The total area of three walls, each 8' feet high; 14, long, and other is 10' long?
1 answer:
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In geometry, it would be always helpful to draw a diagram to illustrate the given problem.
This will also help to identify solutions, or discover missing information.
A figure is drawn for right triangle ABC, right-angled at B.
The altitude is drawn from the right-angled vertex B to the hypotenuse AC, dividing AC into two segments of length x and 4x.
We will be using the first two of the three metric relations of right triangles.
(1) BC^2=CD*CA (similarly, AB^2=AD*AC)
(2) BD^2=CD*DA
(3) CB*BA = BD*AC
Part (A)
From relation (2), we know that
BD^2=CD*DA
substitute values
8^2=x*(4x) => 4x^2=64, x^2=16, x=4
so CD=4, DA=4*4=16 (and AC=16+4=20)
Part (B)
Using relation (1)
AB^2=AD*AC
again, substitute values
AB^2=16*20=320=8^2*5
=>
AB
=sqrt(8^2*5)
=8sqrt(5)
=17.89 (approximately)
2A on 2A + 3B + 3C = 8 pieces
The probability: