Question

⭐In an isosceles triangle ABC of area 42 cm^2,AB=AC.if the base BC =(4x+5)cm and the altitude AD=X cm then find the equal sides of the triangle, correct to two decimal places. ⭐

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CLASS 10 CHAPTER :QUADRATIC EQUATIONS

BEST ANSWER WILL BE MARKED AS BRAINLIEST!

Answer 1

Formulas to be applied : -

$\boxed{\mathsf{Area\:of\:triangle=\dfrac{height\times base}{2}}}$

$\boxed{\mathsf{Heron's\:formula= \sqrt{(s-a)(s-b)(s-c)}}}$

where s is the semi - perimeter and a , b and c are the sides.

    Solution  : -

Given,

Length of base BC = ( 4x + 5 ) cm

Length of altitude( height ) AD = x cm

Area of the triangle = 42 cm^2

   On the basis of the formula given above

⇒ 42 = [ ( 4x + 5 )x ] / 2

⇒ 84 = 4x^2 + 5x

⇒ 4x^2 + 5x - 84 = 0

⇒ 4x^2 + ( 21 - 16 )x  - 84 = 0

⇒ 4x^2 - 16x + 21x - 84 = 0

⇒ 4x( x - 4 ) + 21( x - 4 ) = 0

⇒ ( x - 4 )( 4x + 21 ) = 0

∴ x = 4   OR   x =  - 21 / 4

Sides can't be negative, x = 4

Now,

Base ( BC ) = 4x + 5

                   = 4( 4 ) + 5

                   = 16 + 5

                   = 21 cm

Let the length of AB ( or AC ) be a cm,

So, semi - perimeter = ( 21 + a + a ) / 2

                                  = ( 21 + 2a ) / 2

$\implies 42 = \sqrt{\bigg(\dfrac{21+2a}{2} - a\bigg) \bigg(\dfrac{21+2a}{2} - a\bigg) \bigg(\dfrac{21+2a}{2}-21\bigg)}}\\\\\\\implies 42^2 = \bigg( \dfrac{21+ 2a-2a}{2} \bigg)\bigg( \dfrac{21+2a-2a}{2} \bigg)\bigg( \dfrac{21+2a-42}{2}\bigg)\\\\\\\implies 42 \times 42 = \dfrac{21}{2} \times \dfrac{21}{2} \times \dfrac{2a-21}{2}\\\\\\\implies \dfrac{42\times42\times2\times2\times2}{21\times21} = 2a-21\\\\\\\implies 32 = 2a-21\\\\\implies 32 + 21 = 2a \\\\\implies 53 / 2= a \\\\\implies 26.5 = a$

Therefore the length of equal sides is 26.5 cm

Answer 2

Answer:

11.24cm..............