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Math Expert V
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Difficulty:   25% (medium)

Question Stats: 78% (01:29) correct 22% (01:39) wrong based on 49 sessions

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Competition Mode Question

What is the greatest possible area of a triangular region with one side that corresponds with the diameter of a circle with radius 6, and the other vertex of the triangle on the circle?

(A) 24
(B) 36
(C) 40
(D) 48
(E) 72

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What is the greatest possible area of a triangular region with one sid  [#permalink]

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One side of a triangle has length of 12 (diameter of a circle). This side will be the base.

The maximum height of the triangle is 6 (radius of a circle)

Maximum area=0.5*12*6=36

Originally posted by chondro48 on 17 Jan 2020, 01:47.
Last edited by chondro48 on 22 Jan 2020, 08:34, edited 4 times in total.
##### General Discussion
Director  D
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Re: What is the greatest possible area of a triangular region with one sid  [#permalink]

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What is the greatest possible area of a triangular region with one side that corresponds with the diameter of a circle with radius 6, and the other vertex of the triangle on the circle?

(A) 24
(B) 36
(C) 40
(D) 48
(E) 72

Isosceles triangle will have max area in the circle, with radius = 6 and diameter = 12,
So, 45-45-90 - x:x:x√2, here x√2 = 12, so x= 6√2.
Area of the triangle = 1/2*6√2*6√2 = 36

Ans. B
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Re: What is the greatest possible area of a triangular region with one sid  [#permalink]

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largest possible ∆ ;
two isoscles ∆ ; area ; 1/2 * 6*6 * 2 ; 36
IMO B

What is the greatest possible area of a triangular region with one side that corresponds with the diameter of a circle with radius 6, and the other vertex of the triangle on the circle?

(A) 24
(B) 36
(C) 40
(D) 48
(E) 72
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Posts: 1343
Location: United States
Re: What is the greatest possible area of a triangular region with one sid  [#permalink]

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Quote:
What is the greatest possible area of a triangular region with one side that corresponds with the diameter of a circle with radius 6, and the other vertex of the triangle on the circle?

(A) 24
(B) 36
(C) 40
(D) 48
(E) 72

Triangle with a side as diameter and another vertex on circle, is a right-angle triangle;
Base is the diameter =2r = 12
Area = 12*6/2 = 36

Ans (B)
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Re: What is the greatest possible area of a triangular region with one sid  [#permalink]

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What is the greatest possible area of a triangular region with one side that corresponds with the diameter of a circle with radius 6, and the other vertex of the triangle on the circle?

Area= 1/2 base•height
base = diameter of the circle
.: Base = 2(6) = 12
.: Area = 1/2(12•6)= 36

Hit that B

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Re: What is the greatest possible area of a triangular region with one sid  [#permalink]

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Approach:

- It will be a right angled triangle, Hypotenuse given as(6+6) 12
- For greatest possible area, right angled triangle should be isosceles.
- base and height be 'b' and 'h', and b = h
- equation: 2$$b^2$$ = 144 => $$b^2$$ = 72
- area: $$\frac{1}{2}*b*b = 36$$

IMO Option B!
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Re: What is the greatest possible area of a triangular region with one sid  [#permalink]

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Let the triangle have vertices A, B & C with base BC.
BC = diameter of the circle = 12

A is any point on the circle
—> Let the altitude from A to BC meet at point D.

Area of the triangle is maximum when the height AD is maximum.
—> Maximum value possible for AD = radius = 6 (When triangle ABC is an isosceles right triangle)

—> Maximum area = 1/2*BC*AD = 1/2*12*6 = 36

Option B

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To be the greatest possible area of a triangular region within a circle , the triangle must be isosceles.

Possible maximum area = ½ x base x height = ½ x 12 x 6 = 36(B)
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What is the greatest possible area of a triangular region with one side that corresponds with the diameter of a circle with radius 6, and the other vertex of the triangle on the circle?

(A) 24
(B) 36
(C) 40
(D) 48
(E) 72

The maximum area of triangle would be possible only if the it is an isosceles right angle triangle since the side corresponding to diameter is the largest.
Thus, as per Pythagoras theorem where other two sides are 'a'

$$Diameter^2 = a^2 + a^2$$
$$144 = 2a^2$$
a = 6√2

$$Area = \frac{1}{2} * 6√2 * 6√2$$ = 36

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Re: What is the greatest possible area of a triangular region with one sid  [#permalink]

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As the circle is inscribed in the circle with one side passing through the centre of the circle, the triangle is a right angled.
To maximise the area of the triangle, the other two sides of the triangle must be equal.
Hence hypotenuse is 12, as the radius is 6
the other two sides are equal each of length 6sqrt(2), by using 45-45-90 triangle.
therefore, area of the triangle is (1/2)*6sqrt(2)*6sqrt(2) = 36

B is correct
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Re: What is the greatest possible area of a triangular region with one sid  [#permalink]

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What is the greatest possible area of a triangular region with one side that corresponds with the diameter of a circle with radius 6, and the other vertex of the triangle on the circle?

(A) 24
(B) 36
(C) 40
(D) 48
(E) 72

d=2r
d=2(6)=12

the largest possible area is if the vertexes is on the edge of the circle.
the base of the triangle is 12 the height must be 6(the radius of the circle)
therefore the largest possible area is 1/2(12)(6)=36
B
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If one side of the triangle is diameter of a circle and the other vertex of the triangle is on the circle, that triangle is right-angled triangle.
--> In order the area of that right-angled triangle to be greatest, that triangle should be the isosceles right-angled triangle.

r=6 --> diameter= 12
--> $$a^{2}+ a^{2} = 144$$
$$2*a^{2} =144$$
$$a^{2}= 72$$
--> the area of the triangle =$$\frac{a^{2}}{2}= \frac{72}{2} = 36$$
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Re: What is the greatest possible area of a triangular region with one sid  [#permalink]

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Ans: B

area is = .5*12*6=36 Re: What is the greatest possible area of a triangular region with one sid   [#permalink] 19 Jan 2020, 20:52
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