NCERT Class 9 Solutions: Areas of Parallelograms and Triangles (Chapter 9) Exercise 9.2 – Part 1

the Area of a Parallelogram a=Bh

The area of a parallelogram it is equal to the product of the base the height,A=bh

Q-1 In the figure, ABCD is a parallelogram, A E ⊥ D C and C F ⊥ A D .If A B = 16 c m , A E = 8 c m and C F = 10 c m , find A D .

Segment f Segment f: Segment [A, B] Segment g Segment g: Segment [B, C] Segment j Segment j: Segment [A, D] Segment k Segment k: Segment [D, C] Segment l Segment l: Segment [A, E] Segment m Segment m: Segment [C, F] Point A A = (-0.42, 4.28) Point A A = (-0.42, 4.28) Point A A = (-0.42, 4.28) Point B B = (-2.16, 1.78) Point B B = (-2.16, 1.78) Point B B = (-2.16, 1.78) Point C C = (3.22, 1.78) Point C C = (3.22, 1.78) Point C C = (3.22, 1.78) Point D Point D: Intersection point of h, i Point D Point D: Intersection point of h, i Point D Point D: Intersection point of h, i Point E Point E: Point on g Point E Point E: Point on g Point E Point E: Point on g Point F Point F: Point on f Point F Point F: Point on f Point F Point F: Point on f Parallelomgram ABCD

Parallelomgram ABCD also AE⊥DC andCF⊥AD.

Solution:

Given,

Now, A r e a o f p a r a l l e l o g r a m = B a s e × A l t i t i d e

Q-2 If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar ( E F G H ) = 1 2 a r ( A B C D ) .

Solution:

Segment k Segment k: Segment [E, H] Segment l Segment l: Segment [H, G] Segment m Segment m: Segment [E, F] Segment n Segment n: Segment [F, G] Segment p Segment p: Segment [A, B] Segment q Segment q: Segment [B, C] Segment r Segment r: Segment [C, D] Segment s Segment s: Segment [A, D] Segment t Segment t: Segment [H, F] Segment a Segment a: Segment [E, G] Segment a Segment a: Segment [E, G] Point A A = (-1.54, 4.2) Point A A = (-1.54, 4.2) Point A A = (-1.54, 4.2) Point B B = (2.96, 4.2) Point B B = (2.96, 4.2) Point B B = (2.96, 4.2) Point C C = (3.8, 2.06) Point C C = (3.8, 2.06) Point C C = (3.8, 2.06) Point D Point D: Point on i Point D Point D: Point on i Point D Point D: Point on i Point E Point E: Point on f Point E Point E: Point on f Point E Point E: Point on f Point H Point H: Point on j Point H Point H: Point on j Point H Point H: Point on j Point G Point G: Point on i Point G Point G: Point on i Point G Point G: Point on i Point F Point F: Point on h Point F Point F: Point on h Point F Point F: Point on h Parallelogram ABCD

Parallelogram ABCD also E,F,G,H are respectivly the mid-point of the side

Given, E, F, G and H are respectively the mid-point of the sides of a parallelogram ABCD

To prove, a r ( E F G H ) = 1 2 a r ( A B C D )

Construction, H and F are joined.

Proof,

A D ∥ B C and A D = B C (opposite sides of a parallelogram) so 1 2 A D = 1 2 B C

Also, A H ∥ B F and D H ∥ C F therefore A H = B F and D H = C F (H and F are mid points) Thus, ABFH and HFCD are parallelograms.

Now, △ E F H And parallelogram ABFH lie on the same base FH and between the same parallel lines AB and HF.

So, area of E F H = 1 2 a r e a o f A B F H ……….equation (1) (this is because area of triangle is 1 2 b h and area of parallelogram is b h and the triangle and parallelogram share same base and height),

Similarly, area of G H F = 1 2 a r e a o f H F C D ……..equation (2)

Adding equation (1) and (2)

Area of △ E F H + a r e a o f △ G H F = 1 2 area of A B F H + 1 2 a r e a o f H F C D

A r e a o f E F G H = 1 2 [ a r e a o f A B F H + a r e a o f H F C D ]

A r e a ( E F G H ) = 1 2 a r e a ( A B C D ) ( ∵ A B F H + H F C D = A B C D )

a r ( E F G H ) = 1 2 a r ( A B C D )