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What is the largest possible area of a right triangle inscribed in a

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New post 13 Dec 2017, 10:12
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Difficulty:

  45% (medium)

Question Stats:

56% (01:09) correct 44% (01:07) wrong based on 25 sessions

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What is the largest possible area of a right triangle inscribed in a circle with radius 10?

A. 50
B. 50\(\sqrt{3}\)
C. 96
D. 100
E. 100\(\sqrt{3}\)

Source: Experts Global
[Reveal] Spoiler: OA

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New post 13 Dec 2017, 14:04
pushpitkc wrote:
What is the largest possible area of a right triangle inscribed in a circle with radius 10?

A. 50
B. 50\(\sqrt{3}\)
C. 96
D. 100
E. 100\(\sqrt{3}\)

Source: Experts Global

Attachment:
inscribed.png
inscribed.png [ 11 KiB | Viewed 231 times ]

Short answer

Largest possible area is an isosceles right triangle with diameter = hypotenuse

In that case, area = \(r^2 = 10^2 = 100\)

Answer D

Explanation

If a right triangle is inscribed in a circle, the hypotenuse is the circle's diameter

To maximize area, we need to maximize height
--- Height is "highest" when the altitude is perpendicular to the base
--- See diagram, right side: as vertex is farther from the center, triangle height decreases
--- Same base, smaller height? Smaller area

So at maximum height, when altitude is perpendicular to base, height = radius of circle (left side of diagram)

The right triangle is isosceles, in diagram, AB = BC:

AB and BC are hypotenuses of the two smaller triangles
Two smaller triangles are congruent by SAS
-- two sides and their included angles are equal
-- OA = OB = OC
-- Altitude creates two 90° included angles (∠ AOB and ∠COB)

For congruent triangles, corresponding sides are equal:
AB = BC, the legs of equal length of the larger right ∆ ABC

Two ways to figure area:

1. Area \(=\frac{(b * h)}{2} =\frac{(20 * 10)}{2} = 100\)

2. Area of this inscribed right isosceles triangle= \(r^2 = 10^2 = 100\)

base = diameter = \(2r\)
height = radius = \(r\)
Area =\(\frac{b*h}{2} =\frac{2r *r}{2}=\frac{2r^2}{2}= r^2\)

Answer D
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New post 13 Dec 2017, 15:14
The diameter is the hipotenuse (20 = 2r) wich oppose to the 90° angle of course, hence the 2 other sides are equal:
180 - 90 = 90; 90/2 = 45°
and remember the isoceseles right triangle proportions
90° - Hipotenuse = Cathetus√2
45° - Cathetus
So Area of the triangle:
(BxH) /2 = ((20/√2) x (20/√2)) / 2 = 200 / 2 = 100

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What is the largest possible area of a right triangle inscribed in a   [#permalink] 13 Dec 2017, 15:14
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