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Dmitriy789 [7]
3 years ago
12

What is the area of the given trapezoid ABCD?   A. 28 in2   B. 56 in2   C. 110 in2   D. 112 in2

Mathematics
2 answers:
valina [46]3 years ago
8 0
B. 56

because ((6 + 22) x 4) / 2
USPshnik [31]3 years ago
5 0

Answer:  The correct option is (B) 56 in².

Step-by-step explanation:  We are given to find the area of the trapezoid ABCD shown in the figure.

We know that

AREA of a trapezoid with length of the parallel sides a units, b units and height h units is given by

A=\dfrac{1}{2}(a+b)h.

For the given trapezoid ABCD, we have

a = AB = 6 in,  b = CD = 22 in   and    height, h = 4 in.

Therefore, the AREA of trapezoid ABCD is given by

A\\\\\\=\dfrac{1}{2}(a+b)h\\\\\\=\dfrac{1}{2}(6+22)\times4\\\\=2\times28\\\\=56~\textup{in}^2.

Thus, the required area of the trapezoid ABCD is 56 in².

Option (B) is CORRECT.

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vodomira [7]

Answer:

\boxed{x = 7, y = 9, z = 68}

Step-by-step explanation:

We must develop three equations in three unknowns.

I will use these three:

\begin{array}{lrcll}(1) & 8x + 13y +7 & = & 180 & \\(2)& 9x - 7 + 13y +7 & = & 180 & \\(3)& 8x + 5y - 11 + z & = & 180 &\text{We can rearrange these to get:}\\(4)& 8x + 13y & = & 173 &\\(5) & 9x + 13y & = & 180 & \\(6)& 8x + 5y + z & = & 169 & \\(7)& x & = & \mathbf{7} & \text{Subtracted (4) from (5)} \\\end{array}

\begin{array}{lrcll}& 8(7) + 13y & = & 173 & \text{Substituted (7) into (4)} \\& 56 + 13y & = & 173 & \text{Simplified} \\& 13y & = & 117 & \text{Subtracted 56 from each side} \\(8)& y & =& \mathbf{9}&\text{Divided each side by 13}\\& 8(7) + 5(9) + z & = & 169 & \text{Substituted (8) and (7) into (6)} \\& 56 + 45 + z& = & 169 & \text{Simplified} \\& 101 + z& = & 169 & \text{Simplified} \\&z& = & \mathbf{68} & \text{Subtracted 101 from each side}\\\end{array}

\boxed{\mathbf{ x = 7, y = 9, z = 68}}

4 0
3 years ago
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