Question

in the given figure ABCD is a trapezium in which AD is parallel To BC, angel ABC =90o , AD = 16cm , AC =41cm and BC =40cm .find the area of the trapezium.

Answer 1

Answer:

The area of the given trapezium ABCD = 252 cm²

Step-by-step explanation:

For better understanding of the solution see the attached figure of the problem :

Given : m∠ABC = 90°, AD = 16 cm, AC = 41 cm and BC = 40 cm

To find : Area of trapezium ABCD

Solution : Since, m∠ABC is 90°

So, AB is the height of the trapezium ABCD

Now, to find AB : By using Pythagoras theorem in ΔABC

AC² = AB² + BC²

41² = AB² + 40²

⇒ AB² = 41² - 40²

⇒ AB = 9 cm

Now, Area of the trapezium is given by the formula :

$Area = \frac{1}{2}\times Height\times \text{(Sum of parallel sides)}\\\\\implies Area = \frac{1}{2}\times AB\times (AD+BC)\\\\\implies Area = \frac{1}{2}\times 9\times (40+16)\\\\\implies Area = \frac{1}{2}\times 9\times 56\\\\\implies Area =9\times 28\\\\\bf Area = 252\thinspace cm^2$

Hence, The area of the given trapezium ABCD = 252 cm²

Answer 2

Answer:

Step-by-step explanation:

Answer:

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