Question : $ABC$ is a triangle. $AB = 5$ cm, $AC = \sqrt{41}$ cm and $BC = 8$ cm. $AD$ is perpendicular to $BC$. What is the area (in cm2) of $\triangle ABD$?
Option 1: 12
Option 2: 6
Option 3: 10
Option 4: 20
Correct Answer: 6
Solution :
Let the length of $BD=x$ cm, then $CD= (8-x)$ cm.
In $\triangle ADB$
$AB^2=AD^2+BD^2$
⇒ $5^2=AD^2+x^2$
⇒ $AD^2=5^2-x^2$ ... (i)
In $\triangle ADC$
$AC^2=AD^2+DC^2$
⇒ $(\sqrt{41})^2=AD^2+(8-x)^2$
⇒ $AD^2=41-(8-x)^2$ ... (ii)
From (i) and (ii)
$5^2-x^2=41-(8-x)^2$
⇒ $25=-23+16x$
⇒ $16x=48$
⇒ $x=3$
From equation (i),
$ AD^2=5^2-3^2$
⇒ $AD=4$ cm
Area of $\triangle ABD =\frac{1}{2}×AD×BD$
⇒ Area of $\triangle ABD =\frac{1}{2}×4×3$
Area of $\triangle ABD = 6$ cm
2
Hence, the correct answer is 6.
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