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by sana.noor » Tue Aug 13, 2013 8:46 pm
what is the area of quadrilateral with vertices (5,4), (8,8), (5,12) and (2,8)?

I got answer 25 but author says its 24

source: veritas coordinate geometry drill
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by vipulgoyal » Tue Aug 13, 2013 8:53 pm
certainly the ans is 25, by making right triangles with base and hight 3 and 4 we get the squre with side length 5, area = side^2 = 25
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by ganeshrkamath » Tue Aug 13, 2013 9:28 pm
sana.noor wrote:what is the area of quadrilateral with vertices (5,4), (8,8), (5,12) and (2,8)?

I got answer 25 but author says its 24

source: veritas coordinate geometry drill
A(5,4) B(8,8) C(5,12) D(2,8)
Area of triangle ACD = 1/2 * (12 - 4) * (8 - 5) = 12
Area of triangle ABC = 1/2 * (12 - 4) * (5 - 2) = 12
Total area = 24
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by ganeshrkamath » Tue Aug 13, 2013 9:30 pm
vipulgoyal wrote:certainly the ans is 25, by making right triangles with base and hight 3 and 4 we get the squre with side length 5, area = side^2 = 25
The quadrilateral doesn't form a square. If you notice, the diagonals are not equal in length.
One is 12 - 4 = 8 units and the other is 8 - 2 = 6 units in length.

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by vipulgoyal » Tue Aug 13, 2013 9:38 pm
yes i skipped the part, great explanation
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by Sul » Wed Aug 21, 2013 11:04 am
Hi

It is rhombus- diagnols bisect each other at right angles.
Here is the area of the rhombus : 1/2 d1*d2 (d1 is(8-2))(d2 is 12-4))
Area= 1/2*6*8=24
ganeshrkamath wrote:
vipulgoyal wrote:certainly the ans is 25, by making right triangles with base and hight 3 and 4 we get the squre with side length 5, area = side^2 = 25
The quadrilateral doesn't form a square. If you notice, the diagonals are not equal in length.
One is 12 - 4 = 8 units and the other is 8 - 2 = 6 units in length.

Cheers
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