Question

From the adjoining figure, find
i) the area of $△ABC$
ii) length of $BC$
iii) the length of altitude from $A$ to $BC$.

Answer 1

We know Area of triangle$=\frac{1}{2}\times$Base$\times$Height.

$i)$In $△ABC,$Base$=AB=3$cm and Height$=AC=4$cm

So, Area of $△ABC=\frac{1}{2}\times 3\times 4=6{cm}^2$

$ii)$We apply Pythagoras theorem in $△ABC$ and get

${BC}^2={AB}^2+{AC}^2,$ substitute all values we get

${BC}^2=3^2+4^2=16$

$BC=5$cm

$iii)$We apply Pythagoras theorem in $△ABD$ and get

${AB}^2={AD}^2+{BD}^2$

${AD}^2={AB}^2-{BD}^2$,substitute values we get

${AD}^2=3^2-{BD}^2=9-{BD}^2$ .......$\left(1\right)$

Apply Pythagoras theorem to $△ACD$ and get

${AC}^2={AD}^2+{CD}^2$,

${AD}^2={AC}^2-{CD}^2$,

${AD}^2={AC}^2-{(BC-BD)}^2$

${AD}^2={AC}^2-{BC}^2-{BD}^2+2BC.BD$

${AD}^2=3^2-5^2-{BD}^2+2\times 5.BD$

${AD}^2=-16-{BD}^2+10BD$

$9-{BD}^2=-16-{BD}^2+10BD$

$10BD=25$

$BD=2.5$

Substitute the value in eqn$\left(1\right)$, we get

${AD}^2=9-{\left(2.5\right)}^2=9-6.25=2.75$

$\therefore AD=1.65$cm

$\therefore$ The length of altitude from $A$ to $BC$ is $AD=1.65$cm