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

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

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, AEDC and CFAD .If AB=16cm , AE=8cm and CF=10cm , find AD .

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,

  • ABCD is a parallelogram, AEDC and CFAD

  • Also AB=CD=16cm (opposite side of a parallelogram)

  • CF=10cm and AE=8cm

Now, Areaofparallelogram=Base×Altitide

  • CD×AE=AD(base)×CF(Altitide) ( Areaofparallelogram=base×height )

  • 16×8=AD×10 ( AB=CD=16cm)

  • 128=10AD

  • AD=12.8cm

Q-2 If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH)=12ar(ABCD).

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, ar(EFGH)=12ar(ABCD)

  • Construction, H and F are joined.

Proof,

  • ADBC and AD=BC (opposite sides of a parallelogram) so 12AD=12BC

  • Also, AHBF and DHCF therefore AH=BF and DH=CF (H and F are mid points) Thus, ABFH and HFCD are parallelograms.

  • Now, EFH And parallelogram ABFH lie on the same base FH and between the same parallel lines AB and HF.

  • So, area of EFH=12areaofABFH ……….equation (1) (this is because area of triangle is 12bh and area of parallelogram is bh and the triangle and parallelogram share same base and height),

  • Similarly, area of GHF=12areaofHFCD ……..equation (2)

Adding equation (1) and (2)

  • Area of EFH+areaofGHF=12 area of ABFH+12areaofHFCD

  • AreaofEFGH=12[areaofABFH+areaofHFCD]

  • Area(EFGH)=12area(ABCD) (ABFH+HFCD=ABCD)

  • ar(EFGH)=12ar(ABCD)

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