A) \[52\text{ }{{m}^{2}}\]
B) \[46\text{ }{{m}^{2}}\]
C) \[20\text{ }{{m}^{2}}\]
D) \[50\text{ }{{m}^{2}}\]
Correct Answer: B
Solution :
Ans. In \[\Delta \,ABC,\]\[\angle ACB={{90}^{o}}\] So, \[BC=\sqrt{A{{B}^{2}}-A{{C}^{2}}}\] \[=\sqrt{{{5}^{2}}-{{4}^{2}}}\] (AC= ED) \[=\sqrt{25-16}\] =3 cm \[\therefore \] Area of figure = Area of \[\Delta ABC+\] Area of rectangle ACDE \[=\frac{1}{2}\times 3\times 4+4\times 10\] \[=6+40={{46}^{2}}\]You need to login to perform this action.
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