Chapter 6: Triangles

Key Theorems:
• BPT (Thales): Line parallel to one side divides other two proportionally
• AA Similarity: Two angles equal → similar triangles
• SSS Similarity: All corresponding sides proportional → similar
• SAS Similarity: Two sides proportional + included angle equal → similar
• Area ratio: Area(ΔABC)/Area(ΔPQR) = (AB/PQ)²

Exercise 6.1

Q1. Fill in blanks: (i) All circles are ___. (ii) All squares are ___. (iii) All ___ triangles are similar.

Solution

(i) similar (same shape)
(ii) similar (same shape)
(iii) equilateral triangles are similar

Q2. Give two examples: (i) similar figures (ii) non-similar figures

Solution

(i) Similar: Two equilateral triangles, two circles
(ii) Non-similar: A square and a rectangle, a triangle and a parallelogram

Q3. Is ΔABC (6,8,10) similar to ΔPQR (3,4,5)?

Solution

AB/PQ = 6/3 = 2
AC/PR = 8/5 = 1.6
BC/QR = 10/4 = 2.5
Ratios not equal → Not similar

Exercise 6.2

Q1. In ΔABC, DE || BC. AD=3.6cm, DB=5.4cm, AE=2.4cm. Find EC.

Solution

By BPT: AD/DB = AE/EC
3.6/5.4 = 2.4/EC
EC = 2.4 × 5.4/3.6 = 3.6 cm

Q2. DE || BC, AD/DB=3/5, AC=4.8cm. Find AE.

Solution

AD/DB = AE/EC = 3/5
AE=3k, EC=5k
3k+5k=4.8 → 8k=4.8 → k=0.6
AE=3(0.6)=1.8 cm

Q5. DE || BC. AD=x, DB=x-2, AE=x+2, EC=x-1. Find EC.

Solution

x/(x-2) = (x+2)/(x-1)
x(x-1) = (x+2)(x-2)
x²-x = x²-4
-x=-4 → x=4
EC = 4-1 = 3

Q7. AD/AB=AE/AC=3/8. Prove DE || BC.

Solution

AD/AB = AE/AC = 3/8
By converse of Basic Proportionality Theorem,
DE || BC.

Exercise 6.3

Q1. ΔABC: ∠A=60°, ∠B=80°. ΔPQR: ∠Q=60°, ∠R=40°. Are they similar?

Solution

ΔABC: ∠C = 180-60-80 = 40°
∠A=∠Q=60°, ∠C=∠R=40°
By AA similarity: ΔABC ~ ΔPQR

Q2. ΔDEF: DE=3cm, DF=6cm, EF=5cm. ΔPQR: PQ=5cm, PR=10cm, QR=7.5cm. Similar?

Solution

DE/PQ = 3/5 = 0.6
DF/PR = 6/10 = 0.6
EF/QR = 5/7.5 = 2/3 ≠ 0.6
∴ Not similar

Q6. In ΔABC, ∠B=90°, BD ⊥ AC. Prove ΔDBA ~ ΔDBC.

Solution

In ΔABC with ∠B=90°:
Let ∠A=α, then ∠C=90°-α
In ΔABD: ∠ADB=90°, ∠ABD=90°-α
In ΔBDC: ∠BDC=90°, ∠DBC=α
In ΔDBA and ΔDBC:
∠ADB=∠BDC=90°
∠ABD=90°-α=∠C
By AA: ΔDBA ~ ΔDBC

Exercise 6.4

Q1. ΔABC ~ ΔDEF, AB=2.5cm, DE=5cm, perimeter of ΔABC=20cm. Find perimeter of ΔDEF.

Solution

AB/DE = Perimeter(ABC)/Perimeter(DEF)
2.5/5 = 20/P
P = 20 × 5/2.5 = 40 cm

Q3. Similar triangles have sides ratio 3:5. Area of first is 81cm². Find area of second.

Solution

Area ratio = (3/5)² = 9/25
81/Area = 9/25
Area = 81 × 25/9 = 225 cm²

Q4. ΔABC ~ ΔDEF, BC=3cm, EF=4cm. Area(ΔABC)=12cm². Find Area(ΔDEF).

Solution

BC/EF = 3/4
Area(ABC)/Area(DEF) = (3/4)² = 9/16
12/Area = 9/16
Area = 12 × 16/9 = 64/3 cm²

Exercise 6.5

Q1. Which are right-angled? (i) 7,24,25 (ii) 3,8,6 (iii) 50,80,100 (iv) 13,12,5

Solution

(i) 7²+24²=49+576=625=25² ✓ Right-angled
(ii) 3²+6²=45≠64 ✗ Not right-angled
(iii) 50²+80²=8900≠10000 ✗ Not right-angled
(iv) 12²+5²=169=13² ✓ Right-angled

Q3. In ΔABC, AD ⊥ BC. Prove: AB²+CD²=BD²+AC².

Solution

In ΔABD: AB² = AD²+BD² ...(i)
In ΔACD: AC² = AD²+CD² ...(ii)
(i)-(ii): AB²-AC² = BD²-CD²
∴ AB²+CD² = BD²+AC²

Q7. AD ⊥ BC, BD=3CD. Prove 2AB²=2AC²+BC².

Solution

Let CD=x, BD=3x, BC=4x
AB²=AD²+9x² ...(i)
AC²=AD²+x² ...(ii)
(i)-(ii): AB²-AC²=8x²
AB²=AC²+8x²
2AB²=2AC²+16x²
BC²=16x²
∴ 2AB²=2AC²+BC²

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