9th Maths 9.2
NCERT Class 9th solution of Exercise 9.1
NCERT Class 9th solution of Exercise 9.3
Exercise 9.2
Q1. In Figure, ABCD is a parallelogram, AE⊥DC and CF⊥AD. If AB=16 cm, AE=8 cm. and CF=10 cm. Find AD.
Area of a parallelogram = Base×Height
Area of a parallelogram ABCD=AD×AE
Area of a parallelogram ABCD=(16×8)cm2=128cm2______(1)
Area of a parallelogram ABCD=AD×CF
Area of a parallelogram ABCD=(AD×10)cm2______________(2)
From equation (1) and (2)
128=AD×10
AD=12810=12.8cm
Answer:
AD=12.8cm.
Q2. If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar(EFGH)=12ar(ABCD).
Sol. :
Construction:
Join HF.
Proof:
In △HGF and parallelogram HDCF lies on the same base HF and lie between the same parallels HF and DC.
ar(HGF)=12ar(HDCF)_________(1)
In △HEF and parallelogram ABFH lies on the same base HF and lie between the same parallels HF and AB.
ar(HEF)=12ar(ABFH)_________(2)
By Adding equation (1) and (2)
ar(HGE)+ar(HEF)=12[ar(HDCF)+ar(ABFH)]
ar(EFGH)=12ar(ABCD)
Proved.
Q3. 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).
Sol. :
In △APB and parallelogram ABCD lies on the same base AB and lie between the same parallels AB and DC.
ar(APB)=12ar(ABCD)_________(1)
In △(BQC) and parallelogram ABCD lies on the same base AB and lie between the same parallels BC and AD.
ar(BQC)=12ar(ABCD)____(2)
By equation (1) and (2)
ar(APB)=ar(BQC).
Proved.
Q4. In Figure, P is a point in the interior of a parallelogram ABCD
i) ar(APB)+ar(PCD)=12ar(ABCD)
ii) ar(APD)+ar(PBC)=ar(APB)+ar(PCD)
[Hint: Though P, draw a line parallel to AB.]
Proof:
In △APB and parallelogram ABFE lies on the same base AB and lie between the same paralled the same parallels AB and DC.
ar(APB)=12ar(ABFE)________(1)
and
ar(PCD)=12ar(EFCD)_________(2)
By Adding equation (1) and (2)
ar(APB)+ar(PCD)=12ar(ABEF)+12ar(EFCD)
ar(APB)+ar(PCD)=12[ar(ABEF)+(EFCD)]
ar(APB)+ar(PCD)=12ar(ABCD)_________(3)
ii) Proof:
In △APD and parallelogram AGHD lies on the same base AD and lie between the same parallels AD and HG.
ar(APD)=12ar(AGHD)____________(4)
and
ar(PCB)=12ar(GBCH)_____________(5)
By Adding equation (4) and (5)
ar(APD)+ar(PCB)=12ar(AGHD)+12ar(GBCH)
ar(APD)+ar(PCB)=12[ar(AGHD)+ar(GBCH)]
ar(APD)+ar(PCB)=12ar(ABCD)______(6)
from equation (3) and (6)
ar(APD)+ar(PBC)=ar(APB)+ar(PCD).
Proved.
Q5. 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)=12ar(PQRS)
i) Proof:
Parallelogram PQRS and parallelogram ABRS lies on the same base RS and lie between the same parallels SR and PAQB
ar(PQRS)=ar(ABRS)________(1)
ii) Proof:
In △AXS and parallelogram ABRS lies on the same base AS and lie between the same parallels AS and lie between the same parallels AS and RS.
ar(AXS)=12ar(ABRS)_______(2)
From equation (1) and (2)
ar(AXS)=12ar(PQRS)
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?
Sol. :
Parallelogram PQRS
To Prove:
ar(APQ)=ar(APS)+ar(AQR)
Construction:
Draw a line AB parallel to SP.
Proof:
In △APQ and parallelogram PQRS are on the same base PQ and between the same parallels PQ and RS
ar(APQ)=12ar(PQRS
2ar(APQ)=ar(PQRS)
2ar(APQ)=ar(APS)+ar((APQ)+ar(AQR)
2ar(APQ)-ar(APQ)=ar(APS)+ar(AQR)
ar(APQ)=ar(APS)+ar(AQR)
Proved.
Comments