Math

Question

A brace for a shelf has the shape of a right triangle. Its hypotenuse is 16 inches long and the two legs are equal in length. How long are the legs of the
triangle?

MathHelper · Answer

c^2 = 16^2 = 256
c^2 = a^2 + b^2
a = b
 
256 = a^2 + a^2 = 2*a^2
a^2 = 128
a = sqrt(128) = sqrt(64*2) = 8*sqrt(2)
so,
The legs are each 8*sqrt(2) inches.

sokotiko · Answer

exusia · Answer

C^2 = 16^2 = 256
C^2 = a^2 + b^2
a = b
 
256 = a^2 + a^2 = 2*a^2
A^2 = 128
a = sqrt(128) = sqrt(16*8) = 4*sqrt(8)
So,
 
The legs are each 4*square root (8) inches.

Ganesha · Answer

If a, b are the lengths of the two legs of a right angled triangle and c is the length of its hypotenue, then the relation between the legs and the hupotenuse is given by the equation:
a^2 + b^2 = c^2
Here, given a = b each equal to x  say, and c = 16 inches. Then substituting these values in the equation above, we get:
x^2 + x^2 = 16^2
2x^2 = 256
x^2 = 256/2 = 128
x = 128^(1/2)
x = 11.31
Therefore, the legs of the right angle shellf are 11.31 each.

JohnLance · Answer

Use the Pythagorean Theorem. c^2 = 10^2 = 100 c^2 = a^2 + b^2 a = b
100 = a^2 + a^2 = 2*a^2 a^2 = 50 a = sqrt(50) = sqrt(25*2) = 5*sqrt(2)
 So:
 The legs are each 5*sqrt(2) inches.

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A brace for a shelf has the shape of a right triangle. Its hypotenuse is 16 inches long and the two legs are equal in length. How long are the legs of the

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