Question

7. A wire is (7x-2) metres long. A length of (7+4x) metres is cut for use. Now, answer
the following questions:
(A) How much wire is left?
(B) If this left out wire is used for making an equilateral triangle, what is the
of each side of the triangle so formed?​

Answer 1

$Length \:of \:a \: wire = (7x-2) \: m\: --(1)$

$Length \:of \:the \: wire \:cut = (7+4x) \: --(2)$

$A) Length \:of \:the \:wire \:left \\= (1) - (2) \\= (7x-2) - (7+4x) \\= 7x - 2 - 7 - 4x \\= (7-4)x-9 \\=( 3x-9)\: m$

/* According to the problem given */

If this left out wire is used for making an equilateral triangle

$B) Let \: side \: of \:an \: equilateral \: triangle = a \: m$

$Perimeter \:of \: equilateral\: triangle = (3x-9)$

$\implies 3a = 3(x-3)$

$\implies a = \frac{3(x-3)}{3}$

$\implies a = (x-3)\: m$

Therefore.,

$\red{ Length \:of \:each \:side } \green {= (x-3)\: m}$

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