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|First of all, the triangles are equal by ASA the way the diagram has been marked.
B and E are both right angles.
Side BC = Side DE
<BCA =< EDA
So triangle BCA = triangle EDA
Now to the letters.
x = y - 1 Add 1 to both sides.
x + 1 = y (1)
3x - 2 = 2y + 1 Subtract 1 from both sides.
3x -2 - 1 = 2y
3x - 3 = 2y Divide by 2
3x/2 - 3/2 = 2y/2
1.5x - 1.5 = y (2)
Step One
Since (1) and (2) both have y isolated on their respective right sides, they can be equated.
1.5x - 1.5 = x + 1 Take an x from both sides.
0.5x - 1.5 = x - x + 1
0.5x - 1.5 = 1 Add 1.5 to both sides.
0.5x = 1 + 1.5
0.5x = 2.5 Divide 0.5 on both sides.
0.5x/0.5 = 2.5/0.5
x = 5
Now we need a y value.
x = y - 1
5 = y - 1 Add 1 to both sides.
5 + 1 = y - 1 + 1
6 = y
So the 2 sides and the 2 angles are equal when
x = 5
y = 6
C Answer <<<<<<
The main point which will immediately make it clear is that you have to plug in 2 for x so that you can find the linit. So, using this simple method you will get the <span>-1/4 answer. Hope you will agree with me and find this useful! Regards.</span>
The top row of matrix A (1, 2, 1) is multiplied with the first column of matrix B (1,0,-1) and the result is 1x1 + 2x0 + 1x -1 = 0 this is row 1 column 1 of the resultant matrix
The top row of matrix A (1,2,1) is multiplied with the second column of matrix B (-1, -1, 1) and the result is 1 x-1 + 2 x -1 + 1 x 1 = -2 , this is row 1 column 2 of the resultant matrix
Repeat with the second row of matrix A (-1,-1.-2) x (1,0,-1) = 1 this is row 2 column 1 of the resultant matrix, multiply the second row of A (-1,-1,-2) x (-1,-1,1) = 0, this is row 2 column 2 of the resultant
Repeat with the third row of matrix A( -1,1,-2) x (1,0, -1) = 1, this is row 3 column 1 of the resultant
the third row of A (-1,1,-2) x( -1,-1,1) = -2, this is row 3 column 2 of the resultant matrix
Matrix AB ( 0,-2/1,0/1,-2)
The area of the trapezoid is 37.5 square inches.
This is the way to solve the problem
