Find the area $ A $ of a rhombus if side $a = 4$ and diagonal $d_1 = \sqrt{ 6 }$.
Answer
$A = \sqrt{ 87 }$
Explanation
STEP 1: find diagonal $ d2 $
To find diagonal $ d2 $ use formula:
$$ d_1^2 + d_2^2 = 4 \cdot a^2 $$After substituting $ d_1 = \sqrt{ 6 } $ and $ a = 4 $ we have:
$$ \left( \sqrt{ 6 } \right)^{ 2 } + d_2^2 = 4 \cdot 4^2 $$ $$ 6 + d_2^2 = 64 $$ $$ d_2^2 = 64 - 6 $$ $$ d_2^2 = 58 $$ $$ d_2 = \sqrt{ 58 } $$STEP 2: find area $ A $
To find area $ A $ use formula:
$$ A = \dfrac{ d_1 \cdot d_2 }{ 2 } $$After substituting $ d_1 = \sqrt{ 6 } $ and $ d_2 = \sqrt{ 58 } $ we have:
$$ A = \dfrac{ \sqrt{ 6 } \cdot \sqrt{ 58 } }{ 2 }$$$$ A = \dfrac{ 2 \sqrt{ 87 } }{ 2 } $$$$ A = \sqrt{ 87 } $$This page was created using
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