Question

2. The area of the similas Triangles are
81 cm and 49 cm, respectively. of the altitude
of the first triangle is 5.6 cm, Find the
of
the second triangle.
median​

Answer 1

Step-by-step explanation:

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$\red{\bold{\underline{\underline{QUESTION:-}}}}$

Q:-solve and verify the equation

$\frac{1}{3} x - 4 = x - ( \frac{1}{2} + \frac{x}{ 3} )$

$\huge\tt\underline\blue{Answer }$

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$⟹</p><p> \frac{1}{3} x - 4 = x - ( \frac{1}{2} + \frac{x}{3} )$

$⟹</p><p> \frac{x}{3} - 4 = x - ( \frac{3 + 2x}{6} )$

$⟹</p><p> \frac{x - 12}{3} = x - ( \frac{2x + 3}{6} )$

$⟹</p><p> \frac{x - 12}{3} = x - \frac{2x - 3}{6}$

$⟹</p><p> \frac{x - 12}{3} = \frac{6x - 2x - 3}{6}$

$⟹</p><p> \frac{x - 12}{3} = \frac{4x - 3}{6}$

cancelling 6( R.H.S) By 3 From L.H.S

$⟹ \frac{x - 12}{1} = \frac{4x - 3}{2} </p><p>$

$⟹</p><p>2(x - 12) = 4x - 3$

$⟹</p><p>2x - 24 = 4x - 3$

$⟹</p><p> - 24 + 3 = 4x - 2x$

$⟹</p><p> - 21 = 2x$

$⟹</p><p>x = - \frac{21}{2}$

CHECK:-

$⟹ \frac{ - \frac{21}{2} }{3} - 4 = - \frac{21}{2} - ( \frac{1}{2} + ( - ) \frac{ \frac{21}{2} }{3} )</p><p>$

$⟹</p><p> - \frac{21}{6} - 4 = - \frac{21}{2} - ( \frac{1}{2} - \frac{21}{6} )$

$⟹</p><p> - \frac{7}{2} - 4 = - \frac{21}{2} - ( \frac{1}{2} - \frac{7}{2} )$

$⟹</p><p> \frac{ - 7 - 8}{2} = - \frac{21}{2} - ( - \frac{6}{2} )$

$⟹ - \frac{15}{2} = - \frac{21}{2} - ( - 3) </p><p>$

$⟹</p><p> - \frac{15}{2} = - \frac{21}{2} + 3$

$⟹</p><p> - \frac{15}{2} = \frac{ - 21 + 6}{2} = - \frac{15}{2}$

THEREFORE,L.H.S=R.H.S

VERIFIED✔️

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HOPE IT HELPS YOU..

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