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 20 c m , 50 c m a n d 50 c m . 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

Area of the triangular piece

Area of triangular piece

Cloth required for each colour

Q-7 A kite in the shape of a square with a diagonal 32 c m and an isosceles triangle of base 8 c m a n d s i d e s 6 c m 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

Now A r e a o f s h a p e I = A r e a o f s h a p e I I

Thus, 512 2 c m 2 = 256 c m 2 (area of kite is 512 c m 2 )

So, area of paper required of each shade = 256 c m 2

For the section III length of the triangle of sides = 6 c m , 6 c m a n d 8 c m

Semi perimeter of triangle

Finally, area of the triangular III piece

s ( s − a ) ( s − b ) ( s − c )

10 ( 10 − 6 ) ( 10 − 6 ) ( 10 − 8 ) c m 2

10 × 4 × 4 × 2 c m 2

8 6 c m 2