
If $\Delta DEF \sim \Delta MNK. $ If$DE = 2,{\text{ }}MN = 5$, then find the value of$\dfrac{{A\left( {\Delta DEF} \right)}}{{A\left( {\Delta MNK} \right)}}$ by using the property of similar triangles.
Answer
603.6k+ views
Hint: If two triangles are similar then the ratio of their respective sides are same i.e. (SSS congruence).
$ \Rightarrow \dfrac{{DE}}{{MN}} = \dfrac{{EF}}{{NK}} = \dfrac{{FD}}{{KM}}$
Consider the similar triangles DEF and MNK as shown in figure.
By the property of similar triangles, if two triangles are similar then the ratio of their respective sides are same i.e. (SSS congruence).
$ \Rightarrow \dfrac{{DE}}{{MN}} = \dfrac{{EF}}{{NK}} = \dfrac{{FD}}{{KM}}$
Now it is given that $DE = 2,{\text{ }}MN = 5$
$ \Rightarrow \dfrac{{DE}}{{MN}} = \dfrac{{EF}}{{NK}} = \dfrac{{FD}}{{KM}} = \dfrac{2}{5}...............\left( a \right)$
Now, we consider both as right angle triangle
So, the area of the right angle triangle is half multiplied by base time’s perpendicular.
$
\Rightarrow A\left( {\Delta DEF} \right) = \dfrac{1}{2}\left( {EF} \right)\left( {DE} \right).............\left( 1 \right) \\
\Rightarrow A\left( {\Delta MNK} \right) = \dfrac{1}{2}\left( {NK} \right)\left( {MN} \right).............\left( 2 \right) \\
$
So, divide equation (1) from equation (2)
$ \Rightarrow \dfrac{{A\left( {\Delta DEF} \right)}}{{A\left( {\Delta MNK} \right)}} = \dfrac{{\dfrac{1}{2}\left( {EF} \right)\left( {DE} \right)}}{{\dfrac{1}{2}\left( {NK} \right)\left( {MN} \right)}} = \dfrac{{\left( {EF} \right)\left( {DE} \right)}}{{\left( {NK} \right)\left( {MN} \right)}}..........\left( 3 \right)$
Now from equation (a)
$\dfrac{{DE}}{{MN}} = \dfrac{{EF}}{{NK}} = \dfrac{2}{5}$
Therefore from equation (3)
$ \Rightarrow \dfrac{{A\left( {\Delta DEF} \right)}}{{A\left( {\Delta MNK} \right)}} = \dfrac{{\left( {EF} \right)\left( {DE} \right)}}{{\left( {NK} \right)\left( {MN} \right)}} = {\left( {\dfrac{{DE}}{{MN}}} \right)^2}$
$ \Rightarrow \dfrac{{A\left( {\Delta DEF} \right)}}{{A\left( {\Delta MNK} \right)}} = {\left( {\dfrac{{DE}}{{MN}}} \right)^2} = {\left( {\dfrac{2}{5}} \right)^2} = \dfrac{4}{{25}}$
So, this is the required ratio of the area of triangle DEF to triangle MNK.
Note: In such types of questions always remember the key concept which is if two triangles are similar then the ratio of their respective sides are same i.e. (SSS congruence), and always remember the area of the right angle triangle which is stated above then using these properties calculate the ratio of the area of triangle DEF to triangle MNK which is the required answer.
$ \Rightarrow \dfrac{{DE}}{{MN}} = \dfrac{{EF}}{{NK}} = \dfrac{{FD}}{{KM}}$
Consider the similar triangles DEF and MNK as shown in figure.
By the property of similar triangles, if two triangles are similar then the ratio of their respective sides are same i.e. (SSS congruence).
$ \Rightarrow \dfrac{{DE}}{{MN}} = \dfrac{{EF}}{{NK}} = \dfrac{{FD}}{{KM}}$
Now it is given that $DE = 2,{\text{ }}MN = 5$
$ \Rightarrow \dfrac{{DE}}{{MN}} = \dfrac{{EF}}{{NK}} = \dfrac{{FD}}{{KM}} = \dfrac{2}{5}...............\left( a \right)$
Now, we consider both as right angle triangle
So, the area of the right angle triangle is half multiplied by base time’s perpendicular.
$
\Rightarrow A\left( {\Delta DEF} \right) = \dfrac{1}{2}\left( {EF} \right)\left( {DE} \right).............\left( 1 \right) \\
\Rightarrow A\left( {\Delta MNK} \right) = \dfrac{1}{2}\left( {NK} \right)\left( {MN} \right).............\left( 2 \right) \\
$
So, divide equation (1) from equation (2)
$ \Rightarrow \dfrac{{A\left( {\Delta DEF} \right)}}{{A\left( {\Delta MNK} \right)}} = \dfrac{{\dfrac{1}{2}\left( {EF} \right)\left( {DE} \right)}}{{\dfrac{1}{2}\left( {NK} \right)\left( {MN} \right)}} = \dfrac{{\left( {EF} \right)\left( {DE} \right)}}{{\left( {NK} \right)\left( {MN} \right)}}..........\left( 3 \right)$
Now from equation (a)
$\dfrac{{DE}}{{MN}} = \dfrac{{EF}}{{NK}} = \dfrac{2}{5}$
Therefore from equation (3)
$ \Rightarrow \dfrac{{A\left( {\Delta DEF} \right)}}{{A\left( {\Delta MNK} \right)}} = \dfrac{{\left( {EF} \right)\left( {DE} \right)}}{{\left( {NK} \right)\left( {MN} \right)}} = {\left( {\dfrac{{DE}}{{MN}}} \right)^2}$
$ \Rightarrow \dfrac{{A\left( {\Delta DEF} \right)}}{{A\left( {\Delta MNK} \right)}} = {\left( {\dfrac{{DE}}{{MN}}} \right)^2} = {\left( {\dfrac{2}{5}} \right)^2} = \dfrac{4}{{25}}$
So, this is the required ratio of the area of triangle DEF to triangle MNK.
Note: In such types of questions always remember the key concept which is if two triangles are similar then the ratio of their respective sides are same i.e. (SSS congruence), and always remember the area of the right angle triangle which is stated above then using these properties calculate the ratio of the area of triangle DEF to triangle MNK which is the required answer.
Recently Updated Pages
In cricket, what is a "pink ball" primarily used for?

In cricket, what is the "new ball" phase?

In cricket, what is a "death over"?

What is the "Powerplay" in T20 cricket?

In cricket, what is a "super over"?

In cricket, what is a "tail-ender"?

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Write an application to the principal requesting five class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Who Won 36 Oscar Awards? Record Holder Revealed

