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17 May 2021, 07:15
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Difficulty:
65%
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Question Stats:
55% (02:59) correct
45%
(02:48)
wrong
based on 33
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E is the midpoint of AC in right triangle ABC shown above. If the area of ΔABC is 24, what is the area of ΔBED?
A. \(9\sqrt{3}\)
B. \(9\sqrt{2}\)
C. 12
D. 9
E. \(6\sqrt{\frac{2}{3}}\)
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17 May 2021, 07:46
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Bunuel
AB = AC = x
\(BC = \sqrt{2}x\)
AE = EC = x/2
ED = DC = a
\(\sqrt{2}a = x/2\)
\(\implies\) \(a = \frac{x}{2\sqrt{2}}\)
\(\frac{1}{2}x^2 = 24\)
\(x = 4\sqrt{3}\)
Area of ΔABC = \(\frac{1}{2}(\sqrt{2}x - \frac{x}{2\sqrt{2}})*\frac{x}{2\sqrt{2}}\)
= \(\frac{x^2}{4} - \frac{x^2}{16}\)
= 9
17 May 2021, 08:00
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Area of BED = Area of ABC - (Area of BAE + Area of EDC)
As ABC is a 45-45-90 special right triangle lines AB and AC are equal. Let those sides be x.
Solving for x using area of ABC: (x*x)/2 = 24
x^2 = 48
x = √48
Area of BAE: (x*x/2)/2
(√48 * √48/2)/2
(48/2)/2
12
To solve area of EDC the legs of the triangle are need. As √48/2 is the hypotenuse and as once again it is a 45-45-90 special triangle, we can solve for the legs using Pythagoras. Let each leg be y.
y^2+y^2 = 48/4
2y^2 = 12
y^2 = 6
y = √6
Area of EDC:
(√6*√6)/2
6/2
3
Plugging in values into Area of BED = Area of ABC - (Area of BAE + Area of EDC)
Area of BED = 24 - (12 + 3)
Area of BED = 24 - 15
Area of BED = 9
Answer D
As ABC is a 45-45-90 special right triangle lines AB and AC are equal. Let those sides be x.
Solving for x using area of ABC: (x*x)/2 = 24
x^2 = 48
x = √48
Area of BAE: (x*x/2)/2
(√48 * √48/2)/2
(48/2)/2
12
To solve area of EDC the legs of the triangle are need. As √48/2 is the hypotenuse and as once again it is a 45-45-90 special triangle, we can solve for the legs using Pythagoras. Let each leg be y.
y^2+y^2 = 48/4
2y^2 = 12
y^2 = 6
y = √6
Area of EDC:
(√6*√6)/2
6/2
3
Plugging in values into Area of BED = Area of ABC - (Area of BAE + Area of EDC)
Area of BED = 24 - (12 + 3)
Area of BED = 24 - 15
Area of BED = 9
Answer D
