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) \(8\pi\)

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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\(^2\)

\((2\sqrt{2})^2 + (2\sqrt{2})^2 = (PQ)^2\)

\(8 + 8 = PQ^2\)

\(PQ = \sqrt{16}\)

\(PQ = 4\)
_________________

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that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"