Question

In a triangle $$\mathrm{\Delta }ABC$$ a point ‘D’ on side ‘BC’ is such that $$DC=6,BC=15$$ find the ratios
(i) $$A\left(\mathrm{\Delta }ABD\right):A\left(\mathrm{\Delta }ABC\right)$$
(ii) $$A\left(\mathrm{\Delta }ABD\right):A\left(\mathrm{\Delta }ADC\right)$$

Answer 1

Hint: We solve this problem first by drawing the height from vertex ‘A’ to side ‘BC’. The diagram representing the given data is shown below.

Here the length ‘AE’ will be the height for all the triangles that can be seen in the figure because ‘A’ is a common vertex for all triangles and ‘E’ lies on the base of all triangles. then we use the area formula of triangle given as
$$A={\displaystyle \frac{1}{2}}\left(\text{Base}\right)\left(\text{Height}\right)$$
By using the above formula we find the required ratios.

Complete step-by-step answer:
We are given that there is a point ‘D’ on ‘BC’ such that
$$\Rightarrow BC=15$$
$$\Rightarrow DC=6$$
Let us assume that a perpendicular drawn from the vertex ‘A’ to side ‘BC’ as shown in the figure.

Here, we can see that the line ‘AE’ is the height for all three triangles $$\mathrm{\Delta }ABC,\mathrm{\Delta }ABD,\mathrm{\Delta }ADC$$
Now, let us solve for the first part
(i) $$A\left(\mathrm{\Delta }ABD\right):A\left(\mathrm{\Delta }ABC\right)$$
We know that the formula for area of triangle is given as
$$A={\displaystyle \frac{1}{2}}\left(\text{Base}\right)\left(\text{Height}\right)$$
Let us consider the triangle $$\mathrm{\Delta }ABD$$
By using the area formula the area of triangle $$\mathrm{\Delta }ABD$$ we get
$$\Rightarrow A\left(\mathrm{\Delta }ABD\right)={\displaystyle \frac{1}{2}}\left(BD\right)\left(AE\right)..........equation(i)$$
Now, let us consider the triangle $$\mathrm{\Delta }ABC$$
By using the area formula the area of triangle $$\mathrm{\Delta }ABC$$ we get
$$\Rightarrow A\left(\mathrm{\Delta }ABC\right)={\displaystyle \frac{1}{2}}\left(BC\right)\left(AE\right)..........equation(ii)$$
Now, by dividing the equation (i) with equation (ii) we get
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{A\left(\mathrm{\Delta }ABD\right)}{A\left(\mathrm{\Delta }ABC\right)}}={\displaystyle \frac{{\displaystyle \frac{1}{2}}\times BD\times AE}{{\displaystyle \frac{1}{2}}\times BC\times AE}}\\ & \Rightarrow {\displaystyle \frac{A\left(\mathrm{\Delta }ABD\right)}{A\left(\mathrm{\Delta }ABC\right)}}={\displaystyle \frac{BD}{BC}}\end{array}$$
By substituting the required values in above equation we get
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{A\left(\mathrm{\Delta }ABD\right)}{A\left(\mathrm{\Delta }ABC\right)}}={\displaystyle \frac{6}{15}}\\ & \Rightarrow {\displaystyle \frac{A\left(\mathrm{\Delta }ABD\right)}{A\left(\mathrm{\Delta }ABC\right)}}={\displaystyle \frac{2}{5}}\end{array}$$
We know that the ratio definition that is
$$\Rightarrow a:b={\displaystyle \frac{a}{b}}$$
Therefore, by using the ratio theorem to above equation we get
$$\therefore A\left(\mathrm{\Delta }ABD\right):A\left(\mathrm{\Delta }ABC\right)=2:5$$
Now, let us solve for second part
(ii) $$A\left(\mathrm{\Delta }ABD\right):A\left(\mathrm{\Delta }ADC\right)$$
We know that the formula for area of triangle is given as
$$A={\displaystyle \frac{1}{2}}\left(\text{Base}\right)\left(\text{Height}\right)$$
Let us consider the triangle $$\mathrm{\Delta }ABD$$
By using the area formula the area of triangle $$\mathrm{\Delta }ABD$$ we get
$$\Rightarrow A\left(\mathrm{\Delta }ABD\right)={\displaystyle \frac{1}{2}}\left(BD\right)\left(AE\right)..........equation(i)$$
Now, let us consider the triangle $$\mathrm{\Delta }ADC$$
By using the area formula the area of triangle $$\mathrm{\Delta }ADC$$ we get
$$\Rightarrow A\left(\mathrm{\Delta }ADC\right)={\displaystyle \frac{1}{2}}\left(DC\right)\left(AE\right)..........equation(iii)$$
Now, by dividing the equation (i) with equation (iii) we get
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{A\left(\mathrm{\Delta }ABD\right)}{A\left(\mathrm{\Delta }ADC\right)}}={\displaystyle \frac{{\displaystyle \frac{1}{2}}\times BD\times AE}{{\displaystyle \frac{1}{2}}\times DC\times AE}}\\ & \Rightarrow {\displaystyle \frac{A\left(\mathrm{\Delta }ABD\right)}{A\left(\mathrm{\Delta }ADC\right)}}={\displaystyle \frac{BD}{DC}}........equation(iv)\end{array}$$
Here, we know that the point ‘D’ divides the side ‘BC’ in two parts that is
$$\begin{array}{rl}& \Rightarrow BC=BD+DC\\ & \Rightarrow 15=BD+6\\ & \Rightarrow BD=9\end{array}$$
Now, by substituting the required values in equation (iv) we get
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{A\left(\mathrm{\Delta }ABD\right)}{A\left(\mathrm{\Delta }ADC\right)}}={\displaystyle \frac{9}{6}}\\ & \Rightarrow {\displaystyle \frac{A\left(\mathrm{\Delta }ABD\right)}{A\left(\mathrm{\Delta }ADC\right)}}={\displaystyle \frac{3}{2}}\end{array}$$


