In the figure above, if triangles \(ABC, ACD,\) and \(ADE\) are isosceles right triangles and the area of
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A. \(18\)
B. \(24\)
C. \(36\)
D. \(12\sqrt2\)
E. \(24\sqrt2\)
Answer: B
Source: Official Guide
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In general, isosceles right triangles have the following properties.Vincen wrote: ↑Sat Dec 04, 2021 7:34 am2019-04-26_1753.png
In the figure above, if triangles \(ABC, ACD,\) and \(ADE\) are isosceles right triangles and the area of \(\triangle ABC\) is \(6,\) then the area of \(\triangle ADE\) is
A. \(18\)
B. \(24\)
C. \(36\)
D. \(12\sqrt2\)
E. \(24\sqrt2\)
Answer: B
Source: Official Guide
Notice that the hypotenuse = (√2)(length of one leg)
So, let's label the sides of the BLUE triangle as follows:
This means each leg of the RED triangle has length (√2)(x)
The hypotenuse of an isosceles right triangle = (√2)(length of one leg) . . .
. . . so the length of the hypotenuse = (√2)(√2)(x) = 2x
This means each leg of the GREEN triangle has length 2x
What is the area of ΔADE?
Area of triangle = (base)(height)/2
So, area of ΔADE = (2x)(2x)/2 = 2x²
So, what is the value of 2x²?
GIVEN: the area of ΔABC is 6
ΔABC is the BLUE triangle we started with.
We can write: (x)(x)/2 = 6
Simplify: x²/2 = 6, which means x² = 12
This means the area of ΔADE = 2x² = 2(12) = 24
Answer: B
Cheers,
Brent