CBSE Class 10 Mathematics Triangles Assignment Set 10

Exercise 4: Areas of Similar Triangles

Question 1. Let ΔABC ~ ΔDEF and their areas be 64 cm² and 121 cm² respectively. If EF = 15.4 cm, find BC.
The ratio of areas equals the square of the ratio of corresponding sides.
Answer:
Since ΔABC ~ ΔDEF, \( \frac{ar(ABC)}{ar(DEF)}=\frac{BC^2}{EF^2} \).
\[ \frac{64}{121}=\frac{BC^2}{(15.4)^2}\Rightarrow\ BC^2=\frac{(15.4)^2\times64}{121}\Rightarrow\ BC=\frac{15.4\times8}{11}=11.2 \]
BC = 11.2 cm.

Teacher's Note:
1. Write the similarity statement with corresponding vertices in the same order, then use the ratio of areas equal to the square of the ratio of corresponding sides.
2. Give a reason (AA, SAS or SSS) for every similarity you claim.

Question 2. Diagonals of a trapezium ABCD with AB ∥ DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Answer:
In ΔAOB and ΔCOD: ∠1 = ∠2 and ∠3 = ∠4 (alternate angles, AB ∥ DC) and ∠5 = ∠6 (vertically opposite), so ΔAOB ~ ΔCOD (AA).
\[ \frac{ar(AOB)}{ar(COD)}=\frac{AB^2}{CD^2}=\frac{(2CD)^2}{CD^2}=\frac{4}{1} \]
The required ratio is 4 : 1.
In simple words: Alternate angles make the two triangles similar.

Teacher's Note:
1. Write the similarity statement with corresponding vertices in the same order, then use the ratio of areas equal to the square of the ratio of corresponding sides.
2. Give a reason (AA, SAS or SSS) for every similarity you claim.

Question 3. In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \( \frac{ar(ABC)}{ar(DBC)}=\frac{AO}{DO} \).

Answer:
Construction: Draw AP ⊥ BC and DM ⊥ BC.
Area of a triangle = ½ × base × height, so \[ \frac{ar(ABC)}{ar(DBC)}=\frac{\tfrac12\times BC\times AP}{\tfrac12\times BC\times DM}=\frac{AP}{DM}\qquad\dots(i) \]
In ΔAPO and ΔDMO: ∠APO = ∠DMO = 90° and ∠AOP = ∠DOM (vertically opposite), so ΔAPO ~ ΔDMO (AA). Hence \( \frac{AP}{DM}=\frac{AO}{DO} \) ...(ii).
From (i) and (ii), \( \frac{ar(ABC)}{ar(DBC)}=\frac{AO}{DO} \). Hence proved.
In simple words: Triangles on the same base have areas in the ratio of their heights.

Teacher's Note:
1. Write the similarity statement with corresponding vertices in the same order, then use the ratio of areas equal to the square of the ratio of corresponding sides.
2. Give a reason (AA, SAS or SSS) for every similarity you claim.

Question 4. If the areas of two similar triangles are equal, prove that they are congruent.

Answer:
Let ΔABC ~ ΔPQR with ar(ABC) = ar(PQR).
\[ \frac{ar(ABC)}{ar(PQR)}=\frac{BC^2}{QR^2}\Rightarrow\ \frac{BC^2}{QR^2}=1\Rightarrow\ BC=QR \]
Similarly AB = PQ and AC = PR.
Hence ΔABC ≅ ΔPQR (SSS). Hence proved.
In simple words: Equal areas force the ratio of similarity to be 1.

Teacher's Note:
1. Write the similarity statement with corresponding vertices in the same order, then use the ratio of areas equal to the square of the ratio of corresponding sides.
2. Give a reason (AA, SAS or SSS) for every similarity you claim.

Question 5. D, E and F are respectively the mid-points of the sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF to the area of ΔABC.

