NCERT Class 9 Solutions: Heron's Formula (Chapter 12) Exercise 12.2 Part 4

Important formulas for polygons

Polygons Name and Areas

Important formulas for polygons

Q-6 An umbrella is made by stitching 10 triangular pieces of cloth of two different colors (see Figure), each piece measuring 20cm,50cmand50cm . How much cloth of each colour is required for the umbrella?

Triangle poly1 Triangle poly1: Polygon A, B, C Triangle poly2 Triangle poly2: Polygon C, A, D Triangle poly3 Triangle poly3: Polygon D, E, A Triangle poly4 Triangle poly4: Polygon B, F, A Triangle poly5 Triangle poly5: Polygon F, G, A Arc k Arc k: CircumcircularArc[I, J, K] Segment c Segment c: Segment [A, B] of Triangle poly1 Segment a Segment a: Segment [B, C] of Triangle poly1 Segment b Segment b: Segment [C, A] of Triangle poly1 Segment d Segment d: Segment [C, A] of Triangle poly2 Segment a_1 Segment a_1: Segment [D, C] of Triangle poly2 Segment a_2 Segment a_2: Segment [D, E] of Triangle poly3 Segment d_1 Segment d_1: Segment [E, A] of Triangle poly3 Segment e Segment e: Segment [A, D] of Triangle poly3 Segment a_3 Segment a_3: Segment [B, F] of Triangle poly4 Segment b_1 Segment b_1: Segment [F, A] of Triangle poly4 Segment f Segment f: Segment [A, B] of Triangle poly4 Segment a_4 Segment a_4: Segment [F, G] of Triangle poly5 Segment f_1 Segment f_1: Segment [G, A] of Triangle poly5 Segment g Segment g: Segment [A, F] of Triangle poly5 Segment h Segment h: Segment [H, I] Point A A = (1.55, 4.17) Point A A = (1.55, 4.17) 50cm text1 = "50cm" 20cm text2 = "20cm"

an Umbrella

An umbrella is made by stitching 10 triangular pieces of cloth of two different colors

Solution:

Semi perimeter of each triangle

  • 50+50+202cm

  • 1202cm=60cm

Area of the triangular piece

  • s(sa)(sb)(sc)

  • 60(6050)(6050)(6020)cm2

  • 60×10×10×4cm2

  • 2006cm2

Area of triangular piece

  • 10×2006cm2=20006cm2

Cloth required for each colour

  • 200062cm2

  • 10006cm2

Q-7 A kite in the shape of a square with a diagonal 32cm and an isosceles triangle of base 8cmandsides6cm each is to be made of three different shades as shown in Figure. How much paper of each shade has been used in it?

Quadrilateral poly1 Quadrilateral poly1: Polygon C, A, D, B Triangle poly2 Triangle poly2: Polygon E, F, G Triangle poly3 Triangle poly3: Polygon H, I, J Segment f Segment f: Segment [A, B] Segment g Segment g: Segment [C, D] Segment h Segment h: Segment [C, A] Segment i Segment i: Segment [A, D] Segment j Segment j: Segment [D, B] Segment k Segment k: Segment [B, C] Segment c Segment c: Segment [C, A] of Quadrilateral poly1 Segment a Segment a: Segment [A, D] of Quadrilateral poly1 Segment d Segment d: Segment [D, B] of Quadrilateral poly1 Segment b Segment b: Segment [B, C] of Quadrilateral poly1 Segment g_1 Segment g_1: Segment [E, F] of Triangle poly2 Segment e Segment e: Segment [F, G] of Triangle poly2 Segment f_1 Segment f_1: Segment [G, E] of Triangle poly2 Segment j_1 Segment j_1: Segment [H, I] of Triangle poly3 Segment h_1 Segment h_1: Segment [I, J] of Triangle poly3 Segment i_1 Segment i_1: Segment [J, H] of Triangle poly3 I text1 = "I" II text2 = "II" III text3 = "III" 8 cm text4 = "8 cm"

Kite in the Shape of Square

Kite in the shape of square with a diagonal 32cm and isosceles triangle of base 8cmandside6cm

Solution:

We know that the diagonals of a square bisect each other at right angle.

Area of given kite

  • 12(diagonal)2

  • 12×32×32=512cm2 ( diagonal=32 )

Now AreaofshapeI=AreaofshapeII

Thus, 5122cm2=256cm2 (area of kite is 512cm2)

So, area of paper required of each shade =256cm2

For the section III length of the triangle of sides =6cm,6cmand8cm

Semi perimeter of triangle

  • 6+6+82cm

  • 10cm

Finally, area of the triangular III piece

  • s(sa)(sb)(sc)

  • 10(106)(106)(108)cm2

  • 10×4×4×2cm2

  • 86cm2

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