question_answer

Two straight poles of unequal length stand upright on a ground. The length of the shorter pole is 10 m. A pole joins the top of the two vertical poles. The distance between the two tops is 5 m. The distance between the poles along the ground is 4 m. The area thus formed by the three poles with the ground is

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 \[=\frac{1}{2}\times 3\times 4+4\times 10\] \[=6+40={{46}^{2}}\]