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Evgesh-ka [11]
2 years ago
6

Which of the following pairs of triangles can be proven similar through SSS similarity?

Mathematics
1 answer:
Yakvenalex [24]2 years ago
5 0

Based on the SSS similarity theorem, the pair of triangles that can be proven to be similar is the pair shown in the image attached below.

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

The SSS similarity theorem states that two triangle area similar to each other if the ratio of the three corresponding sides of both triangles are equal.

Thus, in the image attached below, the ratio of the three corresponding sides of the pair of triangles are:

10/2.5 = 11/2.75 = 8/2 = 4

Therefore, the pair of triangles that we can prove to be similar using the SSS similarity theorem is the pair shown in the image attached below.

Learn more about the SSS similarity theorem on:

brainly.com/question/4163594

#SPJ1

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B: is the distance between the wall and the ladder (base) = 6 ft

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dB/dt = is the variation of the base of the triangle = 9 ft/s        

First, we need to find the other side of the triangle:  

H^{2} = B^{2} + L^{2}

L = \sqrt{H^{2} - B^{2}} = \sqrt{(10)^{2} - B^{2}} = \sqrt{100 - B^{2}}

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\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} + B\frac{d(\sqrt{100 - B^{2}})}{dt}]        

\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} - \frac{B^{2}}{(\sqrt{100 - B^{2}})}*\frac{dB}{dt}]  

\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - 6^{2}}*9 - \frac{6^{2}}{\sqrt{100 - 6^{2}}}*9]      

\frac{dA}{dt} = 15.75 ft^{2}/s  

     

Therefore, the area is changing at 15.75 square feet per second.

I hope it helps you!                                    

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