Question

please answer this question.....
I will mark it Brainleist who answer it first​

Answer 1

Class - 9

Chapter -  Areas of Parallelograms and Triangles. ##9

Page ## - 159

Exercise -  9.2 Q ## 5

Answer:

(i)PROVED!

(ii)PROVED!

Explanation:

Given Problem:

In fig. 9.17  PQRS and ABRS are parallelogram PQRS  is a parallelogram and X is any point on side BR.

Show that:

(i) ar (PQRS) = ar (ABRS)

(ii) ar (AXS) =$\frac{1}{2}$  ar (PQRS)

Solution:

First of all we have show:

(i) ar (PQRS) = ar (ABRS)

--------------

Method:

Since PQRS is parallelogram

PQ || RS                             (Opposite sides of parallelogram are parallel)

&ABRS is also a parallelogram

So,

AB || RS                       (Opposite sides of parallelogram are parallel)

Since,

PQ || RS & AB || RS

We can say that,

PB || RS

Now,

PQRS & ABRS are two parallelograms with same base RS and between the same parallels  PB & RS

∴ ar (PQRS) = ar (ABRS)

--------------------

★★Parallelograms with the same base and between the same parallels are equal in area.★★

-------------------------------------------------

Now to show:

ar (AXS) =$\frac{1}{2}$  ar (PQRS)

------------------

Method:

Since ABRS is a parallelogram,

AS || BR                      (Opposite sides of the parallelogram are parallel)

Δ AXS and parallelogram ABRS lie on the same base and are between the same parallel lines AS and BR,

∴Area (ΔAXS) = $\frac{1}{2}$ Area (ABRS)

★We have proved in part (i) that,

Area (PQRS) = Area (ABRS)★

⇒Area (ΔAXS) =  $\frac{1}{2}$ Area (ABRS)

★The area of the triangle is half of the parallelogram if they have the same base and parallels.★

★Hence,

PROVED!!★