For the right triangle shown, what is the tangent of angle A?

For the triangle problem, the triangle has to be a 3, 4, 5 triangle.

therefore, tangent A = opposite over adjacent = 3/4 = 0.75 

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First one.....this is a 3-4-5 right triangle

x = 3     x + 1  = 4      x + 2  = 5

tan A  = BC/AB  = x / [ x + 1 ]  =  3 / 4

Second one....this is a little tough

We have a semi-circle  with a radius of 3

So....its perimeter  =  pi * 3   units

Then, we have a four sided figure

Two of sides have a length of 3  =   6   units

Let the bottom vertex of the figure lie at (0,0)

Then the bottom left vertex lies at  (-5, 2)

And the length of the bottom edge  =  sqrt (5^2 + 2^2)  = sqrt(29) units

And the bottom right vertex lies at  (1,2)

So...the length of the bottom right edge  = sqrt( 1^2 + 2^2)  = sqrt(5) units

So....the total perimeter  is    [ 3pi + 6 + sqrt(29) + sqrt(5) ]  = 23.046 units

Last one

Easy 

area  =  base * height  = 14 * 10   = 140 in^2

1)  C

2)  E

3)  B

For a moment I felt that I'm in a 3rd grade again.