Telanagana SCERT Solution Class VIII (8) Math Chapter 9 Area of Plane Figures Exercise 9.1
Area of the triangle ½ X 2 X 4 = 4 cm2
Area of the square = 4X4 =16 cm2
Therefore, Total area =4+16 =20 cm2
(ii)
Area of the trapezium = 18+7/2 X 8
= 25/2 X 8 =100 cm2
Area of the square =18 X 18 =324 cm2
Total area =324+100 =424 cm2
(iii)
Area of the trapezium = 6+15/2 X 8 =21 X 4 =84 cm2
Area of the square = 20 X 15 =300 cm2
Total area = 300+84 =384 cm2
(3)
Area of the quadrilateral = ½ x AC X 5 + ½ X AC X 6
= ½ X 10 X 5 + ½ x 10 X 6
= 25+30
= 55 cm2
(4)
Width =28-24 =4 cm
Width is save from each section
Area of 1 = ½ X (20+28) X 4 =96 cm2
Area of 2 =96 cm2
Area of 3 = ½ X (24+16) X 4= 80 cm2
Area of 4 = 80 cm2
(5)
(i) Area of DEI = ½ 40 X 60= 1200 m2
Area of EFGI = ½ (500+60) X 70 =3850 m2
Area of DHC = ½ X 40 X 80 =1600 m2
Area of AFG = ½ X 50 X 50 =1250 m2
Area of GCBA = ½ (40+30) X 80 =2800 m2
Total area = 1200+3850+1600+1250+2800 m2
= 10700 m2
(iii)
Area of FEH = ½ X 40 X 20= 400 m2
Area of FGIH = ½ (40+20) X 80 = 2400 m2
Area of GAI = ½ X 70 X 40 =1400 m2
Area of EDCI = ½ (50+40) X 80 =3600 m2
Area of BCIJ = ½ (40+30) X 60 =2100 m2
Area of ABK = ½ X 30 X 50 =150 m2
Total Area = 400+2400+1400+3600+2100+750 m2
= 10650 m2
(6) Let the length of the parallel sides of the trapezium 5x and 3x
Then, ½ (5x+3x) X 16 =960
= 8x X 8= 960
= x = 960/8X8
= x = 15
Then the length be 15X5 = 75 cm and 3X15 =45 cm.
(7) Diagonals are 45 cm and 30 cm
Then area of the rhombus = ½ X 45 X 30=675 cm2
= 675/10,000 m2
Area of 3000 tile = 675/10,000 X 3000 =67.5X3 m2
= 202.5 m2
Then total cost will be 202.5X20 = RS 4050
what about 8th solution in this exercise?