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zhuklara [117]
3 years ago
5

Which similarity statement is true?

Mathematics
1 answer:
Likurg_2 [28]3 years ago
5 0
In order for the triangles to be similar, the corresponding angles should be equal, so in the given, use the 90 degree angle as the reference so that you will not be confused. in the big triangle z is the 90 degree and in the two smaller triangle x is the 90 degrees, so the correct answer is the second choice
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Answer:

The area is changing at 15.75 square feet per second.

Step-by-step explanation:

The triangle between the wall, the ground, and the ladder has the following dimensions:

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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}}

Now, the area (A) of the triangle is:            

A = \frac{BL}{2}  

Hence, the rate of change of the area is given by:

\frac{dA}{dt} = \frac{1}{2}[L*\frac{dB}{dt} + B\frac{dL}{dt}]      

\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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