Answer:
D and E are mid-points of AB and BC, so DE ∥ AC and DE = ½ AC. Similarly EF ∥ AB, EF = ½ AB and DF ∥ BC, DF = ½ BC.
Hence \( \frac{DE}{AC}=\frac{EF}{AB}=\frac{DF}{BC}=\frac{1}{2} \), so ΔDEF ~ ΔCAB (SSS).
\[ \frac{ar(DEF)}{ar(CAB)}=\left(\frac12\right)^2=\frac14 \]
The required ratio is 1 : 4.
In simple words: The mid-point triangle is similar to the original with ratio 1 : 2.

Teacher's Note:
1. Write the similarity statement with corresponding vertices in the same order, then use the ratio of areas equal to the square of the ratio of corresponding sides.
2. Give a reason (AA, SAS or SSS) for every similarity you claim.

Question 6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Answer:
Let ΔABC ~ ΔDEF, and let AM and DN be the medians to BC and EF.
Then \( \frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD} \) and ∠B = ∠E.
Since M and N are mid-points, \( \frac{BM}{EN}=\frac{\tfrac12BC}{\tfrac12EF}=\frac{BC}{EF}=\frac{AB}{DE} \).
In ΔABM and ΔDEN: ∠B = ∠E and \( \frac{AB}{DE}=\frac{BM}{EN} \), so ΔABM ~ ΔDEN (SAS). Hence \( \frac{AB}{DE}=\frac{AM}{DN} \).
\[ \frac{ar(ABC)}{ar(DEF)}=\frac{AB^2}{DE^2}=\frac{AM^2}{DN^2} \]
Hence proved.
In simple words: Corresponding medians are in the same ratio as corresponding sides.

Teacher's Note:
1. Write the similarity statement with corresponding vertices in the same order, then use the ratio of areas equal to the square of the ratio of corresponding sides.
2. Give a reason (AA, SAS or SSS) for every similarity you claim.

Question 7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Answer:
Let ABCD be a square of side a, ΔBQC the equilateral triangle on side BC and ΔAPC the equilateral triangle on the diagonal AC.
All equilateral triangles are similar (AAA), so ΔAPC ~ ΔBQC.
The diagonal AC = √2 × side = √2 BC, hence\[ \frac{ar(APC)}{ar(BQC)}=\frac{AC^2}{BC^2}=\frac{2BC^2}{BC^2}=2 \]
So ar(BQC) = ½ ar(APC). Hence proved.
In simple words: The diagonal of a square is √2 times its side, so the areas are in the ratio 2 : 1.

Teacher's Note:
1. Write the similarity statement with corresponding vertices in the same order, then use the ratio of areas equal to the square of the ratio of corresponding sides.
2. Give a reason (AA, SAS or SSS) for every similarity you claim.

Question 8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangles ABC and BDE is
(a) 2 : 1 (b) 1 : 2 (c) 4 : 1 (d) 1 : 4

Answer:
Correct option: (c)
Both triangles are equilateral, so ΔABC ~ ΔBDE. Since D is the mid-point of BC, BD = ½ BC.
\[ \frac{ar(ABC)}{ar(BDE)}=\frac{BC^2}{BD^2}=\frac{BC^2}{\left(\frac{BC}{2}\right)^2}=4 \]
The ratio is 4 : 1.

Teacher's Note:
1. Write the similarity statement with corresponding vertices in the same order, then use the ratio of areas equal to the square of the ratio of corresponding sides.
2. Give a reason (AA, SAS or SSS) for every similarity you claim.

Question 9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(a) 2 : 3 (b) 4 : 9 (c) 81 : 16 (d) 16 : 81
Answer:
Correct option: (d)
The areas of similar triangles are in the ratio of the squares of their sides: \( \left(\frac49\right)^2=\frac{16}{81} \).

Teacher's Note:
1. Write the similarity statement with corresponding vertices in the same order, then use the ratio of areas equal to the square of the ratio of corresponding sides.
2. Give a reason (AA, SAS or SSS) for every similarity you claim.