9th Maths 9.3

NCERT Class 9th solution of Exercise 9.1

NCERT Class 9th solution of Exercise 9.2

NCERT Class 9th solution of Exercise 10.1

Exercise 9.3

Q1. In Figur, E is any point on median AD of a ABC. Show that ar(ABC)=ar(ACE).
Sol. :
Given: 
ABC, AD is a median.
To Prove:
ar(ABE)=ar(ACE).
Proof:
In ABC
AD is median
ar(ABE)=ar(ACE)__________(1)
In EBC,
ED is median.
ar(BDE)=ar(CDE)__________(2)
By equation (1) - (2)
ar(ABE)-ar(BDE)=ar(ACE)-ar(CDE)
ar(ABE)=ar(ACE)
Proved.
Q2. In a triangle ABC,E is the mid-point of median AD. Show that ar(BED)=14(ar(ABC).
Sol. :
Given:
ABCAD is median and E is the midpoint of AD.
To Prove:
ar(BED)=12ar(ABC)
Proof:
In ABC
AD is the median
ar(ADB)=ar(ACD)
ar(ADB)=12(ABC)__________(1)
In ADB
E is the midpoint of AD,
BE is the median of ADB
ar(BED)=12ar(ADB)________(2)
from equation (1) and (2)
ar(BED)=12×12(ABC)
ar(BED)=14ar(ABC).
Proved.
Q3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Sol.
Given:
To Prove:
ar(AOB)=ar(BOC)=ar(COD)=ar(DOA)
Proof:
In triangleABD, OA is the median
ar(AOB)=ar(AOD)_________(1)
In triangleABC, OA is the median
ar(AOB)=ar(BOC)_________(2)
In triangleBCD, OC is the median
ar(BOC)=ar(COD)_________(3)
from equation (1), (2), and (3)
ar(AOB)=ar(BOC)=ar(COD)=ar(DOA)
Proved.
Q4. In Figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ABC)=ar(ABD).
Sol. :
Given:
triangleABC and triangleABD both have same base AB,
OC=OD
To Prove:
ar(ABC)=ar(ABD)
Proof:
OC=OD, O is the midpoint of CD
In triangleACD, AO is the median
ar(AOC)=ar(AOD)_______(1)
In triangleBCD, OB is the median
ar(BOC)=ar(BOD)________(2)
We Add equation (1) and (2)
ar(AOC)+ar(BOC)=ar(AOD)+ar(BOD)
ar(ABC)=ar(ABD)
Proved.
Q5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a triangleABC.
Show that
i) BDEF is a parallelogram.
ii) ar(DEF)=1/4ar(ABC)
iii) ar(BDEF)=1/2 ar(ABC)
Sol. :
Given:
triangleABC, D, E, F are midpoint of side BC, CA, and AB.
To Prove:
i) BDEF ia a parallelogram.
ii) ar(DEF)=1/4ar(ABC).
iii) ar(BDEF)=1/2(ABC).
Proof:
i) In triangleABC, F and E are the midpoint of sides AB and CA.
EF∥BC or FE∥BD
D and E are the midpoint of sides BC and CA
DE∥BA or DE∥BF
BDEF is a parallelogram.
Proved.
ii) Daigonal DF divide parallelogram BDEF into two triangle BDF and DEF of equal area.
similarly 
ar(AEF)+ar(BDF)+ar(CDE)+ar(DEF)=ar(ABC)
4ar(DEF)=ar(ABC)
ar(DEF)=1/4ar(ABC)
Proved. 
iii) Diagonal divide parallelogram into two triangle of equal area.
ar(BDF)=ar(DEF)
ar(BDEF)=ar(BDF)+ar(DEF)
ar(BDEF)=2ar(DEF)
ar(BDEF)=2times1/4ar(ABC)
ar(BDEF)=1/2ar(ABC)
Proved.
Q6. In Figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB=OD. If AB=CD, then show that:
i) ar(DOC)=ar(AOB)
ii) ar(DCB)=ar(ACB)
iii) DA∥CB or AB⊥CD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC]
Sol. :
Given:
▱ABCD, OB=OD,and AB=CD
To prove:
i) ar(DOC)=ar(AOB)
ii) ar(DCB)=ar(ACB)
iii) DA∥CB or AB⊥CD is a parallelogram.
Construaction:
Draw (DE)and (BF) perpendicular from point D, and B on AC.
Proof.    
i) Draw DN⊥AC and BM⊥AC
In triangleDNO and triangleBOM
