NCERT Solutions Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.2

NCERT Solutions Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.2

NCERT Solutions Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.2

?Chapter – 9?

✍Areas of Parallelograms and Triangles✍

?Exercise 9.2?

Question 1. In figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Solution:
ABCD We have, AE ⊥ DC and AB = 16 cm
∵ AB = CD [Opposite sides of parallelogram]
∴ CD = 16 cm
Now, area of parallelogram ABCD = CD x AE
= (16 x 8) cm² = 128 cm² [∵ AE = 8 cm]
Since, CF ⊥ AD
∴ Area of parallelogram ABCD = AD x CF
⇒ AD x CF = 128 cm
⇒ AD x 10 cm = 128 cm² [∵ CF= 10 cm]
⇒ AD = 128/10 cm = 12.8 cm 10
Thus, the required length of AD is 12.8 cm

Question 2. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/2 ar (ABCD).
Solution:
Join GE and HE, where GE || BC || DA and HF || AB || DC
(∵ E, F, G and H are the mid¬points of the sides of a ||^gm ABCD).
If a triangle and a parallelogram are on the same base and between the same parallels, then A E U the area of the triangle is equal to half the area of the parallelogram.

Now, ∆EFG and parallelogram EBCG are on the same base EG and between the same parallels EG and BC.
∴ ar(∆EFG) = 1/2ar(∥gm EBCG) … (1)
Similarly, ar(∆EHG) = 1/2ar(∥gm AEGD) …(2)
Adding (1) and (2), we get
ar(∆EFG) + ar(∆EHG) = 1/2ar(∥gm EBCG)+1/2ar(∥gm AEGD)
= 1/2ar(∥gm ABCD)
Thus, ar(EFGH) = 1/2ar(ABCD)

Question 3. P and Q are any two points lying on the sides DC and AD, respectively of a parallelogram ABCD. Show that ar (APB) = AR (BQC).
Solution:
∵ ABCD is a parallelogram.
∴ AB || CD and BC || AD.
Now, ∆APB and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.
∴ ar(∆APB) = 1/2ar (∥gm  ABCD) …….(1)

Also, ∆BQC and parallelogram ABCD are on the same base BC and between the same parallels BG and AD.
∴ ar(∆BQC) = 1/2ar (∥gm ABCD) …(2)
From (1) and (2), we have ar(∆APB) = ar(∆BQC).

Question 4. In figure, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar (APB) + ar (PCD) = 1/2ar(ABCD)
(ii) ar (APD) + ar(PBC) = ar (APB) + ar (PCD)
Solution:
We have a parallelogram ABCD, i.e., AB || CD and BC || AD. Let us draw EF || AB and HG || AD through P.

(i) ∆APB and ||^gm AEFB are on the same base AB and between the same parallels AB and EF.
∴ ar(∆APB) = 1/2ar (∥gm AEFB) …(1)
Also, ∆PCD and parallelogram CDEF are on the same base CD and between the same parallels CD and EF.
∴ ar(APCD) = 1/2ar (∥gm CDEF) …(2)
Adding (1) and (2), we have
ar(∆APB) + ar(∆PCD) = 1/2ar( ∥gm AEFB)+12ar(∥gm CDEF)
⇒ ar(∆APB) + ar(∆PCD) = 1/2ar (∥gm ABCD) …(3)

(ii) ∆APD and ||^gm ∆DGH are on the same base AD and between the same parallels AD and GH.
∴ ar(∆APD) = 1/2ar(∥gm ADGH) …(4)
Similarly,
ar(∆PBC) = 1/2ar(∥gm BCGH) …(5)
Adding (4) and (5), we have
ar(∆APD) + ar(∆PBC) = = 1/2ar(∥gm ADGH)+1/2ar(∥gm BCGH)
⇒ ar(∆APD) + ar(∆PBC) =1/2ar(∥gm ABCD) …….(6)
From (3) and (6), we have
ar(∆APD) + ar(∆PBC) = ar(∆APB) + ar(∆PCD)

Question 5. In figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = 1/2ar(PQRS)

Solution:
(i) Parallelogram PQRS and parallelogram ABRS are on the same base RS and between the same parallels RS and PB.
∴ ar(PQRS) = ar(ABRS)
(ii) AAXS and ||^gm ABRS are on the same base AS and between the same parallels AS and BR. *
∴ ar(AXS) = 1/2ar(ABRS) …(1)
But ar(PQRS) = ar(ABRS) …(2) [Proved in (i) part]
From (1) and (2), we have
ar(AXS) = 1/2ar(PQRS)

Question 6. 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.
Solution:
The farmer is having the field in the form of parallelogram PQRS and a point A is situated on RS. Join AP and AQ.
Clearly, the field is divided into three parts i.e., in ∆APS, ∆PAQ and ∆QAR.

Since, ∆PAQ and pt.
parallelogram PQRS are on the same base PQ and between the same parallels PQ and RS.
ar(∆PAQ) = 1/2ar(∥gm PQRS) …(1)
⇒ ar(||^gm PQRS) – ar(∆PAQ) = ar(||^gm PQRS) – 1/2ar(∥gm PQRS)
⇒ [ar(∆APS) + ar(∆QAR)] = 1/2ar(∥gm PQRS) …(2)
From (1) and (2), we have
ar(∆PAQ) = ar[(∆APS) + (∆QAR)]
Thus, the farmer can sow wheat in (∆PAQ) and pulses in [(∆APS) + (∆QAR)] or wheat in [(∆APS) + (∆QAR)] and pulses in (∆PAQ).