We know that the ratio definition that is
$$\Rightarrow a:b={\displaystyle \frac{a}{b}}$$
Therefore, by using the ratio theorem to above equation we get
$$\therefore A\left(\mathrm{\Delta }ABD\right):A\left(\mathrm{\Delta }ADC\right)=3:2$$

Note: We can solve the second part in another method also.
From the first part we have the ratio
$$\Rightarrow A\left(\mathrm{\Delta }ABD\right):A\left(\mathrm{\Delta }ABC\right)=2:5$$
$$\Rightarrow A\left(\mathrm{\Delta }ABC\right):A\left(\mathrm{\Delta }ABD\right)=5:2$$
Now by converting ratio to division we get
$$\Rightarrow {\displaystyle \frac{A\left(\mathrm{\Delta }ABC\right)}{A\left(\mathrm{\Delta }ADC\right)}}={\displaystyle \frac{5}{2}}.........equation(v)$$
We know that when a line divides the triangle into two parts then the total area of triangle will be sum of those two parts that is from the figure
$$\Rightarrow A\left(\mathrm{\Delta }ABC\right)=A\left(\mathrm{\Delta }ABD\right)+A\left(\mathrm{\Delta }ADC\right)$$
By substituting the above equation in equation (v) we get
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{A\left(\mathrm{\Delta }ABD\right)+A\left(\mathrm{\Delta }ADC\right)}{A\left(\mathrm{\Delta }ADC\right)}}={\displaystyle \frac{5}{2}}\\ & \Rightarrow {\displaystyle \frac{A\left(\mathrm{\Delta }ABD\right)}{A\left(\mathrm{\Delta }ADC\right)}}+1={\displaystyle \frac{5}{2}}\\ & \Rightarrow {\displaystyle \frac{A\left(\mathrm{\Delta }ABD\right)}{A\left(\mathrm{\Delta }ADC\right)}}={\displaystyle \frac{3}{2}}\end{array}$$
By converting into ratio we get
$$\therefore A\left(\mathrm{\Delta }ABD\right):A\left(\mathrm{\Delta }ADC\right)=3:2$$