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Savatey [412]
2 years ago
10

Ramon wants to make an acute triangle with three pieces of wood. So far, he has cut wood lengths of 7 inches and 3 inches. He st

ill needs to cut the longest side. What length must the longest side be in order for the triangle to be acute? exactly StartRoot 58 EndRoot inches greater than StartRoot 58 EndRoot inches but less than 10 inches less than StartRoot 58 EndRoot inches but greater than 7 inches not enough information given
Mathematics
2 answers:
MrMuchimi2 years ago
7 0

Answer:     less than StartRoot 58 EndRoot inches but greater than 7 inches

Step-by-step explanation

o-na [289]2 years ago
5 0

Answer: Exactly square root 58 inches

Step-by-step explanation: The dimensions given for the right angled triangle are 7 inches and 3 inches respectively. The third side is yet unknown. However what we know is that a right angled triangle can be solved by using the Pythagoras theorem which states that,

AC^2 = AB^2 + BC^2

Where AC is the longest side. The question requires us to calculate the longest side and with the other two sides already known, the Pythagoras theorem now becomes,

AC^2 = 7^2 + 3^2

AC^2 = 49 + 9

AC^2 = 58

Add the square root sign to both sides of the equation

AC = square root 58 inches

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Answer:

The top of the ladder is now at 10 ft.

Step-by-step explanation:

At the start, we have a height H=6, a length L=10 and a base B, that has to be calculated by the Pythagorean theorem:

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The base is moved twice the distance the height moves up.

We called this distance x, so we have:

L^2=(H+x)^2+(B-2x)^2=H^2+2Hx+x^2+B^2-4Bx+4x^2\\\\L^2=(H^2+B^2)+5x^2+(2H-4B)x\\\\L^2=L^2+5x^2+(2H-4B)x\\\\0=5x^2+(2H-4B)x\\\\5x+(2H-4B)=0\\\\x=\dfrac{4B-2H}{5}=\dfrac{4*8-2*6}{5}=\dfrac{32-12}{5}=\dfrac{20}{5}=4

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The base travels 2x=8, so the new base B' is 0.

This means that the ladder is all against the wall (L=H').

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