Question

In a given figure o is the point in the interior of parallelogram pqrs if the area of parallelogram is 80 cm ^2then find ar (∆pqo) +ar(∆rso)

Answer 1

Answer:

SORRY BRO WHERE IS THE FIGURE?

Step-by-step explanation:

PLZZZ AS BRAINLIST PLZZZZZ FOLLOW ME

Answer 2

The ar (∆PQO) + ar(∆RSO) is 40 cm².

Step-by-step explanation:

It is given that

PQRS is a parallelogram

O is a point interior to the parallelogram PQRS

 

Let draw a line AB parallel to sides PQ and RS just as shown in the figure attached below

Step 1:

From the attached figure, we can say that  

The ΔPQO and the parallelogram PQBA have the same base PQ and lie between the same parallel lines PA and QB.

We know that if a triangle and a parallelogram are on the same base and lie between the same parallel lines, then the area of the triangle is equal to half the area of the parallelogram.

∴ Area (∆ PQO) = ½ * Area (parallelogram PQBA) …….. (i)

Similarly, ΔRSO and the parallelogram RSAB have the same base RS and lie between the same parallel lines SA and RB.

∴ Area (∆ RSO) = ½ * Area (parallelogram RSAB) …….. (ii)

Step 2:

Now, Adding eq. (i) & (ii), we get

Area (∆ PQO) + Area (∆ RSO) = ½ * [Area (parallelogram PQBA) + Area (parallelogram RSAB)]

Since //gm PQBA + //gm RSAB = //gm PQRS and Ar(parallelogram PQRS) = 80 cm²

⇒ Area (∆ PQO) + Area (∆ RSO) = ½ * 80 = 40 cm²

Thus, the Area (∆ PQO) + Area (∆ RSO) is 40 cm².  

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