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C
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Difficulty:
85%
(hard)
Question Stats:
54% (02:40) correct
46%
(03:01)
wrong
based on 130
sessions
History
Date
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If the perimeter of a right triangle is 20 units, what is the approximate area of the largest circle that can be inscribed in this triangle?
(A) \(\pi\)
(B) \(6.3 \pi\)
(C) \(4 \pi\)
(D) \(\frac{\pi}{2}\)
(E) \(\frac{3 \pi}{2}\)
(A) \(\pi\)
(B) \(6.3 \pi\)
(C) \(4 \pi\)
(D) \(\frac{\pi}{2}\)
(E) \(\frac{3 \pi}{2}\)
06 Sep 2014, 13:43
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The best strategy to solve this problem is to just dive in.
Largest circle inscribed within a right triangle means that this is a isosceles right triangle. For an isosceles right triangle there are two equal sides a. Using the Pythagoream theorem you can determine that the hypotenuse side:
\(a^2\) + \(a^2\) = \(c^2\)
Thus, c = \(\sqrt{2}a\)
With that in mind, the perimeter of the triangle is 20 units.
\(a + a + \sqrt{2}a\) = 20
a ~ 6
Now, look at the diagram shown below. For a right triangle with sides a, b, and hypotenuse c, the area is \(A = \frac{1}{2}ab\)
The inradius can be found by equating the area of the triangle ABC with sum of the areas of the triangles ACI, BCI and ABI.
The altitudes for these triangles serve as the inradii for the circle. Let's call this radius (shown in red), r. This gives us
\(\frac{1}{2}ab = \frac{1}{2}ra + \frac{1}{2}rb + \frac{1}{2}rc = \frac{1}{2}r(a + b + c)\)
Solving for r gives \(r = \frac{ab}{a + b + c}\)
In this case we already established above that \(a = b\) and \(c = \sqrt{2}a\) and that \(a = 6\)
Thus, using these values we establish that the radius of the circle, \(r = 1.8\)
Plugging this into the formula for area of a circle, \(A = pi (1.8)^2\)
This gives us approx. \(4pi\)
Thus, the answer is C
For more Right Triangle properties, check out: https://mathworld.wolfram.com/RightTriangle.html
Hope this helps! Good luck studying!
Largest circle inscribed within a right triangle means that this is a isosceles right triangle. For an isosceles right triangle there are two equal sides a. Using the Pythagoream theorem you can determine that the hypotenuse side:
\(a^2\) + \(a^2\) = \(c^2\)
Thus, c = \(\sqrt{2}a\)
With that in mind, the perimeter of the triangle is 20 units.
\(a + a + \sqrt{2}a\) = 20
a ~ 6
Now, look at the diagram shown below. For a right triangle with sides a, b, and hypotenuse c, the area is \(A = \frac{1}{2}ab\)
The inradius can be found by equating the area of the triangle ABC with sum of the areas of the triangles ACI, BCI and ABI.
The altitudes for these triangles serve as the inradii for the circle. Let's call this radius (shown in red), r. This gives us
\(\frac{1}{2}ab = \frac{1}{2}ra + \frac{1}{2}rb + \frac{1}{2}rc = \frac{1}{2}r(a + b + c)\)
Solving for r gives \(r = \frac{ab}{a + b + c}\)
In this case we already established above that \(a = b\) and \(c = \sqrt{2}a\) and that \(a = 6\)
Thus, using these values we establish that the radius of the circle, \(r = 1.8\)
Plugging this into the formula for area of a circle, \(A = pi (1.8)^2\)
This gives us approx. \(4pi\)
Thus, the answer is C
For more Right Triangle properties, check out: https://mathworld.wolfram.com/RightTriangle.html
Hope this helps! Good luck studying!
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General Discussion
m3equals333
GMAT 3: 710 Q48 V40
Posts: 141
07 Feb 2014, 22:52
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Largest circle inscribed inside a right triangle is when the circle's diameter is maximized...this is a 45/45/90 right triangle.
With perimeter given as 20
a + a + (a)sqrt2 = 20
a(2 + sqrt2) = 20
a=20/(3.41) ~ 6
Radius of an inscribed circle in a right triangle is the sum of the legs - the hypotenuse divided by 2.
12 - 6(1.41) / 2
=6-3(1.41)
=3(2-1.41)
~1.8
A=pi(1.8)^2
~4pi
C
With perimeter given as 20
a + a + (a)sqrt2 = 20
a(2 + sqrt2) = 20
a=20/(3.41) ~ 6
Radius of an inscribed circle in a right triangle is the sum of the legs - the hypotenuse divided by 2.
12 - 6(1.41) / 2
=6-3(1.41)
=3(2-1.41)
~1.8
A=pi(1.8)^2
~4pi
C
