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Question 60
$$a, b, c$$ are three positive numbers, such that, $$(a + b + c) = 20, a^2 + b^2 + c^2 = 152$$. The value of $$(ab + bc + ca)$$ is equal to:
Solution
$$(a + b + c)^{2}=a^2 + b^2 + c^2+2(ab+bc+ac)$$
$$20^{2}=a^2 + b^2 + c^2+2(ab+bc+ac)$$
$$400=152+2(ab+bc+ac)$$
$$(ab+bc+ac)=\frac{248}{2}=124$$
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