CBSE Class 10 Mathematics Triangles Notes Set 05

Areas of Two Similar Triangles

Theorem 6. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: ΔABC ~ ΔXYZ.
To prove: \( \frac{ar(ABC)}{ar(XYZ)}=\frac{AB^2}{XY^2}=\frac{BC^2}{YZ^2}=\frac{AC^2}{XZ^2} \).
Construction: Draw AM ⊥ BC and XN ⊥ YZ.

Proof: Since ΔABC ~ ΔXYZ, the corresponding angles are equal and the corresponding sides are proportional: ∠A = ∠X, ∠B = ∠Y, ∠C = ∠Z and\[ \frac{AB}{XY}=\frac{BC}{YZ}=\frac{AC}{XZ}\qquad\dots(i) \]In ΔAMB and ΔXNY, ∠AMB = ∠XNY = 90° and ∠B = ∠Y, so ΔAMB ~ ΔXNY (AA criterion). Therefore\[ \frac{AM}{XN}=\frac{AB}{XY}\qquad\dots(ii) \]From (i) and (ii),\[ \frac{AB}{XY}=\frac{BC}{YZ}=\frac{AC}{XZ}=\frac{AM}{XN}\qquad\dots(iii) \]Now\[ \frac{ar(ABC)}{ar(XYZ)}=\frac{\tfrac12\times BC\times AM}{\tfrac12\times YZ\times XN}=\frac{BC}{YZ}\times\frac{AM}{XN}=\frac{BC}{YZ}\times\frac{BC}{YZ}=\frac{BC^2}{YZ^2}\quad[\text{using (iii)}] \]But by (i), \( \frac{BC^2}{YZ^2}=\frac{AB^2}{XY^2}=\frac{AC^2}{XZ^2} \). Hence\[ \frac{ar(ABC)}{ar(XYZ)}=\frac{AB^2}{XY^2}=\frac{BC^2}{YZ^2}=\frac{AC^2}{XZ^2} \]Hence proved.

Knowledge Hub
(i) The areas of two similar triangles are in the ratio of the squares of their corresponding altitudes.
(ii) The areas of two similar triangles are in the ratio of the squares of their corresponding medians.
(iii) The areas of two similar triangles are in the ratio of the squares of their corresponding angle bisector segments.
(iv) If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other. In the figure, ΔABD ~ ΔBCD ~ ΔABC (right angle at B, BD ⊥ AC).

Example 1. D and E are points on the sides AB and AC respectively of a ΔABC such that DE ∥ BC and DE divides ΔABC into two parts of equal area. Find \( \frac{BD}{AB} \).

Answer:
Given: DE ∥ BC and ar(ΔADE) = ar(trapezium BCED). Hence ar(ΔABC) = 2 ar(ΔADE).
In ΔADE and ΔABC: ∠ADE = ∠B (corresponding angles) and ∠A is common, so ΔADE ~ ΔABC (AA).
\[ \frac{ar(ADE)}{ar(ABC)}=\frac{AD^2}{AB^2}\Rightarrow\ \frac12=\frac{AD^2}{AB^2}\Rightarrow\ \frac{AD}{AB}=\frac{1}{\sqrt2} \]
So AB = √2 AD = √2 (AB − BD), that is √2 BD = √2 AB − AB = (√2 − 1) AB.
\[ \frac{BD}{AB}=\frac{\sqrt2-1}{\sqrt2}\ (=1-\tfrac{1}{\sqrt2}\approx 0.293) \]

Example 2. The areas of two similar triangles ABC and DEF are 16 cm² and 36 cm² respectively. If EF = 9 cm, find BC.

Answer:
Since ΔABC ~ ΔDEF, \( \frac{ar(ABC)}{ar(DEF)}=\frac{BC^2}{EF^2} \).
\[ \frac{16}{36}=\frac{BC^2}{9^2}\Rightarrow\ BC^2=\frac{16}{36}\times81=36 \]
Hence BC = 6 cm.

Example 3. Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 81. Find the ratio of their corresponding heights.

Answer:
Let ΔABC and ΔPQR be isosceles with AB = AC, PQ = PR and ∠A = ∠P. Then \( \frac{AB}{AC}=\frac{PQ}{PR}=1 \), so \( \frac{AB}{PQ}=\frac{AC}{PR} \)
In ΔABC and ΔPQR: \( \frac{AB}{PQ}=\frac{AC}{PR} \) and ∠A = ∠P, so ΔABC ~ ΔPQR (SAS). The altitudes AL and PM from the vertices are also medians.
The areas of similar triangles are in the ratio of the squares of the corresponding medians (altitudes):\[ \frac{ar(ABC)}{ar(PQR)}=\frac{AL^2}{PM^2}\Rightarrow\ \frac{AL^2}{PM^2}=\frac{36}{81}\Rightarrow\ \frac{AL}{PM}=\frac{6}{9}=\frac{2}{3} \]
The ratio of the corresponding heights is AL : PM = 6 : 9 = 2 : 3.