Question

A triangle has an area of 90 cm. find the length of the base if the base is 3 cm more than height​

Answer 1

Answer:

$let \: the \: height \: of \: the \: triangle \: be \: x \: and \: base \: be \:( x + 3) \\ according \: to \: given \: condition \\ area \: of \: triangle = \frac{1}{2} \times b \times h \\ 90 = \frac{1}{2} \times (x + 3) \times x \\ 90 = \frac{1}{2} \times {x}^{2} + 3x \\ 180 = {x}^{2} + 3x \\ {x}^{2} + 3x - 180 = 0 \\ x(x + 15) - 12( x + 15) = 0 \\ (x + 15)(x - 12) = 0 \\ so. \:( x + 15) = 0 \: or \: x - 12 = 0 \\ therefore</strong><strong>,</strong><strong> \: x = - 15 \: or \: x = 12 \\ but</strong><strong>,</strong><strong> \: base \: cannot \: be \: negative \: \\ so</strong><strong>,</strong><strong> \: - 15 \: is \: rejected \\ now </strong><strong>,</strong><strong>\:height = 12 \: cm \\ base = x + 3 \\ base = 12 + 3 \\ so</strong><strong>,</strong><strong> \: base = 15cm$

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