Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles NCERT Solutions: All Exercises Solved Step-by-Step

Introduction

Chapter 9 finally brings areas into geometry after all those proofs. The big ideas are simple: parallelograms with same base and height have equal areas, same for triangles. You'll also see how triangles between same parallel lines have equal areas.

Two exercises with diagrams - draw them carefully as heights matter. Key for coordinate geometry next chapter.

Key Formulas

Parallelogram Area: base × corresponding height
Triangle Area: (1/2) × base × height
Key Theorems:

Same base + same height → equal areas
Same parallelograms cut from same triangle → equal areas
Triangles with same base between parallel lines → equal areas
Parallelograms between same parallel lines → equal areas

Exercise 9.1 Solved Questions

Question 1: Parallelograms ABCD, EFGH. AB = EF, heights same. Prove equal areas.

Solution: Area = base × height. AB = EF (base), h1 = h2 (height). Areas equal.

Question 2: Triangles ABC, ABD same base AB, heights from C,D to AB equal. Prove equal areas.

Solution: Area = (1/2) base × height. AB common, heights equal → areas equal.

Question 3: Parallelograms ABCD, PBCQ on same base BC. Prove equal areas.

Solution: Both height from A,D and P,Q to BC same line → equal heights, base BC common.

Question 4: From vertex A of triangle ABC, draw parallels to BC meeting AB,AC. Prove equal areas small triangles.

Solution: All between same parallels BC and line from A → equal areas.

Question 5: Fig 9.17: Show parallelograms equal areas.

Solution: Same base or same height properties apply.

Exercise 9.2 Solved Questions

Question 1: Parallel lines l1,l2. Triangles same base between them equal area?

Solution: Yes. Height between parallels fixed, base same → equal areas.

Question 2: Parallelograms between same parallels, same base equal area.

Solution: Height (distance between parallels) fixed, base same.

Question 3: Fig 9.24: PQ||BC. Prove ar(ΔAPQ) = (1/2) ar(ΔABC).

Solution: Same height from A to BC line, base PQ = (1/2) BC → half area.

Question 4: BE||AC meets DC at E. Prove ar(ΔABE) = ar(ΔCBE).

Solution: Same height from B to AC extended, bases AE=EC → equal.

Question 5: Fig 9.28: DE||CA, EF||PB. Prove ar(ΔDEF) = (1/4) ar(ΔPQR).

Solution: Each parallel cuts half → quarter area.

Question 6: Parallelogram ABCD, E,F on AB,DC. EF||AD prove equal areas triangles.

Solution: Midline theorem → EF parallel, half → equal areas.

Question 7: Fig 9.33: Prove ar(quad ABCD) = ar(ΔPBC) + ar(ΔADC).

Solution: Common vertex areas add up.

Extra Practice Questions

Extra Q1: Triangle base 10cm, height 6cm. Area? 30 cm².

Extra Q2: Parallelogram base 8cm, height 4cm. Area? 32 cm².

Extra Q3: Two triangles same base 12cm, heights 5cm,7cm. Area ratio? 5:7.

Extra Q4: Rectangle 6×4 cm parallelogram? Area 24 cm² same.

15 FAQs

Parallelogram area formula? Base × height.
Triangles same base height? Equal areas.
Parallel lines triangles? Equal areas same base.
Height means? Perpendicular distance.
PQ||BC half base? Half area triangle.
Same parallelograms cut? Equal areas.
Base common, vertex line? Equal areas.
Distance between parallels? Fixed height.
Triangle area half? Parallelogram same base height.
Midline theorem area? Quarter total.
Rectangle special parallelogram? Yes.
Rhombus area formula? Base × height.
Figures between parallels? Proportional areas.
Common vertex property? Areas add directly.
Scale factor areas? Square of linear.

Areas chapter connects all previous geometry. Pop over to www.fuzymathacademy.com for my diagram videos where I draw everything live plus practice worksheets. You'll ace Chapter 9 easily.