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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,  [#permalink]

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New post 04 Feb 2018, 21:45
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In the figure above, if isosceles right triangle PQR has an area of 4,  [#permalink]

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New post 05 Feb 2018, 01:16
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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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New post 05 Feb 2018, 01:55
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Bunuel wrote:
Image
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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New post 05 Feb 2018, 16:49
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Bunuel wrote:
Image
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
2018-02-05_0842ed.png [ 19.54 KiB | Viewed 1554 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 \(x\sqrt{2}= (2\sqrt{2} * \sqrt{2}) = (2 * 2) = 4\)

• 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\(^2\)
\((2\sqrt{2})^2 + (2\sqrt{2})^2 = (PQ)^2\)
\(8 + 8 = PQ^2\)
\(PQ = \sqrt{16}\)
\(PQ = 4\)

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

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New post 31 Mar 2019, 06:43
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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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