I’m rohmbus ABCD, AB= 14 and AC= 26. Find the area of the rhombus to the nearest tenth.
1 answer:
The complete question in the attached figure
we know
In a rhombus ABCD<span> , </span>AC <span> = 26 , </span>AB<span> = 14
</span><span>the diagonals of a rhombus are perpendicular bisectors of each other
</span>AX<span> = </span>XC<span> = 13. triangle </span>ABX<span> is a right triangle with hypotenuse 14.
</span>
</span>
the answer is the option D) 135.1 units ²
we know
In a rhombus ABCD<span> , </span>AC <span> = 26 , </span>AB<span> = 14
</span><span>the diagonals of a rhombus are perpendicular bisectors of each other
</span>AX<span> = </span>XC<span> = 13. triangle </span>ABX<span> is a right triangle with hypotenuse 14.
</span>
From the pythagoras theorem AB² =<span> AX</span>² + BX²
14² = 13² + BX²
196 = 169 + BX²
27 = BX²----- > BX=√27
BX = 5.20
</span>
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
the answer is the option D) 135.1 units ²
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