Ch 6: Triangles (Old Syllabus Topics)
Covers the Areas of Similar Triangles Theorem and Pythagoras Theorem proofs from Old Exercises 6.4 & 6.5.
Exercise 6.4 (Areas of Similar Triangles)
💡 Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Q1. Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Solution:
Given: ∆ABC ~ ∆DEF. Therefore, Area(∆ABC) / Area(∆DEF) = (BC / EF)².
Substituting the values:
64 / 121 = (BC / 15.4)²
Taking square root on both sides:
8 / 11 = BC / 15.4
BC = (8 × 15.4) / 11
BC = 8 × 1.4 = 11.2 cm.
Q2. 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.
Solution:
In ∆AOB and ∆COD:
∠AOB = ∠COD (Vertically opposite angles)
∠OAB = ∠OCD (Alternate interior angles, since AB || DC)
Therefore, by AA similarity, ∆AOB ~ ∆COD.
Using the Area Theorem of similar triangles:
Area(∆AOB) / Area(∆COD) = (AB / CD)²
Given AB = 2CD, so AB/CD = 2/1.
Ratio = (2 / 1)² = 4 / 1.
The ratio of the areas of triangles AOB and COD is 4:1.
Q8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is:
Solution:
All equilateral triangles are similar. Hence, ∆ABC ~ ∆BDE.
Since D is the midpoint of BC, BD = BC / 2.
Therefore, BC / BD = 2 / 1.
Area(∆ABC) / Area(∆BDE) = (BC / BD)²
= (2 / 1)² = 4 / 1
The correct option is 4:1.
Exercise 6.5 (Pythagoras Theorem)
Q4. ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:
Given: In ∆ABC, ∠C = 90° and it is isosceles, which means legs are equal: AC = BC.
By Pythagoras Theorem in right ∆ABC:
Hypotenuse² = Base² + Perpendicular²
AB² = AC² + BC²
Since AC = BC, we replace BC with AC:
AB² = AC² + AC²
AB² = 2AC² (Hence Proved)
Q13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².
Proof:
Apply Pythagoras Theorem in four right triangles formed at C:
In ∆ACE: AE² = AC² + CE² --- (1)
In ∆BCD: BD² = BC² + CD² --- (2)
In ∆ABC: AB² = AC² + BC² --- (3)
In ∆DCE: DE² = CD² + CE² --- (4)
Adding (1) and (2):
AE² + BD² = AC² + CE² + BC² + CD²
Rearranging the terms:
AE² + BD² = (AC² + BC²) + (CD² + CE²)
Substituting from (3) and (4):
AE² + BD² = AB² + DE² (Hence Proved)
Q15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD² = 7AB².
Proof:
Let the side of equilateral ∆ABC be a. (So AB = BC = AC = a).
Draw an altitude AE ⊥ BC. In an equilateral triangle, the altitude bisects the base.
So, BE = EC = a/2.
Given: BD = 1/3 BC = a/3.
Length of DE = BE - BD = a/2 - a/3 = a/6.
By Pythagoras Theorem in right ∆AEB:
AE² = AB² - BE²
AE² = a² - (a/2)² = a² - a²/4 = 3a²/4.
Now, by Pythagoras Theorem in right ∆AED:
AD² = AE² + DE²
AD² = (3a²/4) + (a/6)²
AD² = 3a²/4 + a²/36
Taking LCM of 4 and 36 (which is 36):
AD² = (27a² + a²) / 36
AD² = 28a² / 36
AD² = 7a² / 9
Cross-multiplying by 9:
9AD² = 7a²
Since a = AB, we get: 9AD² = 7AB² (Hence Proved).