in the given figure ABCD is a trapezium in which AD parallel to BC. Angle ABC= 90degree. AD=16cm, AC=41cm and BC=40cm.find the area of the trapezium.
since angle B is a right angle, AB is the height of the trapezoid. So, the area is
(16+40)/2 * √(41^2 - 40^2)
252cm
To find the area of a trapezium, you can use the formula:
Area = (sum of parallel sides / 2) × height
In this case, the sum of the parallel sides is AB + CD. Since AD is parallel to BC, AB = AD and CD = BC.
Given that AD = 16 cm and BC = 40 cm, we have AB = 16 cm and CD = 40 cm.
The height of the trapezium can be found using the Pythagorean Theorem. In triangle ABC, angle ABC is 90 degrees, so triangle ABC is a right triangle.
AC is the hypotenuse, and we are given that AC = 41 cm.
Using the Pythagorean Theorem, we can find the height (BD) of the trapezium as follows:
BD^2 = AC^2 - BC^2
BD^2 = 41^2 - 40^2
BD^2 = 1681 - 1600
BD^2 = 81
BD = 9 cm
Now we have all the values we need to calculate the area:
AB + CD = 16 + 40 = 56 cm
Height (BD) = 9 cm
Using the formula for the area of a trapezium:
Area = (sum of parallel sides / 2) × height
Area = (56 / 2) × 9
Area = 28 × 9
Area = 252 cm^2
Therefore, the area of the trapezium is 252 square centimeters.