Find the area of a quadrilateral ABCD in which AB = 8 cm, BC = 6 cm, CD = 8 cm, DA = 10 cm and AC = 10 cm.
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CDA is an isosceles triangle since AD = AC = 10. If X is the midpoint of CD, then CX = 4 cm and AX = √(AC^2 - CX^2) = V(100 - 16) = √84 = 2V21 . Hence the area of the triangle CDA is 4 x 2√21 = 8√21 cm^2.
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