Question

In the given figure, $BC$ is parallel to $DE$. Area of triangle \[ABC = 25c{m^2}\], Area of trapezium $BCED = 24c{m^2}$ and $DE = 14cm$. Calculate the length of $BC$. Also, find the area of triangle $BCD$.

Answer 1

Hint: We will first prove the triangles $\vartriangle ABC$ and $\vartriangle ADE$ similar. Then, we will write the relation of ratio of their area and the sides. Substitute the given values and find the length of $BC$. Next, find the height of trapezium to find the area of the triangle $BCD$ using the formula $\dfrac{1}{2} \times {\text{base}} \times {\text{height}}$.

Complete step by step answer:

We are given that $BC$ is parallel to $DE$. We will first prove that the triangles $ABC$ and triangle $ADE$ are similar, then we can use the relation of their areas to find the required area.
Now, consider $\vartriangle ABC$ and $\vartriangle ADE$.
We have $\angle BAC$ equal to $\angle DAE$ as they are common angles in both the triangles.
Also, \[\angle ABC = \angle ADE\] as they are the corresponding angles where $BC\parallel DE$.
Similarly, \[\angle ACB = \angle AED\] which are also corresponding angles.
Hence, by AAA similarity criterion rule, then $\vartriangle ABC \sim \vartriangle ADE$
If triangles are similar, then the ratio of the square root of their areas is equal to the ratio of their sides.
That is, $\dfrac{{\sqrt {Area\left( {\vartriangle ABC} \right)} }}{{\sqrt {Area\left( {\vartriangle ADE} \right)} }} = \dfrac{{BC}}{{DE}}$
Also, triangle \[ADE\] is the combination of triangle $ABC$ and trapezium $BCED$
Hence, the above ratio can be written as,
$\dfrac{{\sqrt {Area\left( {\vartriangle ABC} \right)} }}{{\sqrt {Area\left( {\vartriangle ABE} \right) + Area\left( {\square BCED} \right)} }} = \dfrac{{BC}}{{DE}}$
Now, we will substitute the given values, \[ABC = 25c{m^2}\], $BCED = 24c{m^2}$ and $DE = 14cm$ to calculate the length of $BC$.
$
  \dfrac{{\sqrt {25} }}{{\sqrt {25 + 24} }} = \dfrac{{BC}}{{14}} \\
   \Rightarrow \dfrac{{\sqrt {25} }}{{\sqrt {49} }} = \dfrac{{BC}}{{14}} \\
   \Rightarrow \dfrac{5}{7} = \dfrac{{BC}}{{14}} \\
$
Now, we will cross multiply the above equation to find the value of $BC$
$14\left( 5 \right) = 7\left( {BC} \right)$
Which is equal to $70 = 7BC$
Divide the equation throughout by 7,
$BC = 10cm$
We also have to calculate the area of triangle $BCD$

Now, area of trapezium $BCED = 24c{m^2}$, $BC = 10cm$ and $DE = 14cm$.
We will first calculate the height of the trapezium.
We know that the area of the trapezium is given as $A = \dfrac{1}{2}\left( {a + b} \right)h$, where $a$ and $b$are length of trapezium and $h$ is the height of the trapezium.
Then,
 $
  24 = \dfrac{1}{2}\left( {10 + 14} \right)h \\
   \Rightarrow 24 = \dfrac{1}{2}\left( {24} \right)h \\
   \Rightarrow h = 2cm \\
$
As it is know that the area of triangle is given as $\dfrac{1}{2} \times {\text{base}} \times {\text{height}}$
Thus, area of triangle $BCD$ is $\dfrac{1}{2} \times BC \times h$
On substituting the values, we get,
$
  A = \dfrac{1}{2} \times 10 \times 2 \\
  A = 10c{m^2} \\
$
Hence, area of $\vartriangle BCD = 10c{m^2}$.

Note: Many students write the relation $\dfrac{{\sqrt {Area\left( {\vartriangle ABC} \right)} }}{{\sqrt {Area\left( {\vartriangle ADE} \right)} }} = \dfrac{{BC}}{{DE}}$ as $\dfrac{{Area\left( {\vartriangle ABC} \right)}}{{Area\left( {\vartriangle ADE} \right)}} = \dfrac{{BC}}{{DE}}$ which is incorrect. Trapezium is a quadrilateral with a set of parallel lines and a set of non-parallel lines, the height is the perpendicular distance between the set of parallel lines. Also, students must know the formula of area of trapezium and area of triangle.