Areas of Similar Triangles

Areas of Similar Triangles

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Introduction to Similarity:

If two triangles are similar it means that:

  • All corresponding angle pairs are equal
  • All corresponding sides are proportional 

However, in order to be sure that two triangles are similar, we do not necessarily need to have information about all sides and all angles.

enlightenedThink: Two congruent triangles have the same area. What about two similar triangles? What is the relation between their areas? 

Let us discuss this rigorously.


Theorem for Areas of Similar Triangles

It states that "The ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of their corresponding sides".

Consider the following figure, which shows two similar triangles, \(\Delta ABC\) and \(\Delta DEF\):

Areas of similar triangles are proportional

Theorem for Areas of Similar Triangles tells us that

\[\frac{{ar(\Delta ABC)}}{{ar(\Delta DEF)}} = \frac{{A{B^2}}}{{D{E^2}}} = \frac{{B{C^2}}}{{E{F^2}}} = \frac{{A{C^2}}}{{D{F^2}}}\]


Proof of Theorem for Areas of Similar Triangles

Consider two triangles, \(\Delta ABC\) and \(\Delta DEF\),

Given: \(\Delta ABC \sim \Delta DEF\)

To prove: \(\frac{{ar(\Delta ABC)}}{{ar(\Delta DEF)}} = {\left( {\frac{{AB}}{{DE}}} \right)^2} = {\left( {\frac{{BC}}{{EF}}} \right)^2} = {\left( {\frac{{AC}}{{DF}}} \right)^2}\)

Construction: Draw the altitudes AP and DQ, as shown below:

Altitudes of similar triangles are proportional

Proof: Since, \(\angle B = \angle E\), \(\angle APB = \angle DQE\)

We can note that \(\Delta ABP\) and \(\Delta DEQ\) are equi-angular

Hence,

\[\Delta ABP \sim \Delta DEQ\]

Thus,

\[\frac{{AP}}{{DQ}} = \frac{{AB}}{{DE}}\]

This further implies that,

\[\frac{{AP}}{{DQ}} = \frac{{BC}}{{EF}}....(1)\]

Thus,

\[\begin{align}
  \frac{{ar(\Delta ABC)}}{{ar(\Delta DEF)}} &= \frac{{\frac{1}{2} \times BC \times AP}}{{\frac{1}{2} \times EF \times DQ}} \hfill \\
   &= \left( {\frac{{BC}}{{EF}}} \right) \times \left( {\frac{{AP}}{{DQ}}} \right) \hfill \\
   &= \left( {\frac{{BC}}{{EF}}} \right) \times \left( {\frac{{BC}}{{EF}}} \right)....{\text{[from (1)]}} \hfill \\
   \Rightarrow \frac{{ar(\Delta ABC)}}{{ar(\Delta DEF)}} &= {\left( {\frac{{BC}}{{EF}}} \right)^2} \hfill \\ 
\end{align} \]

Similarly, we can show that

\[\boxed{\frac{{ar(\Delta ABC)}}{{ar(\Delta DEF)}} = {{\left( {\frac{{AB}}{{DE}}} \right)}^2} = {{\left( {\frac{{BC}}{{EF}}} \right)}^2} = {{\left( {\frac{{AC}}{{DF}}} \right)}^2}}\]


Solved Examples:

Example 1: Consider two similar triangles, \(\Delta ABC\) and \(\Delta DEF\), as shown below:

Medians of similar triangles are proportional

\(AP\) and \(DQ\) are medians in the two triangles. Show that

\[\frac{{ar\Delta (ABC)}}{{A{P^2}}} = \frac{{ar\Delta (DEF)}}{{D{Q^2}}}\]

Solution: Since \(\Delta ABC \sim \Delta DEF\),

\[\begin{align}
  \frac{{AB}}{{DE}} &= \frac{{BC}}{{EF}} \hfill \\
   \Rightarrow \frac{{AB}}{{DE}} &= \frac{{\frac{1}{2}BC}}{{\frac{1}{2}EF}} \hfill \\
   \Rightarrow \frac{{AB}}{{DE}} &= \frac{{BP}}{{EQ}}....(1) \hfill \\ 
\end{align} \]

Also,

\[\angle B = \angle E....(2)\]

From (1) and (2) and by SAS similarity criterion, We can note that, 

\[\begin{align}
  \Delta ABP &\sim \Delta DEQ \hfill \\
   \Rightarrow \frac{{AB}}{{DE}} &= \frac{{AP}}{{DQ}}....(3) \hfill \\ 
\end{align} \]

Now, By Theorem for Areas of Similar Triangles,

\[\begin{align}
  \frac{{ar\Delta (ABC)}}{{ar\Delta (DEF)}} &= \frac{{A{B^2}}}{{D{E^2}}} = \frac{{A{P^2}}}{{D{Q^2}}}....[{\text{from (3)}}] \hfill \\
   \Rightarrow \frac{{ar\Delta (ABC)}}{{A{P^2}}} &= \frac{{ar\Delta (DEF)}}{{D{Q^2}}} \hfill \\ 
\end{align} \]


Example 2: Consider the following figure:

Line parallel to base of triangle

It is given that \(XY\parallel AC\) and divides the triangle into two parts of equal areas. Find the ratio \(AX:XB\).

Solution: Since \(XY\parallel AC\), \(\Delta AXY\) must be similar to \(\Delta ABC\).

Now, By Theorem for Areas of Similar Triangles,

\[\frac{{ar(\Delta ABC)}}{{ar\left( {\Delta AXY} \right)}} = \frac{{A{B^2}}}{{A{X^2}}}....(1)\]

Also, \(XY\) divides the triangle into two parts of equal areas. Thus,

\[\frac{{ar(\Delta ABC)}}{{ar\left( {\Delta AXY} \right)}} = 2{\text{  }}....(2)\]

From (1) and (2), we have,

\[\begin{align}
  \frac{{A{B^2}}}{{A{X^2}}} &= 2 \hfill \\
   \Rightarrow \frac{{AB}}{{AX}} &= \sqrt 2  \hfill \\
   \Rightarrow \frac{{AB}}{{AX}} - 1 &= \sqrt 2  - 1 \hfill \\
   \Rightarrow \frac{{XB}}{{AX}} &= \sqrt 2  - 1 \hfill \\
   \Rightarrow \frac{{AX}}{{XB}} &= \frac{1}{{\sqrt 2  - 1}} \hfill \\ 
\end{align} \]

yesChallenge: It is given that \(\Delta ABC \sim \Delta XYZ\). The area of \(\Delta ABC\) is 45 sq units and the area of \(\Delta XYZ\) is 80 sq units. \(YZ = 12\) units.

Find \(BC\)?

⚡Tip: Use Theorem for Areas of Similar Triangles.


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