Question

The isosceles triangle below has height \( A Q \) of length 9 and base \( B C \) of length 4. A point \( P \) may be placed anywhere along the line segment \( A Q \). Let \( x \) be the length of QP . (a) What is the expression for the sum of the lengths of \( A P, B P \), and \( C P \) in terms of \( x \) ? (b) What is the minimum value of the sum of the lengths of \( A P, B P \), and \( C P \) ? The isosceles triangle below has height AQ of length 9 and base BC of length 14 . A point P may be placed anywhere along the line segment AQ . Let \( x \) be the length of QP . (a) What is the expression for the sum of the lengths of AP, BP, and CP in terms of \( x \) ? (b) What is the minimum value of the sum of the lengths of \( A P, B P \), and \( C P \) ?

          The isosceles triangle below has height \( A Q \) of length 9 and base \( B C \) of length 4. A point \( P \) may be placed anywhere along the line segment \( A Q \). Let \( x \) be the length of QP . (a) What is the expression for the sum of the lengths of \( A P, B P \), and \( C P \) in terms of \( x \) ? (b) What is the minimum value of the sum of the lengths of \( A P, B P \), and \( C P \) ? The isosceles triangle below has height AQ of length 9 and base BC of length 14 . A point P may be placed anywhere along the line segment AQ . Let \( x \) be the length of QP .
(a) What is the expression for the sum of the lengths of AP, BP, and CP in terms of \( x \) ?
(b) What is the minimum value of the sum of the lengths of \( A P, B P \), and \( C P \) ?

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the isosceles triangle below has height a q of length 9 and base b c of length 4 a point p may be placed anywhere along the line segment a q let x be the length of qp a what is the expressio 05363

Added by Holly M.

Geometry A Common Core Curriculum

Ron Larson, Laurie Boswell 1st Edition

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Solved by Expert Gregory Higby

03/03/2024

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A point P may be placed anywhere along the line segment AQ. Let $x$ be the length of QP. P B C Q (a) What is the expression for the sum of the lengths of AP, BP, and CP in terms of $x$? (b) What is the minimum value of the sum of the lengths of AP, BP, and CP? Show more…

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The isosceles triangle below has height \( A Q \) of length 9 and base \( B C \) of length 4. A point \( P \) may be placed anywhere along the line segment \( A Q \). Let \( x \) be the length of QP . (a) What is the expression for the sum of the lengths of \( A P, B P \), and \( C P \) in terms of \( x \) ? (b) What is the minimum value of the sum of the lengths of \( A P, B P \), and \( C P \) ? The isosceles triangle below has height AQ of length 9 and base BC of length 14 . A point P may be placed anywhere along the line segment AQ . Let \( x \) be the length of QP . (a) What is the expression for the sum of the lengths of AP, BP, and CP in terms of \( x \) ? (b) What is the minimum value of the sum of the lengths of \( A P, B P \), and \( C P \) ?

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Transcript

00:01 All right, so what i would do in this problem is just identify that this has a height of 3.

00:06 Let me write that off to the side.

00:09 And then with all the way across being 2, that means each individual piece is 1 on either side.

00:16 Let me just put point p somewhere on here, and let me just identify what's going on.

00:22 It's called this x, which means this little piece up here needs to be 3 minus x.

00:31 And what did they want? so they wanted to connect b and ap and bp and cp.

00:42 So let's figure out, i forgot which one was which now.

00:47 So a was at the top, b was over here, and c was over here.

00:52 So i need to do the pythagorean theorem to figure out what pc is, which would be one squared plus plus x squared, and then you square root that.

01:00 And that's actually going to be the same on this side.

01:03 So then the sum, i'm going to use x as my variable, is we have two of those square roots of 1 plus x squared, but then we have to add to it that 3 minus x.

01:16 And i'm setting this up like this, so then we can take the derivative.

01:22 And remember, the square root's the 1 half power.

01:25 Let me write that right here.

01:26 Here.

01:28 And i multiply the exponent by the coefficient, which brings me to 1.

01:32 And now it's to the negative 1 half power, which puts it in the denominator, times the derivative of the inside, which is 2x.

01:39 Derivative of 3 is 0.

01:40 Derivative of negative x is negative 1...

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