Question

$ABCD$ is a trapezium $\overline{AB}\parallel \overline{CD}$ If $AB=20cmBC=8cm,CD=10cm$ and $AD=6cm.$ Find the area of $ABCD.$

Answer 1

R.E.F. Image.

ABCD is trapezium. Draw CM.|| AD.

In $\Delta$ CMB ; 6, 8, 10 cm.

$8=\frac{a+b+c}{2}=\frac{6+8+10}{2}=12cm$ a = 6 b = 8 c = 10

$\Delta =\frac{1}{2}\times b\times h=\sqrt{S(S-a)(S-b)(S-c)}$

$=\frac{1}{2}\times 8\times h=\sqrt{12(12-6)(12-8)(12-10)}$

$=4h=\sqrt{\mathrm{12.6.4.2}}=\sqrt{576}=24$

$x=h=6$

Area of trapezium = Area of AMCD + Area of $\Delta$ MBC

$=10\times 6+24=84c{m}^2$