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

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

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

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

yes i skipped the part, great explanation

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