sqrt{3}}{4}\)
Shaded region's area = Area of ∆ OPX + ∆ OPY = 2*AREA of triangles
\((\frac{r^2 \sqrt{3}}{4})*2 =\)
\(\frac{2r^2 \sqrt{3}}{4} =\)
\(\frac {r^2 \sqrt{3}}{2}\) = \((\frac{\sqrt{3}}{2})(r^2)\)
Answer B
II. Use 30:60:90 triangles to find area of shaded region
Shaded region's area = Area of ∆ OPX + ∆ OPY
Use 30:60:90 triangle to find area of OPX
∆ OPX = ∆ OPY and they are equilateral
(diagram, left side, two triangles in red outline all sides =\(r\))
1) Divide shaded region into ∆ OPX and ∆ OPY, and
2) drop an altitude from vertex X to base OP, which creates
two 30-60-90 triangles
• altitude is a perpendicular bisector of the base
• altitude bisects the opposite angle
• thus the altitude divides the equilateral triangle
into two 30-60-90 triangles
3) 30-60-90 triangles have corresponding sides
opposite those angles in ratio
\(x : x\sqrt{3} : 2x\)
or in this case \(r\) corresponds with \(2x\), so the ratio is halved
\(\frac{r}{2}:\frac{r\sqrt{3}}{2}: r\)
4) use \(\frac{b*h}{2}\) to find area of ∆ OPX
Double that area
Area of ∆ OPX = \(\frac{b*h}{2}\)
base = OP =\(r\)
height =\(\frac{r\sqrt{3}}{2}\)
Area of ∆ OPX + ∆ OPX = \((r*\frac{r\sqrt{3}}{2}* \frac{1}{2}) * 2=\)
\(\frac{2r^2\sqrt{3}}{4}=\frac {r^2 \sqrt{3}}{2}=(\frac{\sqrt{3}}{2})(r^2)\)
Answer B
