NCERT Solutions for Exercise 9.2 Class 9 Maths Chapter 9 - Areas Of Parallelograms And Triangles

Q2 If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that

$\small ar(EFGH)=\frac{1}{2}ar(ABCD)$

Answer:

Join GE and HE,
Since F, F, G, H are the mid-points of the parallelogram ABCD. Therefore, GE || BC ||AD and HF || DC || AB.
It is known that if a triangle and the parallelogram are on the same base and between the same parallel lines. then the area of the triangle is equal to the half of the area of the parallelogram.

Now, $\Delta$ EFG and ||gm BEGC are on the same base and between the same parallels EG and BC.
Therefore, ar ( $\Delta$ EFG) = $1/2$ ar (||gm BEGC)...............(i)
Similarly, ar ( $\Delta$ EHG) = 1/2 . ar(||gm AEGD)..................(ii)
By adding eq (i) and eq (ii), we get

ar ( $\Delta$ EFG) + ar ( $\Delta$ EHG) = 1/2 (ar (||gm BEGC) + ar(||gm AEGD))
ar (EFGH) = 1/2 ar(ABCD)
Hence proved

Q3 P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that $\small ar(APB)=ar(BCQ)$ .

Answer:

We have,
ABCD is a parallelogram, therefore AB || CD and BC || AD.

Now, $\Delta$ APB and ||gm ABCD are on the same base AB and between two parallels AB and DC.
Therefore, ar ( $\Delta$ APB) = 1/2 . ar(||gm ABCD)...........(i)

Also, $\Delta$ BQC and ||gm ABCD are on the same base BC and between two parallels BC and AD.
Therefore, ar( $\Delta$ BQC) = 1/2 . ar(||gmABCD)...........(ii)

From eq(i) and eq (ii), we get,
ar ( $\Delta$ APB) = ar( $\Delta$ BQC)
Hence proved.

Q4 (i) In Fig. $9.16$ , P is a point in the interior of a parallelogram ABCD. Show that
$\small ar(APB)+ar(PCD) = \frac{1}{2}ar(ABCD)$

[ Hint : Through P, draw a line parallel to AB.]

Answer:

We have a ||gm ABCD and AB || CD, AD || BC. Through P, draw a line parallel to AB
Now, $\Delta$ APB and ||gm ABEFare on the same base AB and between the same parallels EF and AB.
Therefore, ar ( $\Delta$ APB) = 1/2 . ar(ABEF)...............(i)
Similarly, ar ( $\Delta$ PCD ) = 1/2 . ar (EFDC) ..............(ii)
Now, by adding both equations, we get
$\small ar(APB)+ar(PCD) = \frac{1}{2}ar(ABCD)$

Hence proved.

Q4 (ii) In Fig. $\small 9.16$ , P is a point in the interior of a parallelogram ABCD. Show that

$\small ar(APD)+ar(PBC)=ar(APB)+ar(PCD)$

[ Hint: Through P, draw a line parallel to AB.]

Answer:

We have a ||gm ABCD and AB || CD, AD || BC. Through P, draw a line parallel to AB
Now, $\Delta$ APD and ||gm ADGHare on the same base AD and between the same parallels GH and AD.
Therefore, ar ( $\Delta$ APD) = 1/2 . ar(||gm ADGH).............(i)

Similarily, ar ( $\Delta$ PBC) = 1/2 . ar(||gm BCGH)............(ii)

By adding the equation i and eq (ii), we get

$\small ar(APD)+ar(PBC)=ar(APB)+ar(PCD)$

Hence proved.

Q5 In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

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

Answer:

(i) Parallelogram PQRS and ABRS are on the same base RS and between the same parallels RS and PB.
Therefore, $\small ar(PQRS)=ar(ABRS)$ ............(i)
Hence proved

(ii) $\Delta$ AXS and ||gm ABRS are on the same base AS and between same parallels AS and RB.
Therefore, ar ( $\Delta$ AXS) = 1/2 . ar(||gm ABRS)............(ii)

Now, from equation (i) and equation (ii), we get

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

Hence proved.

Q6 A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Answer:

We have a field in the form of parallelogram PQRS and a point A is on the side RS. Join AP and AQ. The field is divided into three parts i.e, $\Delta$ APS, $\Delta$ QAR and $\Delta$ PAQ.

Since $\Delta$ APQ and parallelogram, PQRS is on the same base PQ and between same parallels RS and PQ.
Therefore, $ar(\Delta APQ) = \frac{1}{2}ar(PQRS)$ ............(i)
We can write above equation as,

ar (||gm PQRS) - [ar ( $\Delta$ APS) + ar( $\Delta$ QAR)] = 1/2 .ar(PQRS)
$\Rightarrow ar(\Delta APS)+ar(\Delta QAR) = \frac{1}{2}ar(PQRS)$
from equation (i),
$\Rightarrow ar(\Delta APS)+ar(\Delta QAR) =ar(\Delta APQ)$

Hence, she can sow wheat in $\Delta$ APQ and pulses in [ $\Delta$ APS + $\Delta$ QAR] or wheat in [ $\Delta$ APS + $\Delta$ QAR] and pulses in $\Delta$ APQ.