Bunuel wrote:
In the figure above, if isosceles right triangle PQR has an area of 4, what is the area of the shaded portion of the figure?
(A) \(\pi\)
(B) \(2\pi\)
(C) \(2\sqrt{2}\pi\)
(D) \(4\pi\)
(E) \(8\pi\)
Attachment:
The attachment 2018-02-05_0842.png is no longer available
Attachment:
2018-02-05_0842ed.png [ 19.54 KiB | Viewed 1314 times ]
The shaded region is a sector of the circle.
Use hypotenuse PQ to find the radius of circle.
The radius plus the central angle of sector PQR will yield shaded region's area.
Draw a line from R to PQ: RX = Height of ∆ PQR = radius
• From given area, find side length of ∆ PQR, then find hypotenuse
Side lengthArea of ∆ PQR* =
\(\frac{s^2}{2} = 4\)
\(s^2 = 8\)
\(\sqrt{s^2} = \sqrt{4*2}\)
\(s = 2\sqrt{2}\)Hypotenuse PQ**
∆ PQR has angle measures 45-45-90.
Sides opposite those angles are in ratio
\(x : x: x\sqrt{2}\)\(x = 2\sqrt{2}\)PQ corresponds with
\(x\sqrt{2}= (2\sqrt{2} * \sqrt{2}) = (2 * 2) = 4\)• Find radius
Area of ∆ PQR = Base
\(\frac{PQ * h}{2}\), where
\(PQ = 4\)\(\frac{4 * h}{2} = 4\)
\(4 * h = 8\)
\(h = 2\)radius \(RX = 2\)• Use radius plus central angle to find area of sector
The central angle of the sector is 90°
The sector as a fraction of the circle is
\(\frac{Part}{Whole}=\frac{SectorAngle}{360}=\frac{90}{360}=\frac{1}{4}\)Sector Area =
\(\frac{1}{4}\) Circle AreaCircle area =
\(\pi r^2 = 4\pi\)Sector Area =
\(\frac{1}{4} * 4\pi = \pi\)Sector area = shaded region = \(π\)
Answer A
*You can use Area = \(\frac{b*h}{2}\)
Isosceles triangles have legs of equal lengths.
Base = \(x\), height = \(x\)
\(A = \frac{x * x}{2} = 4\)
\(x^2 = 8\)
\(x = 2\sqrt{2}\)
**Or use Pythagorean theorem: leg\(^2\) + leg\(^2\) = hypotenuse\(^2\)
\((2\sqrt{2})^2 + (2\sqrt{2})^2 = (PQ)^2\)
\(8 + 8 = PQ^2\)
\(PQ = \sqrt{16}\)
\(PQ = 4\)
_________________
SC Butler has resumed!
Get two SC questions to practice, whose links you can find by date, here.