Question

What is the length $x$ of the line segment CD in the triangle drawn given?

A. 4
B. 5
C. 6
D. 8

Answer 1

Hint: In order to solve this question we first draw a perpendicular from C on AD and then we will find the height of the perpendicular using the formula of area of triangle i.e., $\text{area of } \vartriangle \, = \dfrac{1}{2} \times \text{base} \times \text{height}$ and then we will use the Pythagoras theorem which states that ${(hypotensue)^2} = {(perpendicular)^2} + {(base)^2}$

Complete step by step answer:
Here, we have a triangle $CAB$ and we have to find the length of the line segment $CD$ i.e., $x$. Now, draw a perpendicular from C on AD as shown in the figure below:

Now we have to find the length of the perpendicular $CM$.Using the formula of area of triangle i.e.,
$\text{area of } \vartriangle \, = \dfrac{1}{2} \times \text{base} \times \text{height}$
We can find the area of triangle $ACB$ firstly by taking $AC$ as base and $CB$ as height and also by taking \[AB\] as base and \[CM\] as height. On equating these areas we can find the height of the perpendicular $CM$. So,
$ \Rightarrow \dfrac{1}{2} \times 6 \times 8 = \dfrac{1}{2} \times 10 \times CM$
On cancelling the equal terms and multiplying the numbers. We get,
$ \Rightarrow 48 = 10 \times CM$
Shifting $10$ to the left side of the equation in the denominator. We get,
$ \Rightarrow CM = \dfrac{{48}}{{10}}$
On further simplifying, we get
$ \Rightarrow CM = \dfrac{{24}}{5}$
Therefore, the height of the perpendicular $CM = \dfrac{{24}}{5}$

Now, we will find the length of the line segment CD.First consider $\vartriangle ACM$. As it is a right- angled triangle we can use Pythagoras theorem i.e.,
${(hypotensue)^2} = {(perpendicular)^2} + {(base)^2}$
So, $A{C^2} = C{M^2} + A{M^2}$
Putting $AC = 6$ and $CM = \dfrac{{24}}{5}$. We get,
$ \Rightarrow {(6)^2} = {\left( {\dfrac{{24}}{5}} \right)^2} + A{M^2}$
Solving squares of the numbers. We get,
$ \Rightarrow 36 = \dfrac{{576}}{{25}} + A{M^2}$
Shifting $\dfrac{{576}}{{25}}$ to left side of the equation. We get,
$ \Rightarrow A{M^2} = 36 - \dfrac{{576}}{{25}}$
Solving the equation. We get,
$ \Rightarrow A{M^2} = \dfrac{{325}}{{25}}$
$ \Rightarrow AM = \dfrac{{18}}{5}$
Therefore, $MD = 5 - \dfrac{{18}}{5}$
Solving the above equation. We get,
$ \Rightarrow MD = \dfrac{7}{5}$

Now, consider $\vartriangle CMD$ and apply Pythagoras theorem in it. So,
$ \Rightarrow C{D^2} = C{M^2} + M{D^2}$
Putting $CM = \dfrac{{24}}{5}$ and $MD = \dfrac{7}{5}$. We get,
$ \Rightarrow C{D^2} = {\left( {\dfrac{{24}}{5}} \right)^2} + {\left( {\dfrac{7}{5}} \right)^2}$
Solving the squares of the numbers. We get,
$ \Rightarrow C{D^2} = \dfrac{{576}}{{25}} + \dfrac{{49}}{{25}}$
On adding the numbers. We get
$ \Rightarrow C{D^2} = \dfrac{{625}}{{25}}$
On dividing the numbers. We get
$ \Rightarrow C{D^2} = 25$
Solving the square root. We get,
$ \therefore CD = 5$
Therefore, the length of the $CD = 5$.

Hence, option (B) is the correct answer.

Note: Here, in this question we can’t term CD as altitude or median so we cannot apply Pythagoras theorem directly. First we have to construct a perpendicular and then we will apply the Pythagoras theorem into it. Note that we can only apply Pythagoras theorem in a right- angled triangle where we have one angle of $90^\circ $.