Question

The parallel sides of a trapezium are 20cm and 41cm and the remaining two sides are 10cm and 17cm . if the area of the trapezium =61K,then K=?

Answer 1

Area of trapezium = 1/2 h (a +b) where a and b are the parallel sides.
61 K = 1/2 h ( 20+ 41)
61 K = 1/2 h (61)
61 K /61 = 1/2 h
K = 1/2 h.

now, we need to find the height, 'h' by using the pythogoras theorum for the triangles AED and BFC.

Answer 2

Given:

The parallel sides of a trapezium are 20cm and 41cm

The remaining two sides are 10cm and 17cm

Area of the trapezium =61K

To Find:

The value of K

Solution:

Area of the Trapezium is Height×(sum of its parallel sides)/2

Putting the values we get

Area of Trapezium= $\frac{H(20+41)}{2}$

where H is the height of the Trapezium.

⇒ $\frac{H(20+41)}{2}$=61K

⇒$\frac{61H}{2}$=61K

⇒K=$\frac{H}{2}$

Now, we will find the height of the Trapezium using the Pythagoras theorem.

Height of the trapezium,

⇒$\sqrt{17^{2} -x^{2} }$=$\sqrt{10^{2}-(21-x)^{2} }$

On solving we get,

⇒$x$= 15

⇒Height=$\sqrt{17^{2} -x^{2} }$

⇒height=$\sqrt{17^{2} -15^{2} }$

⇒Height = 8

Now,

⇒K=H/2

⇒K=8/2

⇒K=4

Hence, the value of K is 4