kruttikaaggarwal wrote:
If the coordinates of point B are (-3, -4) and the coordinates of point C are (-7, -7), what is the area of the parallelogram?
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A. 1
B. 2√7
C. 7
D. 8
E. 7√2
.
Here,
Let join the sides from Point C to point G on Y axis and point F on X axis.
The distance from C to G = distance from C to F = 7
The Area of the Square AFCG = \(7^2\) = \(49\)
Point D will be the mirror image of point B and co ordinates of Point D = -4,3
Now let consider the \(\triangle FDA\), \(\triangle FDC\), \(\triangle CGB\) and \(\triangle BGA\)
Area of \(\triangle FDA\) =\(\frac{1}{2}* 7 * 3 = \frac{21}{2}\)
Area of \(\triangle FDC\) = \(\frac{1}{2}* 7 * 3 = \frac{21}{2}\)
Area of \(\triangle CGB\) = \(\frac{1}{2} * 7 * 3 = \frac{21}{2}\)
Area of \(\triangle BGA\) = \(\frac{1}{2} * 7 * 3 = \frac{21}{2}\)
Sum of all the \(\triangle\) = \(\frac{21}{2} * 4 = 42\)
Therefore the area of parallelogram = Area of the Square - Area of the sum of All triangles = \(49 - 42 = 7\)
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