Question

an altitude of a triangle is five-thirds the length of its corresponding base if the altitude be increased by force and them and the best decreased by 2 cm the area of a triangle remain the same find the base and the altitude of the triangle

Answer 1

$let \: the \: base \: of \: the \: triangle \: be \: x \: cm$
$then \: the \: altitude \: of \: the \: triangle \: will \: be \: \frac{5}{3}x \: cm$
$therefore \: the \: area \: of \: the \: given \: triangle = \frac{1}{2} \times base \times altitude$
$\: \: \: \: \: \: \: \: \: \: = \frac{5}{6} {x}^{2}$





$we \: now \: have \: area \: of \: the \: new \: triangle = area \: of \: the \: given \: triangle \:$
$therefore \\ \\ area \: of \: the \: new \: triangle \: = \frac{1}{2} \times new \: base \times new \: altitude$
$\frac{5}{6} {x}^{2} = \frac{1}{2}(x - 2)( \frac{5}{3} x + 4)$
$\frac{5}{6} {x}^{2} = \frac{1}{2} x( \frac{5}{3} + 4) - 2( \frac{5}{3} x + 4)$
$\frac{5}{6} {x}^{2} = \frac{1}{2}( \frac{5}{3} {x}^{2} + 4x - \frac{10}{3}x - 8)$
$\frac{5}{6} {x}^{2} = \frac{1}{2} \frac{5}{3} {x}^{2} + \frac{12x - 10x}{3} - 8$
$\frac{5}{6} {x}^{2} = \frac{1}{2} ( \frac{5}{3} {x }^{2} + \frac{2}{3}x - 8)$
$\frac{5}{6} {x}^{2} = \frac{5}{6} {x}^{2} + \frac{1}{3} x - 4$
$0 = \frac{1}{3}x - 4 \: \: \: \: \: (like \: terms \: \frac{5}{6} {x}^{2} gets \: cncelled)$
$0 = \frac{1}{3}x = 4$
$3x \frac{1}{3} = 3 \times 4 \: \: \: \: \: (multiplying \: by \: 3 \: on \: both \: the \: sides)$

$altitude \: of \: the \: given \: tringle = \frac{5}{3} \times 12 = 20cm$
$area \: of \: the \: given \: triangle = \frac{1}{3} \times 12 \times 20 = 120 {cm}^{2}$
$area \: of \: the \: new \: triangle = \frac{1}{2} \times (12 - 2) \times (20 + 4)$
$= \frac{1}{2} \times 10 \times 24 = 120 {cm}^{2}$
Thus,the base of the triangle=12cm



Area of the new triangle=Area of the given triangle=
$120 {cm}^{2}$



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