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Question Number 185924 by Mingma last updated on 29/Jan/23

	Answered by HeferH last updated on 29/Jan/23

	Commented by HeferH last updated on 30/Jan/23

	$i . \frac{D E}{b} = \frac{a}{c} \Rightarrow D E = \frac{a b}{c}$ $\frac{C E}{l} = \frac{b}{c} \Rightarrow C E = \frac{b l}{c}$ $\frac{C D}{C E} = \frac{D B}{E B} \Rightarrow \frac{b}{\frac{b l}{c}} = \frac{a}{E B} \Rightarrow E B = \frac{a l}{c}$ $i i . C E + E B = C B \Rightarrow$ $\frac{b l}{c} + \frac{a l}{c} = l$ $a + b = c$ $a^{2} + b^{2} + 2 a b = c^{2}$ $i i i . C E \cdot E B = A E \cdot E D \Rightarrow \frac{{a b l}^{2}}{c^{2}} = \frac{(c^{2} - a b) a b}{c^{2}}$ $c^{2} - l^{2} = a b$ $i v . a^{2} + b^{2} + 2 a b = c^{2}$ $a^{2} + b^{2} + 2 (c^{2} - l^{2}) = c^{2}$ $a^{2} + b^{2} + 2 c^{2} - 2 l^{2} = c^{2}$ $l^{2} = \frac{a^{2} + b^{2} + c^{2}}{2}$
	Answered by mr W last updated on 30/Jan/23

	$l^{2} = a^{2} + b^{2} + a b . . . (i)$ $l^{2} = b^{2} + c^{2} - b c . . . (i i)$ $l^{2} = a^{2} + c^{2} - a c . . . (i i i)$ $(i i) - (i i i) :$ $(a - b) c = (a - b) (a + b)$ $\Rightarrow c = a + b$ $2 \times (i) :$ $2 l^{2} = a^{2} + b^{2} + a^{2} + b^{2} + 2 a b = a^{2} + b^{2} + {(a + b)}^{2}$ $2 l^{2} = a^{2} + b^{2} + c^{2}$ $\Rightarrow l^{2} = \frac{a^{2} + b^{2} + c^{2}}{2}$
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