In the figure above, if isosceles right triangle PQR has an area of 4,

Question

https://gmatclub.com/forum/download/file.php?id=39844 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 2018-02-05_0842.png

pushpitkc · Answer

https://gmatclub.com/forum/download/file.php?id=39844

As the isosceles right-angled triangle has an area of 4,  \frac{1}{2}*x^2 = 4  
From this we can calculate the side of the isosceles right-angled triangle  x = 2\sqrt{2}

The hypotenuse of this isosceles right-angled triangle is  \sqrt{2*(2\sqrt2)^2}  = 4

The circumradius of the isosceles triangle will be  \frac{1}{2} *Hypotenuse =  \frac{1}{2}*4 = 2

Therefore, the area of the shaded region is  \frac{1}{4}*\pi*2^2 = \pi  (Option A)

chetan2u · Answer

Hi...
If you get the question with same choices, you can answer it in 5 seconds..
Area of triangle is 4,   so the area of shaded portion has to be less than 4

Only choice is π, so A

Otherwise as pushpit has shown....
 area of isosceles triangle is equal to HALF of altitude*hypotenuse, where altitude is same as radius of circle..

generis · Answer

2018-02-05_0842ed.png 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 length Area 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 Area Circle 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

G-Man · Answer

Triangle AOQ forms an isosceles triangle
Then area 1/2*b*h = h^2 = 4
Implies, h = 2

Here h = radius of the circle

Hence area of sector = 90/360 * pi * r^2
= 1/4*pi*4
= pi

In the figure above, if isosceles right triangle PQR has an area of 4,

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In the figure above, if isosceles right triangle PQR has an area of 4,

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