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 GlobalAttachment:
inscribed.png [ 11 KiB | Viewed 12208 times ]
Short answerLargest possible area is an isosceles right triangle with diameter = hypotenuse
In that case, area =
\(r^2 = 10^2 = 100\)Answer D
ExplanationIf 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