Question

In $\Delta ABC,$ AD is the median, find the length of AD if AC =7 cm, BC =8 cm and AB =9 cm.

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• A. $7\sqrt{5}cm$
• B. 7 cm
• C. $8\sqrt{5}cm$
• D. 8 cm

Answer 1

The correct option is B

7 cm

Construction: Draw $AE\perp DC$

Given: AB = 9 cm

BD =DC =4 cm [AD is median] AC =7 cm

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

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

Here, a=7 b=8 c=9, $s=\frac{7+8+9}{2}=12$

$\therefore$ Area $\left(\Delta ABC\right)=\sqrt{12(12-7)(12-8)(12-9)}$

$=\sqrt{12\times 5\times 4\times 3}$

$=\sqrt{12\times 12\times 5}$

$=12\sqrt{5}c{m}^2$

So, area $\left(\Delta ADB\right)$ =area $\left(\Delta ADC\right)=6\sqrt{5}c{m}^2$

[Since median of a triangle divides it in two equal parts].

Area $\left(\Delta ADC\right)=6\sqrt{5}c{m}^2$

$\frac{1}{2}\times DC\times AE=6\sqrt{5}$

$\frac{1}{2}\times 4\times AE=6\sqrt{5}$

$AE=3\sqrt{5}cm$

In $\Delta AEC$, using Pythagoras theorem,

$A{E}^2+E{C}^2=A{C}^2$

$\therefore E{C}^2=A{C}^2-A{E}^2$

$\Rightarrow EC=\sqrt{7^2-(3\sqrt{5}{)}^2}$

$=\sqrt{49-45}$

$=\sqrt{4}$

=2

or EC =2 cm.

DC =4 cm, EC =2 cm $\Rightarrow DE=2cm.$

This means, $\Delta ADC$ must be an isosceles triangle. Since, perpendicular AE bisects the base CD.

Hence AC =AD =7 cm.