angleDNO=angleBMO=90^circ [By construction]
angleDON=angleBOM [Vertically opposite angles]
OD=OB    [Given]
triangleDONcongtriangleBOM____(1) [By AAS rule]
DN=BM[By CPCT]
In triangleDNC and triangleBMA
angleDNC=angleBMA=90^circ [By construction]
DC=AB  [ Given]
DN=BM [from proved (i)]
triangleDNCcongtriangleBAM_______(2)[By RHS rule]
We Add equation (1) and (2)
ar(DON)+ar(DCN)=ar(BOM)+ar(BAM)
ar(DOC)=ar(AOB)
Proved.
ii) ar(DOC)=ar(AOB)
We Add ar(BOC) both sides
ar(DOC)+ar(BOC)=ar(BOM)+ar(BOC)
ar(DCB)=ar(ACB).
Proved.
iii) triangleDCB and triangleACB both have equal areas and the same base BC. So that triangles
are lie between the same parallels AD and BC.
then DA∥CB so that ABCD is a parallelogram.
Proved.
Q7. D and E are points on sides AB and AC respectively of triangleABC such that ar(DBC)=ar(EBC). Prove that DE∥BC.
Sol. :
triangleDCB and triangleEBC both have same base BC and equal area. So that they will lies between the same parallels DE and BC.
DE∥BC
Proved.
Q8. XY is a line parallel to side BC of a triangle ABC. If  BE∥AC and CF∥AB meet XY at E and F respectively, show that ar(ABE)=ar(ACF)
Sol. :
Given:
XY∥BC, BE∥AC and CF∥AB
To Prove:
ar(ABE)=ar(ACF)
Proof:
In quadrilateral ACBE
AE∥CB [ BC∥XY]
AC∥BE  [ Given ]
ACBE is a parallelogram.
Similarly
ABCF is a parallelogram.
parallelogram ABCF and parallelogram ACBE both have same base BC and between the same parallel BC and XY
ar(ACBE)=ar(ABCF)
ar(ABE)+ar(ABC)=ar(ABC)+ar(ACF)
ar(ABE)=ar(ACF)
Proved.
Q9. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Figure). Show that ar(ABCD)=ar(PBQR).
[Hint: Join AC and PQ. Now compare ar(ACQ) and ar(APQ).]
Sol. :
Given:
ABCD and BPRQ are parallelograms,
CP∥AQ
To Prove:
ar(ABCD)=ar(BPRQ)
Construction:
Join AC and PQ.
Proof:
triangleACQ and triangleAPQ both have same base AQ and are between the same parallels AQ and CP.
ar(ACQ)=ar(APQ)
ar(ABQ)+ar(ABC)=ar(ABQ)+ar(BQC)
ar(ABC)=ar(BQC)________(1)
triangleABC and ▱ABCD both have same base AC and are between the same parallels DC and AB.
ar(ABC)=1/2ar(ABCD)______(2)
triangleBQC and ▱PBQR both have same base BQ and are between the same parallels BQ and PR
ar(BQC)=1/2ar(PBQR)______(3)
from equation (1), (2) and (3)
1/2ar(ABCD)=1/2ar(PBQR)
ar(ABCD)=ar(PBQR)
Proved.
Or
Alternative Method
Given:
ABCD and BPRQ are parallelograms,
CP∥AQ
To Prove:
ar(ABCD)=ar(BPRQ)
Construction:
Join AC and PQ.
Proof:
triangleACQ and triangleAPQ both have same base AQ and are between the same parallels AQ and CP.
ar(ACQ)=ar(APQ)
ar(ABQ)+ar(ABC)=ar(ABQ)+ar(BQP)
ar(ABC)=ar(BQP)________(1)
We multiply by 2 both sides
ar(ABC)=ar(BQP)
ar(ABCD)=ar(PBQR) [Both are parallelogram]
Proved.
Q10. Diagonals AC and BD of a trapezium ABCD with AB∥DC intersect each other at O. Prove that ar(AOD)=ar(BOC).
Sol. :
Given:
ABCD is a trapezium, AB∥DC
To Prove:
ar(AOD)=ar(BOC)
Proof:
triangleADC and triangleBCD both have same base CD and between the same parallels AB and CD.
ar(ADC)=ar(BCD)
Subtracting ar(DOC) from both sides
ar(ADC)-ar(DOC)=ar(BCD)-ar(DOC)
ar(AOD)=ar(BOC)
Proved.
