Question

Find the area enclosed by the following figures
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Answer 1

Answer:

Thanks for free points sister or brother

Answer 2

Answer:

Consider a trapezium of ABCD. Let AB and DC be the parallel sides as shown in the figure.

Now, CM will be the distance between the two parallel sides or the height of the trapezium.

We know,

Area of trapezium = ½ × sum of parallel sides × height.

So, height has to be found.

In the diagram, draw CL || AD

Now, ALCD is a parallelogram ⇒ AL = CD = 20 cm and CL = AD = 26 cm

As AD = CB,

CL = CB ⇒ ΔCLB is an isosceles triangle with CB as its height.

Here, BL = AB – AL = (40 – 20) = 20 cm. So,

LM = MB = ½ BL = ½ × 20 = 10 cm

Now, in ΔCLM,

CL² = CM² + LM² (Pythagoras Theorem)

26² = CM² + 10²

CM² = 262 – 102

Using algebraic identities, we get; 26² – 10² = (26 – 10) (26 + 10)

hence,

CM2 = (26 – 10) (26 + 10) = 16 × 36 = 576

CM = √576 = 24 cm

Now, the area of trapezium can be calculated.

Area of trapezium, ABCD = ½ × (AB + CD) × CM

= ½ × (20 + 40) × 24

Or, Area of trapezium ABCD = 720 cm²