Question

A lamina is in the shape of an isosceles trapezium. the parallel sides are 5m apart and their lengths are 3m and 2m The distance of the centre of the gravity of the trapezium form the longer of the parallel sides is :

• A. $\frac{8}{3}m$
• B. $\frac{8}{15}$
• C. 6m
• D. $\frac{7}{3}$

Answer 1

The correct option is D $\frac{7}{3}$

y co-ordinate of centre of man is $y_{cm}=\frac{A_1{y}_1+A_2{y}_2+A_3{y}_3}{A_1+A_2+A_3}$

Area of triangle ABC$\left(A_1\right)=$Area of triangle DEF$\left(A_2\right)=\frac{1}{2}\times \frac{1}{2}\times 5=\frac{5}{4}{m}^2$

Area of rectangle ACDE$\left(A_3\right)$ is $=2\times 5=10m^2$

y co-ordinate centre of mass of triangle is $=\frac{h}{3}$ [h is perpendicular dropped from one at its vertex on opposite base]

$h=5$m

$y_1=y_2=\frac{5}{3}$

$y_3=\frac{5}{2}$

$y=\frac{A_1{y}_1+A_2{y}_2+A_3{y}_3}{A_1+A_2+A_3}$

$=\frac{\frac{5}{4}\times \frac{5}{3}+\frac{5}{4}\times \frac{5}{3}+10\times \frac{5}{2}}{\frac{5}{4}+\frac{5}{4}+10}$

$=\frac{\frac{5}{4}+25}{\frac{10}{4}+10}$

$\therefore \frac{135}{75}=\frac{7}{3}m$

$\therefore y=\frac{7}{3}m$.