Q11. In Figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that 
i) `ar(ACB)=ar(ACF)`
ii) ar(AEDF)=ar(ABCDE)
Given:
ABCDE is pentagon.
To Prove:
i) `ar(ACB)=ar(ACF)`
ii) ar(AEDF)=ar(ABCDE)
Proof:
i) triangleACB and triangleACF both have same base AC and between the same parallels AC and BF.
ar(ACB)=ar(ACF)
Proved.
ii) We Add ar(AEDC) on both sides
ar(ACB)+ar(AEDC)=ar(ACF)+ar(AEDC)
ar(ABCDE)=ar(AEDF)
Proved.
Q12. A villager Itwaari has a plot of the shape pf a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amouint of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Sol. :
Given:
ABCD is a quadrilateral.
Construction:
Draw DE∥CA which intersect BA produced at E.
To Prove:
ar(ABCD)=ar(EBC)
Proof:
triangleADC and triangleACE both have same base AC and between the same parallels AC and DE
ar(ADC)=ar(ACE)
We Add ar(ABC) both sides.
ar(ADC)+ar(ABC)=ar(ACE)+ar(ABC)
ar(ABCD)=ar(EBC)
Proved.
Q13. ABCD is a trapezium with AB∥DC.  A line parallel to AC intersects AB at X and BC at Y. Prove that ar(ADX)=ar(ACY).
[Hint: Join CX.]
Sol. :
Given:
ABCD is a trapezium, AB∥DC and AC∥XY.
To Prove:
ar(ADX)=ar(ACY)
Construction:
Join CX.
Proof:
triangleACX and triangleADX both have same base AX and between the same parallel AB and DC.
ar(ACX)=ar(ADX)________(1)
triangleACX and triangleACY both have same base AC and between the same parallel AC and XY.
ar(ACX)=ar(ACY)________(2)
from equation (1) and (2)
ar(ADX)=ar(ACY)
Proved.
Q14. In Figure, AP∥BQ∥CR. Prove that ar(AQC)=ar(PBR).
Given:
AP∥BQ∥CR
To Prove:
ar(AQC)=ar(PBR)
Proof:
triangleABQ and trianglePBQ both have same base BQ and between the same parallels AP and BQ.
ar(ABQ)=ar(PBQ)____________(1)
triangleBCQ and triangleBQR both have same base BQ and between the same parallels CR and BQ
ar(BCQ)=ar(BQR)____________(2)
We Add equation (1) and (2)
ar(ABQ)+ar(BCQ)=ar(PBQ)+ar(BQR)
ar(AQC)=ar(PBQ)
Proved.
Q15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(BDP)=ar(ARC). Prove that ABCD is a trapezium.
Sol. :
Given:
ar(AOD)=ar(BOC)
To Proved:
ABCD is a trapezium.
Proof:
ar(AOD)=ar(BOC)  [Given ]
We Add ar(AOB) both sides
ar(AOD)+ar(AOB)=ar(BOC)+ar(AOB)
ar(BAD)=ar(ABC)
triangleBAD and triangleABC both have same base AB and both have equal area.
these triangles lie between the same parallels
AB∥DC
So that ABCD is a trapezium.
Proved.
Q16. In Figure, ar(DRC)=ar(DPC) and ar(BDP)=ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapezium.
Given:
ar(DRC)=ar(DPC) and
ar(BDP)=ar(ARC)
To Prove:
ABCD and DCPR are trapezium.
Proof:
triangleDCR and triangleDPC both have the same base DC and equal area
these triangles lie between the same parallels
DC∥RP
So that DCPR is a trapezium.
ar(BDP)=ar(ACR)_______(1) [Given]
ar(DPC)=ar(DRC)_______(2) [Given]
Subtracting equation (1) and (2)
ar(BDP)-ar(DPC)=ar(ACR)-ar(DRC)
ar(BDC)=ar(ADC)
triangleADC and triangleBDC both have same base DC and have equal area.
these triangles lie between the same parallels
AB∥DC
So that ABCD is a trapezium.

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