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In the right triangle above, what is the length of AC?
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20 Sep 2018, 16:55
20 Sep 2018, 16:55
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In the right triangle above, what is the length of AC?
(A) 9
(B) 10
(C) 12
(D) 13
(E) 15
Re: In the right triangle above, what is the length of AC?
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25 Sep 2018, 20:02
25 Sep 2018, 20:02
2
use the pythagoras theorem to find the value of x
\((x-1)^2 + (x+2)^2 = (x+5)^2\)
\(x^2 + 1 -2x + x^2 + 4 + 4x = x^2 +25 + 10x\)
simplify to get,
\(x^2 -8x -20 = 0\)
factorize to get
\((x+2)(x-10) = 0\)
since x can only be +ve x = 10
for this only option c becomes valid
\((x-1)^2 + (x+2)^2 = (x+5)^2\)
\(x^2 + 1 -2x + x^2 + 4 + 4x = x^2 +25 + 10x\)
simplify to get,
\(x^2 -8x -20 = 0\)
factorize to get
\((x+2)(x-10) = 0\)
since x can only be +ve x = 10
for this only option c becomes valid
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Re: In the right triangle above, what is the length of AC?
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28 Apr 2021, 05:51
28 Apr 2021, 05:51
sandy wrote:
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Capture.JPG
In the right triangle above, what is the length of AC?
(A) 9
(B) 10
(C) 12
(D) 13
(E) 15
Since we have a right triangle, we can apply the Pythagorean Theorem
When we do so we get: (x - 1)² + (x + 2)² = (x + 5)²
Expand each square to get: (x² - 2x + 1) + (x² + 4x + 4) = x² + 10x + 25
Simplify left side to get: 2x² + 2x + 5 = x² + 10x + 25
Subtract x² from both sides to get: x² + 2x + 5 = 10x + 25
Subtract 10x from both sides to get: x² - 8x + 5 = 25
Subtract 25 from both sides to get: x² - 8x - 20 = 0
Factor: (x - 10)(x + 2) = 0
So, EITHER x = 10 OR x = -2
We can rule out the solution x = -2, since that would make some of the lengths NEGATIVE
So, it MUST be the case that x = 10
AC has length x + 2
So, AC = 10 + 2 = 12
Answer: C
Cheers,
Brent
