Areas of parallelograms and triangles-CLASS IX

1. Areas of parallelograms and triangles-CLASS IX PREPARED BY P.E.VENUGOPALAN, K.V.KELTRON NAGAR

2. Part of the plane enclosed by a simple closed figure is called a planar region Magnitude or measure of this planar region is called its area Area is expressed with the help of a number in some units Commonly used units of area are sq.cm (cm2) sq.m (m2), hectares etc Area

3. Two figures are called congruent if they have the same shape and same size Congruent figures B A

4. If two figures A and B are congruent ,they have equal area If A and B are two congruent figures, then ar(A) = ar(B) Two figures having equal area may not be congruent Areas of congruent figures B A

5. If a planar region is formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q then ar(T) = ar(P)+ar(Q) Area of a planar region P Q T

6. Area of a parallelogram Area = length x breadth Area = base x height height Base Area of a parallelogram = base x height

7. Area of a Triangle = ½ x base x height Area of a triangle height base height base

8. Figures on the same base

9. Figures between the same parallels

10. Figures on the same base and between the same parallels Rectangle and trapezium on the same base base base Rectangle and triangle on the same base

11. Parallelograms on the same base and between the same parallels are equal in area Theorem 9.1 A E B F D base C

12. Given :ABCD and EFCD are two parallelograms on the same base DC and between the same parallels To prove: ar(ABCD) = ar(EFCD) Proof: In ΔADE and ΔBCF ∟DAE =∟CBF (why) ∟AED = ∟BFC (why) therefore ∟ADE = ∟BCF (why) Also AD = BC (why) so ΔADE ≡ΔBCF (ASA) Now ar(ABCD) = ar(ADE) +ar(EDCB) = ar(BCF) + ar(EDCB) (why) = ar(EFCD) So parallelograms ABCD and EFCD are equal in area proof

13. Two triangles on the same base and between the same parallels are equal in area Theorem 9.2 base Area of a Triangle = ½ x base x height