Looking for various approaches to this Algebra problem disguised as a Geometry question.
In right triangle ABC, BC is the hypotenuse. If BC is 13 and AB + AC = 15, what is the area of the triangle?
A. 2√7
B. 2√14
C. 14
D. 28
E. 56
Source: Veritas Prep
OA: C
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Area of a Triangle
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freyesinsb
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Let AB = x and AC = yIn right triangle ABC, BC is the hypotenuse. If BC is 13 and AB + AC = 15, what is the area of the triangle?
A. 2√7
B. 2√14
C. 14
D. 28
E. 56
Given AB + AC = 15 => x+y=15.......(1)
Hypotenuse = 13
Therefore x^2+y^2 = 169 .......(2)
Now solve eq 1
(x+y)^2=225
x^2 + y^2 + 2xy = 225
x^2 + y^2 = 225 - 2xy
We know from equation 2 that x^2+y^2 = 169
so 169 = 225 - 2xy
xy = 28
Area of the triangle 1/2 base * Height
1/2*28=14.....Answer c
Hope this helps
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(AB + AC)² = 15²freyesinsb wrote:Looking for various approaches to this Algebra problem disguised as a Geometry question.
In right triangle ABC, BC is the hypotenuse. If BC is 13 and AB + AC = 15, what is the area of the triangle?
A. 2√7
B. 2√14
C. 14
D. 28
E. 56
Source: Veritas Prep
OA: C
AB² + 2(AB)(AC) + AC² = 225.
AB² + AC² + 2(AB)(AC) = 225.
Since ∆ABC is a right triangle:
AB² + AC² = BC²
AB² + AC² = 13²
AB² + AC² = 169.
Substituting AB² + AC² = 169 into AB² + AC² + 2(AB)(AC) = 225, we get:
169 + 2(AB)(AC) = 225
(AB)(AC) = 28.
Since A = (1/2)(AB)(AC), we get:
A = (1/2)(28) = 14.
The correct answer is C.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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As a tutor, I don't simply teach you how I would approach problems.
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freyesinsb
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