Question

The sides of a triangle are $21m,20m$ and $13m$. Find the areas of the triangles into which it is divided by the perpendicular upon the longest side from the opposite angular point

• A. $12m^2,16m^2$
• B. $96m^2,30m^2$
• C. $72m^2,30m^2$
• D. $72m^2,16m^2$

Answer 1

Answer: (B)

$96m^2,30m^2$

From figure,

$Areaof△ABC=\sqrt{s(s-a)(s-b)(s-c)}$

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

$s=\frac{21+20+13}{2}$

$s=27sq.m$

$\therefore Areaof△ABC=\sqrt{27(27-13)(27-20)(27-21)}$

$Areaof△ABC=126sq.m$

Also, $Areaof△ABC=\frac{1}{2}\times base\times height$

$126=\frac{1}{2}\times 21\times CM$

$CM=12m$

In $△ACM$, by Pythagoras theorem

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

${20}^2={12}^2+A{M}^2$

$AM=16m$

$⟹MB=AM-AM=21-16=5m$

$Areaof△AMC=\frac{1}{2}\times AM\times CM$

$=\frac{1}{2}\times 16\times 12$

$Area=96sq.m$

$Areaof△MCB=\frac{1}{2}\times MB\times CM$

$=\frac{1}{2}\times 5\times 12$

$Area=30sq.m$

Option B.