question_answer

In a trapezium ABCD, AB and DC are parallel sides and \[\angle ADC=90{}^\circ .\]If AB = 15 cm, CD = 40 cm and diagonal AC =41 cm, then the area of the trapezium ABCD is [SSC CGL Tier II, 2017]

A) \[245\text{ }c{{m}^{2}}\]

B) \[240\text{ }c{{m}^{2}}\]

C) \[247.5\text{ }c{{m}^{2}}\]

D) \[250\text{ }c{{m}^{2}}\]

Correct Answer: C

Solution :

[c] In the adjoining figure,.                         In right angled \[\Delta ADC,\]\[AC=41\,cm,\]\[DC=40\,cm\] \[\because \]       \[{{(AC)}^{2}}={{(AD)}^{2}}+{{(DC)}^{2}}\] \[\therefore \]      \[{{(AD)}^{2}}={{(AD)}^{2}}+{{(DC)}^{2}}\] \[{{(AD)}^{2}}={{(41)}^{2}}-{{(40)}^{2}}=1681-1600\] \[{{(AD)}^{2}}=81\]\[\Rightarrow \]\[AD=\sqrt{81}\] \[\Rightarrow \]\[AD=9\,cm\] (which is height of trapezium ABCD) \[\therefore \]Area of trapezium \[ABCD=\frac{1}{2}\times \] (Sum of parallel lines) \[\times \](Height) \[=\frac{1}{2}\times (AB+CD)\times AD\] \[=\frac{1}{2}\times (15+40)\times 9\] \[=\frac{1}{2}\times 55\times 9=247.5\,c{{m}^{2}}\]