Question

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.

Answer 1

Hint:Here first we will draw the diagrams of two similar triangles and then apply the formula of area of triangle in both the triangles to get the desired ratio as the corresponding sides of similar triangles are in same ratio

Complete step-by-step answer:
Let there be two triangles ABC and DEF such that
\[\Delta ABC \sim \Delta DEF\] and AM and DN be the altitudes of respective triangles.

Now as we know that the area of a triangle is given by:
\[Area = \dfrac{1}{2} \times base \times height\]
Therefore, the area of \[\Delta ABC\] is given by:
\[Ar\left( {\Delta ABC} \right) = \dfrac{1}{2} \times BC \times AM...........................\left( 1 \right)\]
Where AM is the height and BC is the base.
Similarly, the area of \[\Delta DEF\] is given by:-
\[Ar\left( {\Delta DEF} \right) = \dfrac{1}{2} \times EF \times DN...........................\left( 2 \right)\]
In order to get the ratio of areas of both the triangles we need to divide equation 1 by equation 2
Therefore,
\[
  \dfrac{{Ar\left( {\Delta ABC} \right)}}{{Ar\left( {\Delta DEF} \right)}} = \dfrac{{\dfrac{1}{2} \times BC \times AM}}{{\dfrac{1}{2} \times EF \times DN}} \\
  \dfrac{{Ar\left( {\Delta ABC} \right)}}{{Ar\left( {\Delta DEF} \right)}} = \dfrac{{BC \times AM}}{{EF \times DN}}........................\left( 3 \right) \\
 \]
Now in \[\Delta ABM\]and \[\Delta DEN\]
\[
  \angle AMB = \angle DNE....................\left( {{{90}^ \circ }} \right) \\
  \angle ABM = \angle DEN.................\left( {{\text{\Delta ABC and \Delta DEF are similar triangles}}} \right) \\
 \]
Therefore, by AA criterion of similar triangles \[\Delta ABM \sim \Delta DEN\]
Hence the ratio of corresponding sides are equal
Therefore,
\[\dfrac{{AB}}{{DE}} = \dfrac{{BM}}{{EN}} = \dfrac{{AM}}{{DN}}...................\left( 4 \right)\]
Similarly, in \[\Delta ACM\] and \[\Delta DFN\]
\[
  \angle AMC = \angle DNF....................\left( {{{90}^ \circ }} \right) \\
  \angle ACM = \angle DFN.................\left( {{\text{\Delta ABC and \Delta DEF are similar triangles}}} \right) \\
 \]
Therefore, by AA criterion of similar triangles \[\Delta ACM \sim \Delta DFN\]
Hence the ratio of corresponding sides are equal
Therefore,
\[\dfrac{{AC}}{{DF}} = \dfrac{{CM}}{{FN}} = \dfrac{{AM}}{{DN}}...................\left( 5 \right)\]
Now, since
\[\dfrac{{BC}}{{EF}} = \dfrac{{BM + MC}}{{EN + NF}}\]
Therefore from equations 4 and 5 and the property of ratios we get:-
\[\dfrac{{BC}}{{EF}} = \dfrac{{BM + MC}}{{EN + NF}} = \dfrac{{AM}}{{DN}}\]
Putting this value in equation 3 we get:
\[
  \dfrac{{Ar\left( {\Delta ABC} \right)}}{{Ar\left( {\Delta DEF} \right)}} = \dfrac{{AM \times AM}}{{DN \times DN}} \\
  \dfrac{{Ar\left( {\Delta ABC} \right)}}{{Ar\left( {\Delta DEF} \right)}} = {\left( {\dfrac{{AM}}{{DN}}} \right)^2} \\
 \]
And since it is given that
\[\dfrac{{AM}}{{DN}} = \dfrac{6}{9}\]
Therefore, putting the values we get:-
\[
  \dfrac{{Ar\left( {\Delta ABC} \right)}}{{Ar\left( {\Delta DEF} \right)}} = {\left( {\dfrac{6}{9}} \right)^2} \\
  \dfrac{{Ar\left( {\Delta ABC} \right)}}{{Ar\left( {\Delta DEF} \right)}} = \dfrac{{36}}{{81}} \\
  \dfrac{{Ar\left( {\Delta ABC} \right)}}{{Ar\left( {\Delta DEF} \right)}} = \dfrac{4}{9} \\
 \]
Therefore the ratio of areas is 4:9

Note: the similarity of triangles has several criterions:
AAA (angle angle angle) criterion in which the angles of the similar triangles are equal
SAS(side angle side) criterion in which two sides of similar triangles are in same ratio and the angle between them is equal.
ASA(angle side angle) criterion in which two angles of the similar triangles are equal and angle between them is equal.
SSS(side side side) criterion in which the ratio of the sides of the similar triangles are equal.