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Senior PS Moderator V
Joined: 26 Feb 2016
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What is the largest possible area of a right triangle inscribed in a  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 56% (01:37) correct 44% (01:41) wrong based on 32 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

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Senior SC Moderator V
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What is the largest possible area of a right triangle inscribed in a  [#permalink]

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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 [ 11 KiB | Viewed 2139 times ]

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

In that case, area = $$r^2 = 10^2 = 100$$

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 =$$\frac{b*h}{2} =\frac{2r *r}{2}=\frac{2r^2}{2}= r^2$$

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GMAT 1: 560 Q38 V29 What is the largest possible area of a right triangle inscribed in a  [#permalink]

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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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Re: What is the largest possible area of a right triangle inscribed in a  [#permalink]

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_________________ Re: What is the largest possible area of a right triangle inscribed in a   [#permalink] 06 Feb 2019, 08:59
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