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brilliants [131]
2 years ago
6

The figure shows triangle PQR and line segment AB, which is parallel to QR:

Mathematics
2 answers:
Darya [45]2 years ago
7 0

Based on the triangle similarity theorem:

A. ΔPQR ~ ΔPAB

B. AB corresponds to  QR

C. ∠A corresponds to ∠Q.

<h3>What is the Triangle Similarity Theorem?</h3>

According to the triangle similarity theorem, when a segment that is parallel to one side of a triangle intersects the other two sides, the new triangle formed is therefore similar to the original triangle. This also implies that the two sides would be divided by the line segment proportionally.

Part A:

Based on the diagram given, we have:

AB is parallel to QR, therefore, the new triangle formed, triangle PAB is similar to the original triangle PQR, based on the triangle similarity theorem.

Part B:

Line segment AB corresponds to line segment QR because they lie in the same position facing angle P.

Part C:

Angle A lies in similar position with angle Q, so, angle A corresponds tp angle Q.

Learn more about the triangle similarity theorem on:

brainly.com/question/21247688

#SPJ1

Mashutka [201]2 years ago
5 0

Based on the triangle similarity theorem:

A. ΔPQR ~ ΔPAB

B. AB corresponds to  QR

C. ∠A corresponds to ∠Q.

<h3>What is the Triangle Similarity Theorem?</h3>

Triangle similarity theorem, when a segment that is parallel to one side of a triangle intersects the other two sides, the new triangle formed is therefore similar to the original triangle. This also implies that the two sides would be divided by the line segment proportionally.

Part A:

AB is parallel to QR, therefore, the new triangle formed, ΔPAB is similar to the original ΔPQR, based on the triangle similarity theorem.

Part B:

Line segment AB corresponds to line segment QR because they lie in the same position facing angle P.

Part C:

Angle A lies in similar position with angle Q, so, angle A corresponds to angle Q.

Learn more about triangle similarity theorem from:

brainly.com/question/21247688

#SPJ1

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The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati
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Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\

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{\displaystyle A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51 }

And if you rotate the second region around the "y" axis you get that

{\displaystyle A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188 }

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If you revolve just the outer curve you get

If you rotate the first  region around the x axis you get that

{\displaystyle A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708 }

And if you rotate the second region around the x axis you get that

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