NCERT Class 9 Solutions: Heron's Formula (Chapter 12) Exercise 12.2 Part 1

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Herons formula with semi-perimeter

Herens Formula in Two Parts

Herons formula with semi-perimeter

Q-1 A park, in the shape of a quadrilateral ABCD, has B=90°,CA=9m,AB=12m,BD=5mandCD=8m . How much area does it occupy?

Solution:

Angle ? Angle ?: Angle between D, B, A Angle ? Angle ?: Angle between D, B, A Angle ? Angle ?: Angle between D, B, A Segment f Segment f: Segment [A, B] Segment g Segment g: Segment [A, C] Segment h Segment h: Segment [C, D] Segment i Segment i: Segment [D, B] Segment j Segment j: Segment [D, A] Point A A = (-1.2, 0.84) Point A A = (-1.2, 0.84) Point A A = (-1.2, 0.84) Point B B = (3.04, 0.84) Point B B = (3.04, 0.84) Point B B = (3.04, 0.84) Point C C = (-0.5, 4.24) Point C C = (-0.5, 4.24) Point C C = (-0.5, 4.24) Point D D = (3.04, 3.08) Point D D = (3.04, 3.08) Point D D = (3.04, 3.08) 12m text1 = "12m" 9m text2 = "9m" 8m text3 = "8m" 5m text4 = "5m"

Quadrilateral ABCD

Quadrilateral ABCD, ∠B=90°,CA=9m,AB=12m,BD=5mandCD=8

Given a quadrilateral ABCD with, B=90°,CA=9m,AB=12m,BD=5mandCD=8m .

To find the area of quadrilateral we can add the areas of two trianlges. In ΔABD , by applying Pythagoras theorem,

  • AD2=AB2+BD2

  • AD2=122+52

  • AD2=169

  • AD=13m

Also since triangle ΔABD is right triangle, Area of ΔABD=12×12×5=30m2 Now, semi perimeter of ΔCAD 8+9+132m=302m=15m

Using heron's formula, area of ΔCAD

  • s(sa)(sb)(sc)

  • 15(1513)(159)(158)m2

  • 15×2×6×7m2

  • 635m2

  • 35.49m2 (approx.)

Area of quadrilateral ABCD=AreaofΔABD+AreaofΔCAD=30m2+35.49m2=65.49m2

Q-2 Find the area of a quadrilateral ABCD in which CD=3cm,CB=4cm,CD=4cm,AD=5cmandBD=5cm.

Solution:

Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, 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, D] Segment g Segment g: Segment [A, C] Point A A = (-2.06, 4.18) Point A A = (-2.06, 4.18) Point A A = (-2.06, 4.18) Point B B = (2.58, 4.18) Point B B = (2.58, 4.18) Point B B = (2.58, 4.18) Point C C = (2.58, 0.14) Point C C = (2.58, 0.14) Point C C = (2.58, 0.14) Point D D = (-0.78, 0.12) Point D D = (-0.78, 0.12) Point D D = (-0.78, 0.12) 4cm text1 = "4cm" 4cm text2 = "4cm" 3cm text3 = "3cm" 5cm text4 = "5cm" 5cm text5 = "5cm"

Quadrilateral ABCD

Quadrilateral ABCD with CD=3cm,CB=4cm,CD=4cm,AD=5cmandBD=5cm.

Given a quadrilateral ABCD, CD=3cm,CB=4cm,CD=4cm,AD=5cmandBD=5cm.

We can calculate the area of the two triangles by using heroes’ formula on the two triangles. However we can make things simple by first proving that that ΔDCB is a right triangle. Let’s see if it satisfies Pythagoras theorem,

  • BD2=DC2+CB2

  • 52=32+42

  • 25=9+16

  • 25=25

Thus, ΔDCB is a right angled at C. Area of ΔCBA

  • 12×3×4=6cm2

Now, semi perimeter of ΔDBA =5+5+42cm=142cm=7m

Using heron's formula, area of ΔDCA

  • =s(sa)(sb)(sc)

  • =7(75)(75)(74)cm2

  • =7×2×2×3cm2

  • =221cm2

  • 9.17cm2 (approx.)

Finally, area of quadrilateral

  • ABCD=AreaofΔDCB+AreaofΔDCA=6cm2+9.17cm2=15.17cm2

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