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 if In a Δ ABD, C is the midpoint of BD &  AB=10, AD=12, AC=9

Step-by-step explanation:

Let say side BD = 2a

Then find Area of Δ ABD

s = (10 + 12 + 2a)/2 = 11 + a

Area = √(11 + a)(a + 1)(a - 1)(11 - a)

= √( 121 - a²)(a² - 1)

as AC is median

Area of ΔABC = Area of ΔACD = (1/2) Area of Δ ABD

find  Area of ΔABC

s = (10 + 9 + a)/2 = (19 + a)/2

Area of ΔABC  = √((19 + a)(a - 1)(a + 1)(19 - a) /(2 * 2 * 2 * 2))

= √(361 - a²)(a² - 1)/16

√(361 - a²)(a² - 1)/16 = (1/2) √( 121 - a²)(a² - 1)

Squaring both sides

=> (361 - a²)(a² - 1)/16 = (1/4) ( 121 - a²)(a² - 1)

=> 361 - a² = 4( 121 - a²)

=> 3a² = 123

=> a² = 41

=> a = √41

=> 2a = 2√41

BD = 2√41 = 12.8 cm

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Answer 2

The value of $BD$ is $12.8\ units$

Step-by-step explanation:

Given,

In $\triangle ABD$ where $C$ is the midpoint of $BD$,$AB=10,AD=12\ and\ AC=9$

Lets,

$BD=2a$

Area of $\triangle ABD=\sqrt{s(s-a)(s-b)(s-c)}$

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

⇒Area of $\triangle ABD=\sqrt{s(s-a)(s-b)(s-c)}$  

                            $=\sqrt{(11+a)(11+a-10)(11+a-12)(11+a-2a)}=\sqrt{(11+a)(1+a)(a-1)(11-a)}=\sqrt{(121-a^{2})(a^{2}-1)}$And,

Area of $\triangle ABC=\sqrt{s(s-a)(s-b)(s-c)}$

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

∴Area of $\triangle ABC=\frac{\sqrt{(361-a^{2})(a^{2}-1)} }{4}$

∴ $2\times$ Area of $\triangle ABC=$Area of $\triangle ABD$

⇒$2\times \frac{\sqrt{(361-a^{2})(a^{2}-1)} }{4}=\sqrt{(121-a^{2})(a^{2}-1) }$

By squaring both sides,

⇒$\frac{(361-a^{2})(a^{2}-1)}{4}=(121-a^{2})(a^{2}-1)$

⇒$361-a^{2} =484-4a^{2}$

⇒$3a^{2}=123$

⇒$a^{2}=41$

⇒$a=6.4$

∴ $2a=12.8$  

So, The value of $BD$ is $12.8\ units$