Question

(3x
1) CII. 11 ILS area 15
Сп, ппа со репппсіс.
13. Each of the equal sides of an isosceles triangle is 2 cm more than its height and the base
of the triangle is 12 cm. Find the area of the triangle.
triangle, each of whose equal sides is 13 cm and base 24 cm.​

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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