NCERT Class 9 Solutions: Heron's Formula (Chapter 12) Exercise 12.2 Part 3

Area formula of rectangle,parallelogram and triangle

Area Formula of Rectangle,Parallelogram and Triangle

Area formula of rectangle,parallelogram and triangle

Q-4 A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26cm,28cmand30cm , and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Solution:

Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, D Triangle poly2 Triangle poly2: Polygon A, F, G Angle ? Angle ?: Angle between B, E, D Angle ? Angle ?: Angle between B, E, D Angle ? Angle ?: Angle between B, E, D Segment a Segment a: Segment [A, B] of Quadrilateral poly1 Segment b Segment b: Segment [B, C] of Quadrilateral poly1 Segment c Segment c: Segment [C, D] of Quadrilateral poly1 Segment d Segment d: Segment [D, A] of Quadrilateral poly1 Segment f Segment f: Segment [B, E] Segment g Segment g: Segment [A, F] of Triangle poly2 Segment a_1 Segment a_1: Segment [F, G] of Triangle poly2 Segment f_1 Segment f_1: Segment [G, A] of Triangle poly2 Point B B = (-1.2, 2.55) Point B B = (-1.2, 2.55) Point B B = (-1.2, 2.55) Point C C = (3.2, 2.58) Point C C = (3.2, 2.58) Point C C = (3.2, 2.58) 30cm text1 = "30cm" 26cm text2 = "26cm" h text3 = "h" 28cm text4 = "28cm"

Triangle and Parallelogram

Triangle and parallelogram. Side of the triangle are 26cm, 28cm and 30cm. Parallelogram base is on 28cm side.

Let’s find the area of the triangle, and then use the fact that triangle and base have same are to find the height of the parallelogram.

The sides of triangle are 26cm,28cmand30cm.

Therefore, semi perimeter = 842cm=42cm

Using Heron’s formula, its area is

  • =s(sa)(sb)(sc)cm2

  • =42(4226)(4228)(4230)cm2

  • =42×16×14×12cm2

  • =112896cm2

  • =336cm2

Given that, Areaofparallelogram=Areaoftriangle

  • b×h=336cm2

  • 28cm×h=336cm2

  • h=33628cm

  • h=12cm

Therefore, height of the parallelogram is 12cm.

Q-5 A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Solution:

Quadrilateral poly1 Quadrilateral poly1: Polygon P, Q, R, S Segment a Segment a: Segment [P, Q] of Quadrilateral poly1 Segment b Segment b: Segment [Q, R] of Quadrilateral poly1 Segment c Segment c: Segment [R, S] of Quadrilateral poly1 Segment d Segment d: Segment [S, P] of Quadrilateral poly1 Segment f Segment f: Segment [P, R] Segment g Segment g: Segment [Q, S] Point P P = (-1.36, 3.82) Point P P = (-1.36, 3.82) Point P P = (-1.36, 3.82) Point Q Q = (2.04, 3.84) Point Q Q = (2.04, 3.84) Point Q Q = (2.04, 3.84) Point R R = (1.44, 0.72) Point R R = (1.44, 0.72) Point R R = (1.44, 0.72) Point S S = (-1.98, 0.7) Point S S = (-1.98, 0.7) Point S S = (-1.98, 0.7) 30m text2 = "30m" 30m text1 = "30m" 30m text3 = "30m" 30m text4 = "30m" 48m text5 = "48m"

Rhombus PQRS With Given Dimensions

Rhombus PQRS, also two congruent triangles of equal area formed by its diagonals

Note that, diagonal PR divides the rhombus PQRS into two congruent triangles of equal areas. We find the area of rhombus by finding areas of each of the triangles.

Semi perimeter of ΔPQR = 30+30+482m=54m

Using heron's formula, area of the ΔPQR

  • =s(sa)(sb)(sc)

  • =54(5430)(5430)(5448)m2

  • =54×14×14×6m2

  • =432m2

Since the two triangles are congruent area of field =2×areaoftheΔPQR = (2×432)m2=864m2

Finally there are 18 cows, therefore area of grass field for each cow = 86418m2=48m2

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