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Math Expert V
Joined: 02 Sep 2009
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In the figure above, if isosceles right triangle PQR has an area of 4,  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 73% (01:52) correct 27% (02:35) wrong based on 48 sessions

### HideShow timer Statistics 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: 2018-02-05_0842.png [ 18.83 KiB | Viewed 2155 times ]

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Math Expert V
Joined: 02 Aug 2009
Posts: 8309
Re: In the figure above, if isosceles right triangle PQR has an area of 4,  [#permalink]

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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:
2018-02-05_0842.png

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..
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In the figure above, if isosceles right triangle PQR has an area of 4,  [#permalink]

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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 1614 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 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 $$x\sqrt{2}= (2\sqrt{2} * \sqrt{2}) = (2 * 2) = 4$$

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 = $$π$$

*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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Senior PS Moderator V
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In the figure above, if isosceles right triangle PQR has an area of 4,  [#permalink]

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2 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)
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Re: In the figure above, if isosceles right triangle PQR has an area of 4,  [#permalink]

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_________________ Re: In the figure above, if isosceles right triangle PQR has an area of 4,   [#permalink] 31 Mar 2019, 06:43
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