Question

In a triangle ABD, C is the midpoint of BD. If AB=10, AD=12, AC=9, find the value of BD?

Answer 1

BD = 2√41 = 12.8 cm

Step-by-step explanation:

Consider the provided information.

Let the length of side BD is 2x

According to herons formula, the area of Δ ABD is:

$s=\frac{a+b+c}{2}$

$A=\sqrt{s(s-a)(s-b)(s-c)}$

Where a, b and c are the sides.

Substitute a=10 b=12 and c=2x

$s=\frac{10+12+2x}{2}=11+x$

$A=\sqrt{(11+x)(11+x-10)(11+x-12)(11+x-2x)}$

$A=\sqrt{(11+x)(1+x)(x-1)(11-x)}$

$A=\sqrt{(121-x^2)(1-x^2)}$

It is given that AC is median

Area of ΔABC =(1/2) Area of Δ ABD

Similarly

Area of ΔABC:

$s=\frac{10+9+x}{2}=\frac{19+x}{2}$

$A=\sqrt{\frac{19+x}{2}(\frac{19+x}{2}-10)(\frac{19+x}{2}-9)(\frac{19+x}{2}-x)}=\frac{1}{2}\sqrt{(121-x^2)(1-x^2)}$

$A=\sqrt{\frac{(361-x^2)(x^2-1)}{16}}=\frac{1}{2}\sqrt{(121-x^2)(1-x^2)}$

$\frac{(361-x^2)(x^2-1)}{16}=\frac{1}{4}\sqrt{(121-x^2)(1-x^2)}$

$361-x^2=4(121-x^2)$

$x^2=41$

$x=\sqrt{41}$

$2x=2\sqrt{41}$

Hence, BD = 2√41 = 12.8 cm

#Learn more

Heron'S formula's formula

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