Area of a on parallelogram area theorem

Good evening my friends... what is Theorem on Area of Triangle? This is the topic for the day.

Firstly tell me if you remember what is Pythagoras theorem? By now you are aware of it right. I have also explained you what is binomial theorem also earlier.


Theorem of area of triangle can be explained with the help of a problem below.

Lets look at this theorem

ABRS and PQRS are parallelograms. X is any point on BR.

a) Show that area of ABRS = area of PQRS
b) Show that area of ASX = ½ (PQRS)
Solution:
a) Parallelogram ABRS and parallelogram PQRS lie on the same base SR and between the same parallel lines and hence they have equal area. Hence, area of ABRS = area of PQRS
b)Area of ΔASX = ½ × AS × height = ½ (area of ASRB)
If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.
But area of ASRB = area of PSRQ, as they are on the same base and between the same parallel lines.
Hence, area of ASX = ½ (PQRS)
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