Question

In △ABC, D∈ AB so that AD:DB=1:3. If ACDB=18m2. Find AACD and AABC.

Answer 1

Note that

$A_{CDB}=\dfrac{1}{2}\cdot DB\cdot h,$ where h is the height drawn from the vertex C to the side AB.

Now

$A_{ACD}=\dfrac{1}{2}\cdot AD\cdot h.$

Since $AD:DB=1:3,$ you have that $AD=\dfrac{1}{3}DB$ and consequently

$A_{ACD}=\dfrac{1}{2}\cdot AD\cdot h=\dfrac{1}{2}\cdot \dfrac{1}{3}DB\cdot h=\dfrac{1}{3}\cdot \dfrac{1}{2}DB\cdot h=\dfrac{1}{3}\cdot A_{CDB}=\dfrac{1}{3}\cdot 18=6\ m^2.$

The area of triangle ABC is

$A_{ABC}=A_{CDB}+A_{ACD}=18+6=24\ m^2.$

Answer: $A_{ACD}=6\ m^2,$ $A_{ABC}=24\ m^2$

Answer 2

The area of Δ$ACD\;and \;$ Δ$ABC$ is $6m^2 \;and\; 24 m^2$ respectively.

Area of Δ$CDB = 18\;m^2$

$\dfrac{AD}{DB}=\dfrac{1}{3}$

$\dfrac{Area\; of ACD}{Area\; of CDB}=\dfrac{\dfrac{1}{2}\times h \times AD}{\dfrac{1}{2}\times h \times AB}$

$\dfrac{Area\; of ACD}{Area\; of CDB}=\dfrac{AD}{DB}$

$\dfrac{Area\; of ACD}{18}=\dfrac{1}{3}$

$Area\; of ACD=\dfrac{18}{3}$

$Area\; of ACD=6\;m^2$

$Area\; of\; ABC=Area\; of \;ACD+Area\; of \;CBD\\\\\\Area\; of \;ABC=6m^2 + 18m^2\\\\Area\; of \;ABC = 24m^2$

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