Question

Let $\Delta ABC \sim \Delta DEF$ and their areas be, respectively, $64c{m^2}$ and $121c{m^2}$, if $EF = 15.4cm$, find $BC$.

Answer 1

Hint: We have learnt that in two similar triangles, the ratio of their corresponding sides is same. There is a relationship between the ratio of their areas and the ratio of the corresponding sides. We know that area is measured in square units. So, you may expect that this ratio is one of their corresponding sides. We have a therme The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Complete step-by-step answer:
Step-1 Here, we already given triangle $ABC$ and triangle $DEF$ is similar triangle i.e. $\Delta ABC \sim \Delta DEF$ The area of $\Delta ABC = 64c{m^2}$ and area of $\Delta DEF = 121c{m^2}$ The side of triangle $\Delta DEF$ is i.e. $15.4cm$ i.e. $EF = 15.4cm$. We have to find the value of $BC$ corresponding sides of $EF$ from triangle $ABC$.
We know that ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding sides
So, it is given that $\Delta ABC \sim \Delta DEF$.
Step-2 $\dfrac{{area(\Delta ABC)}}{{area(\Delta DEF)}} = {\left( {\dfrac{{AB}}{{DE}}} \right)^2} = {\left( {\dfrac{{BC}}{{EF}}} \right)^2} = {\left( {\dfrac{{BC}}{{DF}}} \right)^2}$
Given that $EF = 15.4cm$, $area(\Delta ABC) = 64c{m^2}$, $area(\Delta DEF) = 121c{m^2}$ Therefore, $\dfrac{{area(\Delta ABC)}}{{area(\Delta DEF)}} = {\left( {\dfrac{{BC}}{{EF}}} \right)^2}$
Then implies; $\left( {\dfrac{{64c{m^2}}}{{121c{m^2}}}} \right) = \dfrac{{B{C^2}}}{{{{(15.4cm)}^2}}}$
The implies $\sqrt {\dfrac{{64c{m^2}}}{{121c{m^2}}}} = \dfrac{{B{C^2}}}{{15.4}}$
Square root of $64$ is $8$ and Square root of $121$ is $11$.
We get,
$\dfrac{8}{{11}} = \dfrac{{BC}}{{15.4}}$
Cross multiplication, we get
$\dfrac{{8 \times 15.4}}{{11}} = BC$
After solving,
$\dfrac{{8 \times 15.4}}{{11}} = BC$
$\dfrac{{112}}{{10}} = BC$
i.e. $11.2cm = BC$

Hence the value of side $BC$ is $11.2cm$.

Note: If two angles of a triangle have measure equal to the measure of two angles of another triangle, then the triangles are similar. Corresponding sides of similar polygons are in proportion and corresponding angles of similar polygons have the same measure.