question_answer

Find the area of quadrilateral ABCD in which AB = 9 cm, 6C = 40 cm, CD = 28 cm, DA = 15 cm and \[\angle ABC={{90}^{o}}.\]

A) \[~300\,c{{m}^{2}}\]                         

B)        \[~180\,c{{m}^{2}}\]             

C)        \[~126\,c{{m}^{2}}\]             

D)        \[306\,c{{m}^{2}}\]

Correct Answer: D

Solution :

In right angled\[\Delta \Alpha \Beta C,\]using Pythagoras theorem \[{{(AC)}^{2}}={{(9)}^{2}}+{{(40)}^{2}}=1681\] \[\Rightarrow \]\[AC=41\,cm\] \[\therefore \]Area of \[\Delta \Alpha \Beta C=\frac{1}{2}\times 9\times 40=180\,c{{m}^{2}}\] Now, in \[\Delta \Alpha DC\] \[s=\frac{15+28+41}{2}=42\,cm\] \[\therefore \]Area of \[\Delta \Alpha DC=\sqrt{42\times 27\times 14\times 1}\] \[=126\,c{{m}^{2}}\] Hence, area of quadrilateral ABCD \[=(180+126)c{{m}^{2}}=306\,c{{m}^{2}}\]