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What would be the measure of the diagonal of a square whose area is equal to 882 $$cm^{2}$$ ?
- A38 cm
- B32 cm
- C42 cm
- D48 cm
Answer not set
Discussion 1
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Solution
Let the side of the square be a
$a^2 = 882$
a = $\sqrt{882}$ =$21\sqrt{2}$
Diagonal =$\sqrt{2}a$ = $21\sqrt{2}\sqrt{2}$ = 42
Let the side of the square be a
$a^2 = 882$
a = $\sqrt{882}$ =$21\sqrt{2}$
Diagonal =$\sqrt{2}a$ = $21\sqrt{2}\sqrt{2}$ = 42