A trapezium ABCD is such that AB//CD and the perpendicular from A to CD is 40cm.if AB=20cm,DA=50cm and CD=60cm.calculate the (i)area of the trapezium (ii)<BCD.pls help me with (ii)
solution so far.(i)area=1/2(20 60)40
=1600cm^2
Let P be the base of the perpendicular from A, and Q the base of the perpendicular from B.
APD is a 30-40-50 right triangle.
So, CD=30, making AP=30 and QD=10.
Triangle CQB thus has height 40 and leg CQ=50. So, tan(BCD) = 4/5
To find the measure of angle BCD, we can use the fact that opposite angles in a trapezium are supplementary.
First, let's draw a diagram to visualize the trapezium ABCD:
A _________ B
| |
| |
| |
|__________|
D C
Since AB is parallel to CD, we know that angle BCD is equal to angle ACD (denoted as x).
Now, let's use the given information to set up an equation:
angle BCD + angle ACD = 180 degrees
Since angle ACD is x, we have:
angle BCD + x = 180
We can also use the fact that the sum of all angles in a quadrilateral is 360 degrees:
angle BCD + angle BAC + angle CDA + angle ACD = 360 degrees
Substituting the known values, we have:
angle BCD + 90 + 90 + x = 360
Combining like terms, we get:
angle BCD + x = 180
We can now solve for angle BCD:
angle BCD + x = 180
angle BCD + x - x = 180 - x
angle BCD = 180 - x
Thus, angle BCD in the trapezium ABCD is equal to 180 minus the measure of angle ACD (x).