Dear Student, In right angle triangle ADB, Using pythagorus theoremBD2= AD2+AB2AD=412-402= 9 cm Now since AD is the perpendicular distance between AB and CD , so here area of trapezium ABCD= 12×ADperpendicular distance between AB and CD×sum of parallel sides AB and CD= 12×9×55=247 5 cm2Area of triangle BCD =area of trapezium ABCD- area of right triangle ADB=247 5 -12×ADheight×ABbase=247 5 -12×40×9=247 5 -180=67 5 cm2

Please solve no. c

Dear Student, In right angle triangle ADB, Using pythagorus theorem {BD}^{2} = {AD}^{2} + {AB}^{2} AD = \sqrt{41^{2} - 40^{2}} = 9 cm Now since AD is the p e r p e n d i c u l a r distance between AB and CD, so here area of trapezium ABCD = \frac{1}{2} \times AD (p e r p e n d i c u l a r distance between AB and CD) \times (sum of parallel sides AB and CD) = \frac{1}{2} \times 9 \times 55 = 247.5 {cm}^{2} Area of triangle BCD = area of trapezium ABCD - area of right triangle ADB = 247.5 - \frac{1}{2} \times AD (height) \times AB (base) = 247.5 - \frac{1}{2} \times 40 \times 9 = 247.5 - 180 = 67.5 {cm}^{2}

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