SOLUTION: Please help me solve:
In a rhombus ABCD, AB=10 and AC=16. Find the area of the rhombus to the nearest tenth.
Thanks !!
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Question 576508: Please help me solve:
In a rhombus ABCD, AB=10 and AC=16. Find the area of the rhombus to the nearest tenth.
Thanks !!
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Draw both diagonals, namely AC and BD and label the point of intersection X. The diagonals of a rhombus are perpendicular bisectors of each other, hence AX = XC = 8. Triangle ABX is then a right triangle with a hypotenuse of 10 and a long leg of 8 which is proportional to a 3-4-5 right triangle. 2 X 5 is 10 and 2 X 4 is 8, so the third side, namely BX must measure 2 X 3 = 6. The area of this triangle is 6 X 8 divided by 2 or 24. The four triangles formed by construction of the diagonals are all congruent since the diagonals are perpendicular bisectors (you can show SAS where the included angle is the right angle in each case) hence the total area of the rhombus is 4 times 24 or 96 square units.
John
My calculator said it, I believe it, that settles it
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