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In a rhombus ABCD, AB=18 and AC=28. Find the area of the rhombus to the nearest tenth.

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In a rhombus ABCD, AB=18 and AC=28. Find the area of the rhombus to the nearest tenth.

asked Feb 24, 2014 in GEOMETRY by dkinz Apprentice

1 Answer

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In a rhombus ABCD , AC  = 28 , AB  = 18.

Diagonals of rambus AC  and BD .

Label the point of intersection of diagonals is X.

The diagonals of a rhombus are perpendicular bisectors of each other.

AX  = XC  = 14. triangle ABX  is a right triangle with hypotenuse 18.

Draw the diagram of the rhombus.

From the pythagoras theorem AB2 = AX 2 + BX 2

182 = 142 + BX 2

324 = 196 + BX 2

324 - 196 = BX 2

128 = BX 2

BX  = 11.31

Area of ABX  triangle = 1/2(11.31*14) = (158.34)/2 = 79.17

The four triangles formed by construction of the diagonals are all congruent since the diagonals are perpendicular bisectors.

Total area of the rhombus is 4 times of ABX  triangle.

Area = 4* 79.17 = 316.68 squre units.

Area of the rhambus = 316.7 square units.

answered Apr 4, 2014 by david Expert

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