NO LINKS. Answer the 33(b).
(b) The figure (ii) given below is a trapezium. Find 
(1) AB (ii) area of trapezium ABCD.​

 

Dear Student,



$$\begin{gathered} Draw a line from C to meet AD at E. \\ So, ABCE\hspace{0.17em}is a rec\tan gle with Length = 2 unit \\ And DEC\hspace{0.17em}is a right angle triangle. \\ Where, \\ DE = AD-AE \\ DE =8-2 =6 \\ Hypotenuse DC = 10 unit \\ So by pythagoras theorem, \\ Hypotenus{e}^2 = Bas{e}^2+Perpendicula{r}^2 \\ D{C}^2=C{E}^2+D{E}^2 \\ {10}^2=C{E}^2+6^2 \\ C{E}^2=100-36 \\ CE = \sqrt{64} \\ CE = 8 unit \\ Since ABCE is a rec\tan gle so, CE=AB \\ So, AB = 8 unit \\ Area of rec\tan gle\left(ABCE\right) = 8\times 2 = 16 uni{t}^2 \\ Area of triangle \left(DEC\right) = \frac{1}{2}\times base\times height \\ =\frac{1}{2}\times 8\times 6 = 24 uni{t}^2 \\ Area of trapezium = Area of rec\tan gle + Area of triangle \\ Area of trapezium = 24+16 \\ Area of trapezium = 40uni{t}^2 \end{gathered}$$

Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards