Mathematics Class IX Questions Bank

Question 1 of 41

1. Question
1 point(s)
PQRS is a parallelogram whose area is $180\ cm^2 $ and A is any point on the diagonal QS. The area of $\triangle $ASR is $90\ cm^2 $. Is is true ?
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Hint

False
Area of parallelogram $PQRS=180\ cm^2$ and $QS$ is it diagonal which divides it into two triangles of equal area.
Question 2 of 41

2. Question
1 point(s)
In the figure, PQRS and EFRS are two parallelogram. Is area of $\triangle $MFR equal to $\dfrac{1}{2} $ area of || gm PQRS?
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Hint

$PQRS$ and $EFRS$ are on the same base $SR$ and between the same parallels $EF$ and $SR$ .
So, $ar\ (PQRS)=ar\ (EFRS)$ …(1)
$\triangle MFR=\dfrac{1}{2}\ ar\ (EFRS)$ …(2)
$ar\ (\triangle MFR)=\dfrac{1}{2}\ ar\ PQRS$
Question 3 of 41

3. Question
1 point(s)
BD is one of the diagonals of a quadrilateral ABCD. AM and CN are the perpendiculars from A and C respectively on BD. Show that ar $\mathrm{(ABCD)=\dfrac{1}{2}BD(AM+CN)} $
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Hint

$ar\ (\triangle ABD)=\dfrac{1}{2}\times BD\times AM\\ \\ ar\ (\triangle BCD)=\dfrac{1}{2}\times BD\times CN\\ \\ ar\ (quad\ ABCD)=\dfrac{1}{2}\times BD\times AM+\dfrac{1}{2}\times BD\times CN\\ \\ =\dfrac{1}{2}\times BN\times (AM+CN)$
Question 4 of 41

4. Question
1 point(s)
Check whether the following statement is true. PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If $PS = 5\ cm $, then $ar\ (PAS) = 30\ cm^2 $.
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Hint

Since $A$ is any point on $PQ$ then $ar\ (PAS)\ne30\ cm^2$
But the statement can be true of $PA=PS$
Justification
$PA<PQ\\ \\ ar\ \triangle PQR=\dfrac{1}{2}\times PQ\times PR\\ \\ =\dfrac{1}{2}\times12\times5=30\ cm^2$
Suppose $PA[latex]ar\ (\triangle PAS)<(ar\ \triangle PQR)\\ \\ ar\ (\triangle PAS)<30\ cm^2$
Suppose $PA=PQ$
$ar\ (\triangle PAS)=\dfrac{1}{2}\times PQ\times PS=\dfrac{1}{2}\times12\times5=30\ cm^2$
Question 5 of 41

5. Question
1 point(s)
In $\triangle\ ABC,O$ is any point on its median $AD$. Show that $ar\ (\triangle ABO)=ar\ (\triangle ACO) $
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Hint

In $\triangle ABC$
$AD$ is the median
$ar(\triangle ABD)=ar(\triangle ADC)$ …(1)
In $\triangle OBC$
$OD$ is the median
$ar(\triangle OBD)=ar(\triangle ODC)$ ….(2)
$ar\ (\triangle ABD)=ar(\triangle OBD)=ar\ (\triangle ABC)-ar(\triangle ODC)\\ \\ ar(\triangle ABO)=ar(\triangle ACO)$ (Proved)
Question 6 of 41

6. Question
1 point(s)
In the figure, $LM=\dfrac{3}{4}QR,LM||QR $ and distance between $LM $ and $OR $ is $3\ cm $. If length of $QR=6\ cm $, find the area of $LQRM $.
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Hint

$LM=\dfrac{3}{4}\times QR\\ \\ =\dfrac{3}{4}\times6=\dfrac{9}{2}=4.5$
Area $=\dfrac{1}{2}\times(6+4.5)\times3$
$=15.75\ cm^2$
Question 7 of 41

7. Question
1 point(s)
ABC and BDE are two equilateral triangle such that D is mid–point of BC. Show that ar (BDE) = $\dfrac{1}{4} $ ar (ABC).
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Hint

Ratio of area of triangle is equal to the ratio of square of corresponding sides
$\dfrac{area\ of \ \triangle ABC}{area\ of\ \triangle BDE}=\dfrac{BC^2}{BD^2}=\dfrac{BC^2}{\dfrac{BC^2}{2}}=\dfrac{1}{4}$
area of $\triangle BDE=\dfrac{1}{4}\ ar\ (\triangle ABC)$ (Proved)
Question 8 of 41

8. Question
1 point(s)
Show that the segment joining the mid-points of a pair of opposite sides of a parallelgoram divides it into two equal parallelograms.
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Hint

$AB=DC$ and $AB||DC$ ( $\because\ ABCD$ is a parallelogram)
$\dfrac{1}{2}AB=\dfrac{1}{2}DC$
$AE=DF$ and $AE||DF$
So, $EBCF$ is a parallelogram
$AEFD$ and $EBCF$ are two parallelogram on the base $EF$ and between the same parallels $AD$ and $BC$
$ar\ (AEFD)=ar\ (EBCF)$
Question 9 of 41

9. Question
1 point(s)
In the figure, ABCD and EFGD are two parallegorams and G is the mid-point of CD. Check whether area of $\triangle $PDC is equal to half of area EFGD.
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Hint

The given statement is false
$ar(\triangle DPC)=\dfrac{1}{2}\ (ar||gm\ ABCD)$
( $\because\ \triangle PDC$ and parallelogram $ABCD$ are on the same base $DC$ )
$G$ is the mid point of $DC$
$DG=\dfrac{1}{2}DC\\ \\ ar\ (EFGD)=\dfrac{1}{2}ar\ (ABCD)\\ \\ ar\ \triangle DPC=ar\ EFGD$
Question 10 of 41

10. Question
1 point(s)
In the figure, PQRS is square and T and U are re-spectively the mid-points of PS and QR. Find the area of $\triangle $OTS, if PQ = 8 cm.
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Hint

$TO||PQ\\ \\ \Rightarrow TO||PQ$
In $\triangle PQS,\ T$ is the mid point of $PS$ and $TO||PQ$
$TO=\dfrac{1}{2}PQ=4\ cm\\ \\ TS=\dfrac{1}{2}PS=4 \ cm\\ \\ ar\ (\triangle OTS)=\dfrac{1}{2}(TO\times TS)=\dfrac{1}{2}(4\times4)=8\ cm^2$
Question 11 of 41

11. Question
1 point(s)
In the figure, AD is a median of $\triangle $ABCE is any point on AD. Show that at ($\triangle $BED) = ar ($\triangle $CED)
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Hint

$ar\ (\triangle ABD)=ar\ (\triangle ACD)$
( $\because\ AD$ is the median) …(1)
$ar\ (BED)=ar\ (\triangle CED)$
( $\because\ DE$ is median) …(2)
Subtracting equation (2) from (1)
$ar\ (\triangle ABD)-ar\ (\triangle BED)=ar\ (\triangle ACD)-ar\ (\triangle ABE)=ar\ (\triangle ACE)\ ar(\triangle CED)$
Question 12 of 41

12. Question
1 point(s)
PQRS is a trapezium with PQ||SR. A line parallel to PR intersects PQ at L and QR at M. Prove that $\mathrm{ar\ (\triangle PSL) = ar\ (\triangle PRM)} $.
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Hint

$ar\ (\triangle PSL)=ar\ (\triangle PRL)$ …(1)
(Two triangles having same base and between the same parallels)
$ar\ (\triangle PRL)=ar\ (\triangle PRM)$ …(2)
From (1) and (2)
$ar\ (\triangle PSL)=ar\ (\triangle PRM)$ (Proved)
Question 13 of 41

13. Question
1 point(s)
P and Q are any two points lying on the sides DC and AD respectively of a paralleogram ABCD. Show that ar ($\triangle $APB = ar ($\triangle $BQC)
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Hint

$ar\ (\triangle APB)=\dfrac{1}{2}ar\ (ABCD)$ …(1)
$ar\ (\triangle BQC)=\dfrac{1}{2}\ (ar\ ABCD)$ …(2)
From (1) and (2)
$ar\ (\triangle APB)=ar\ (\triangle BQC)$
Question 14 of 41

14. Question
1 point(s)
Diagonals AC and BD of a trapezium ABCD with AD||CD intersect each other at O. Prove that ar ($\triangle $AOD) = ar ($\triangle $BOC)
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Hint

In $\triangle AOD$ and $\triangle BOC$
$\angle AOD=\angle BOC$ (VOA)
$\angle ADO=\angle CBO$ (alt angles)
$\angle DAO=\angle BCO$ (alt $\angle S$ )
$\triangle AOD\cong\triangle BOC$ (AAA)
Question 15 of 41

15. Question
1 point(s)
In the figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar (BPC) = ar (DPQ).
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Hint

$ar\ (\triangle APC)=ar\ (\triangle BCP)$ ..(1)
( $\because$ Triangles on the same base and between the same parallels)
Similarly $ar\ (\triangle ADQ)=ar\ (\triangle ADC)$ …(2)
$\triangle (ADQ)-ar\ (\triangle ADP)=ar(\triangle ADC)-ar\ (\triangle ADP)$
$ar\ \triangle DPQ=ar\ (\triangle ACP)$ ….(3)
From (1) and (3)
$ar\ (\triangle BCP)=ar\ (\triangle DPQ)$ or $ar\ (\triangle BPC)=ar\ (\triangle DPQ)$ (Proved)
Question 16 of 41

16. Question
1 point(s)
D is the mid–point of side BC of a $\triangle $ABC and E is the mid point of BD. If O is the mid point of AE, then show that ar (BOE) = $\dfrac{1}{8} $ ar (ABC).
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Hint

$ar\ (\triangle ABD)=\dfrac{1}{2}\ ar\ (\triangle ADC)$ ..(1)
( $\because\ AD$ and $AE$ are medians)
$ar\ (\triangle ABE)=\dfrac{1}{2}ar\ (\triangle\ ABD)$ …(2)
$ar\ (\triangle BOE)=\dfrac{1}{2}\ ar\ (\triangle ABE)$ …(3)
From (1), (2) and (3)
$ar\ \triangle BOE=\dfrac{1}{2}ar\ (\triangle ABC)$
Question 17 of 41

17. Question
1 point(s)
In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ=QR = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD)
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Hint

In Parallelogram $PABQ$ ,
$PA||QB$
So $PABQ$ is a parallelogram
$PQ=AB$ …(1)
Similarly $QR=BC$ …(2) and $RS=CD$ ….(3)
Now $PQ=QR=RS$ …(4)
From equation (1), (2), (3) and (4)
$PQ||AB$
In $\triangle PQE$ and $\triangle CFD$
$PQ=CD$
$\angle PRC=\angle FCD$ (alt. $\angle S$ )
$\triangle PQE\cong\triangle DCF$ (ASA)
$ar\ \triangle PQR=ar\ (\triangle CFD)$
( $\because$ congruent triangle have equal areas)
Question 18 of 41

18. Question
1 point(s)
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
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Hint

Diagonal of a parallelogram but it each other
$AO=OC\\ \\ DO=OB$
$ar\ (\triangle AOC)=ar\ (\triangle BOC)$ … (1) ( $\because\ OC$ is median)
$ar\ (\triangle COB)=ar\ (\triangle BOD)$ …(2) ( $\because\ OB$ is median)
$ar\ (\triangle BOD)=ar(\triangle AOD)$ ….(3) ( $\because\ OD$ is median)
$ar\ (\triangle AOC)=ar\ (\triangle BOC)=ar\ (\triangle BOD)=ar\ (\triangle AOD)$
(Proved)
Question 19 of 41

19. Question
1 point(s)
In the figure, ABCD is a square. E and F are respectively the mid–points of BC and CD. If R is the mid–point of EF, show that ar (AER) = ar (AFR)
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Hint

$ar\ (\triangle AER)\\ \\ =\dfrac{1}{2}\times ER\times AN\\ \\ =\dfrac{1}{2}\times FR\times AN$
(Since $R$ is the mid point of $EF$ so, $EF=FR$ )
$=ar\ (\triangle AFR)$ (Proved)
Question 20 of 41

20. Question
1 point(s)
In the figure, M, N are points on sides PQ and PR respectively of $\triangle $PQR , Such that ar ($\triangle $QRN) = ar ($\triangle $QRM) .Show that MN||QR.
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Hint

$\triangle QRN$ and $\triangle QRM$ are on the same base $QR$ and having equal areas
$\Rightarrow$ They lie between the same parallel lines
$\Rightarrow MN||QR$
Question 21 of 41

21. Question
1 point(s)
In the figure, O is any point on the diagonal PR of a parallelogram PQRS. Show that ar (PSO) = ar (PQO).
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Hint

$T$ is the mid point of $QS\ ar\ (\triangle PTS)=ar\ (\triangle PQT)$ ….(1)
$ar\ (\triangle STO)=ar\ (QTO)$ ….(2)
Adding equation (1) and (2)
$ar\ (\triangle PTS)+ar\ (\triangle STO)=ar(\triangle PQT)\\ \\ \Rightarrow ar(\triangle PSO)=ar(\triangle PQO)$ (Proved)
Question 22 of 41

22. Question
1 point(s)
In the figure, ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE
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Hint

In $\triangle ADF$ and $\triangle ECF$
$\angle ADF=\angle ECF$ (alt. int $\angle S$ )
$AD=EC\ (\because\ AD=BC\ \text{and }\ BC=EC)\\ \\ \angle DFA=\angle CFE\ \text{(VOA)}\\ \\ \triangle ADF\cong\triangle ECF\ \text{(AAS)}\\ \\ DF=CF\ \text{(cpct)}\\ \\ ar\ (\triangle ADF)=ar\ (\triangle ECF)\\ \\ ar(\triangle BDC)=2\times ar\ (\triangle DFB)\\ \\ =2\times3=6\ cm^2\\ \\ ar(||gm\ ABCD)=2\ ar(\triangle BDC)\\ \\ =2\times6=12\ cm^2$
Question 23 of 41

23. Question
1 point(s)
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. E and F are the mid-points of non-parallel sides. Show that ar (ABFE) : ar (EFCD) = (3a + b) : (a + 3b).
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Hint

$EM=\dfrac{1}{2}AB\\ \\ MF=\dfrac{1}{2}CD\\ \\ EM+MF=\dfrac{1}{2}\ (a+b)$
Ratio $=\dfrac{\frac{1}{4}\times(3a+b)h}{\frac{1}{4}(3b+a)h}=\dfrac{3a+b}{3b+a}$
Question 24 of 41

24. Question
1 point(s)
Prove that a rectangle and a parallelogram on the same base and between the same parallels, the perimeter of the parallelogram is greater than the perimeter of the rectangle.
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Hint

Let $ABCD$ be a parallelogram and $ABEF$ be the rectangle
Perimeter of parallelogram $ABCD=2(AB+AD)$
Perimeter of rectangle $ABEF=2(AB+AF)$
$\angle AFD=90^o\ \therefore\ \angle ADF<90^o\\ \\ \angle AFD>\angle ADF\ \therefore\ AD>AF\\ \\ AB+AD>AB+AF\\ \\ 2(AB+AD)>2(AB+AF)$ (Proved)
Question 25 of 41

25. Question
1 point(s)
If medians of a triangle $ABC $ intersects at $G $, show that $ar\ \triangle AGB=ar(\triangle AGC)=ar\ (\triangle BGC)=\dfrac{1}{2}ar\ (\triangle ABC) $
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Hint

$ar\ (\triangle ABD)=ar\ (\triangle ACD)$ ( $\because\ AD$ is median) ….(1)
$ar\ (\triangle GBD)=ar\ (\triangle GCD)$ ….(2)
( $\because\ GD$ is median)
$ar\ (\triangle ABD)-ar(\triangle GBD)=ar\ (\triangle ACD)-ar\ (\triangle GCD)\\ \\ ar\ (\triangle AGB)=ar\ (\triangle AGC)$ ….(3)
Similarly $ar\ (\triangle AGB)=ar\ (\triangle BGC)$ ….(4)
From equation (3) and (4)
$ar\ (\triangle AGB)=ar\ (\triangle BGC)=ar\ (\triangle AGC)$ …(5)
$ar\ (\triangle ABC)=ar\ (\triangle AGB)+ar\ (\triangle BGC)+ar\ (\triangle AGC)\\ \\ ar\ ABC=3(ar\ \triangle AGB)\\ \\ ar\ \triangle AGB=\dfrac{1}{3}\ ar\ \triangle ABC$ ….(6)
$ar\ \triangle BGC=\dfrac{1}{3}ar\ \triangle ABC$
(From 5 and 6)
$ar\ \triangle AGC=\dfrac{1}{3}\ ar\ (\triangle ABC)$ (Proved)
Question 26 of 41

26. Question
1 point(s)
In the figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
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Hint

$ar\ (\triangle DRC)=ar\ (DPC)$ …(1)
$ar\ (\triangle BDP)=ar\ (\triangle ARC)$ …(2)
Subtracting equation (1) from equation (2)
$ar\ (\triangle BDP)-ar\ (\triangle DPC)=ar\ (\triangle ARC)-ar(\triangle DRC)\\ \\ ar(\triangle BDC)=ar\ (\triangle ADC)$
Since these two triangle have the same base $DC$ . They must be lying between the same parallel lines $AB||CD$
$ABCD$ is a trapezium
Also $ar\ (DRC)=ar\ (DPC)$
$DC||PR$
Hence $DCPR$ is a trapezium
Question 27 of 41

27. Question
1 point(s)
In the figure, ABCD is a quadrilateral. A line through D parallel to AC meets BC produced at E. Prove that ar ($\triangle $ABE) = ar qard (ABCD)
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Hint

$ar\ (\triangle AEC)=ar\ (\triangle ADC)$ (Triangles on the same base)
Adding $\triangle ABC$ on both the sides
$ar\ (\triangle AEC)+ar\ (\triangle ABC)=ar\ (\triangle ADC)+ar\ (\triangle ABC)\\ \\ ar\ ABE=ar\ \text{quad}\ ABCD$
Question 28 of 41

28. Question
1 point(s)
In the figure, $ABCD $ is a parallelogram. Points $P $ and $Q $ on $BC $ trisect $BC $. Show that $ar\ (APQ) = ar\ (DPQ)=\dfrac{1}{6}\ ar\ (ABCD)$.
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Hint

$\triangle APQ$ and $\triangle DPQ$ are on the same base $PQ$ and between the same parallels $AD$ and $BC$
$ar\ (\triangle APQ)=ar\ (\triangle DPQ)$ ….(1)
$ar\ (\triangle APQ)=\dfrac{1}{2}\ ar\ (||gm\ PQRS)$ ….(2)
$\dfrac{ar\ (||gm\ ABCD)}{ar\ (||gm\ PQRS)}=\dfrac{BC\times height}{PQ\ height}=\dfrac{3PQ}{PQ}\\ \\ ar\ (||gm\ PQRS)=\dfrac{1}{3}\ ar\ (||gm\ ABCD)\\ \\ ar\ \triangle APQ=ar\ (\triangle DPQ)=\dfrac{1}{2}\times\dfrac{1}{3}\ ar||gm\ ABCD\\ \\ =\dfrac{1}{6}\ ar\ (||gm\ ABCD)$
Question 29 of 41

29. Question
1 point(s)
In $\triangle $ABC , if L and M are the points on AB and AC respectively, such that LM||BC, prove that ar (LOB) = ar (MOC)
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Hint

$\triangle\ LBC$ and $\triangle MBC$ lie on the same base and between the same parallels $BC$ and $LM$
$ar\ (\triangle LBC)=ar\ (\triangle MBC)\\ \\ ar(\triangle LOB)+ar(\triangle BOC)=ar(MOC)+ar\ (\triangle BOC)\\ \\ ar\ (\triangle LOB)=ar\ (\triangle MOC)$
(Proved)
Question 30 of 41

30. Question
1 point(s)
In the figure, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD).
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Hint

In $\triangle PEA$ and $\triangle QFD$
$\angle APE=\angle DQP$ (Corresponding angles)
$AE=DF$ (opposite sides)
$\angle AEP=\angle DFQ$ (corresponding angles)
$\triangle PEA\cong\triangle QFD$ (AAS)
$ar\ (\triangle PEA)=ar\ (\triangle QFD)$
( $\because$ congruent triangles have equal area)
Question 31 of 41

31. Question
1 point(s)
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. Show that ar (ABCD) = ar (PBQR).
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Hint

$ar\ (\triangle ACQ)=ar\ (\triangle APQ)$
( $\because\ \triangle ACQ$ and $\triangle APQ$ are on the same base $AQ$
$ar\ (\triangle ACQ)-ar(\triangle ABQ)=ar(\triangle APQ)-ar\ (\triangle ABC)=ar(\triangle QBP)$ …(1)
$ar\ (\triangle ABC)=\dfrac{1}{2}\ (||gm\ ABCD)$ …(2)
$ar\ (\triangle QBP)=\dfrac{1}{2}\ ar\ (||gm\ PBQR)$ ….(3)
From (2) and (3)
$ar\ ||gm\ ABCD=ar\ (||gm\ PBQR)$
Question 32 of 41

32. Question
1 point(s)
In the figure, ABCD is a parallelograms in which BC is produced to E such that CE = BC. AE intersects CD at F. Show that $\mathrm{(\triangle BDF)=\dfrac{1}{4}ar\ (ABCD)} $.
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Hint

$ar\ (\triangle BDF)=ar\ (\triangle ADF)\\ \\ ar(\triangle BAF)=\dfrac{1}{2}\ ar(ABCD)$ …(1)
$ar\ (\triangle ADF)=ar(BCF)$ …(2)
In $\triangle ADF$ and $\triangle CFE$
$CE=AD\ (CE=BC=AD)\\ \\ \angle CFE=\angle AFD$ (VOA)
$\angle CEF=\angle FAD$ (Alt. $\angle S$ )
$\triangle ADF\cong\triangle CFE$ (ASA)
$CF=DF$ (cpct)
$ar\ (\triangle ADF)+ar(\triangle BCF)+ar(\triangle DFA)=ar(ABCD)\\ \\ 2ar\ (\triangle ADF)+\dfrac{1}{2}\ ar(ABCD)=ar\ (ABCD)\\ \\ \Rightarrow 2\ ar\ (\triangle ADF)=\dfrac{1}{2}\ ar \ (ABCD)\\ \\ \Rightarrow\ ar\ BDF=\dfrac{1}{2}\times\dfrac{1}{2}\times(ar\ ABCD)\\ \\ =\dfrac{1}{4}\ ar\ (ABCD)$
Question 33 of 41

33. Question
1 point(s)
The figure, ABCDE is a pentagon and a line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (ABCDE) = ar (AEDF)
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Hint

$\triangle ACB$ and $\triangle ACF$ lie on the same base $AC$ and between the same parallels $AC$ and $BF$
$ar\ (\triangle ACB)=ar\ (\triangle ACF)\\ \\ ar\ (\triangle ACB)+ar\ (ACDE)=ar\ (\triangle ACF)+ar\ (ACDE)\\ \\ ar\ ABCDE=ar\ (AEDF)$
Question 34 of 41

34. Question
1 point(s)
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).
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Hint

$ar\ (\triangle ACX)=ar\ (\triangle ACY)$
( $\because$ They lie on the same base $AC$ and between parallel lines $AC$ and $XY$ )
$ar\ (\triangle ACX)=ar\ (\triangle ADX)$ …(2)
(They lie on the same base $AX$ and between parallel lines $AX$ and $DC$ )
From (1) and (2)
$ar\ (\triangle ACY)=ar\ (\triangle ADX)$
Question 35 of 41

35. Question
1 point(s)
Diagonals PR and QS of quadrilateral PQRS intersect at T such that PT = TR. If PS = QR, show that ar ($\triangle $PTS) = ar ($\triangle $RTQ)
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Hint

In $\triangle PET$ and $\triangle RTF$
$TP=TR$ (given)
$\angle PTE=\angle RTF$ (VOA)
$\angle PET=\angle RFT\ (90^o)$
$\triangle PET\cong\triangle RFT$ (AAS)
$\angle PTE=\angle RTF$ (cpct)
$PE=RF$ (cpct)
$ar\ (\triangle PTE)=ar\ (RTF)$ ….(1)
In $\triangle PSE$ and $\triangle RQF$
$\angle PES=\angle RFQ$ (each $90^o$ )
$PE=RF$ (Proved above)
$PS=QR$ (given)
$\triangle PSE\cong\triangle RQF$ (RHS)
$ar\ (PSE)=ar\ (RQF)$ …(2)
Adding (1) and (2)
$ar\ (PTE)+ar(PSE)=ar\ (RTF)+ar\ (RQF)\\ \\ =ar(PTS)=ar(RTQ)$
Question 36 of 41

36. Question
1 point(s)
Diagonal AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
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Hint

$ar\ (\triangle AOD)=ar\ (\triangle BOC)$ (given)
$ar\ (\triangle AOD)+ar\ (ODC)\\ \\ =ar\ (BOC)+ar\ (ODC)\\ \\ \Rightarrow ar\ (ADC)=ar\ BDC$
$\triangle ADC$ and $\triangle BDC$ lie on the same base $DC$ and equal in area $AB||DC$
In $ABCD$
$AB||DC$
So one pair of opposite side is parallel $ABCD$ is a trapezium
Question 37 of 41

37. Question
1 point(s)
PQRS and ABRS are parallelograms and X is any point on side BR. Show that :
(i) area PQRS = area ABRS
(ii) area AXS = $\dfrac{1}{2} $ area PQRS.
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Hint

Since $PQRS$ is a parallelogram $PQ||RS$ and $ADRS$ is also a parallelogram
So $AB||RS$
So $PQ||RS$
$PQRS$ and $ABRS$ are two parallelograms with some base $RS$
$ar\ (PQRS)=ar\ (ABRS)$
area of $\triangle AXS=\dfrac{1}{2}\ ar\ (ABRS)$
area of $\triangle AXS=\dfrac{1}{2}\ ar\ (PQRS)\\ \\ (\because\ ar\ (PQRS)=ar\ (ABRS))$
Question 38 of 41

38. Question
1 point(s)
In the given figure, AP||BQ||CR. Prove that $ar\ (\triangle AQC)=ar\ (\triangle PBR) $
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Hint

Since $\triangle ABQ$ and $\triangle PBQ$ lie on the same base $BQ$ and are between parallel lines $BQ$ and $AP$
$ar\ (\triangle ABQ)=ar(PBQ)$ …(1)
Also $ar\ (\triangle CBQ)=ar(RBQ)$ …(2)
Adding (1) and (2)
$ar\ (\triangle AQB)+ar\ (BQC)=ar\ (PBQ)+ar\ (QBR)\\ \\ ar\ (AQC)=ar\ (PBR)$
Question 39 of 41

39. Question
1 point(s)
A point E is taken as the midpoint of the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that $ar\ (\triangle ADF)=ar\ (ABFC) $
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Hint

$DC||AB\\ \\ \Rightarrow CF|| AB\\ \\ ar\ (\triangle ACF)=ar\ (\triangle BCF)$ …(1)
Also $ar(\triangle ADC)=ar\ (\triangle ABC)$ ….(2)
Adding (1) and (2)
$(\triangle ADC)+(\triangle ACF)=ar\ (\triangle ABC)+ar\ (\triangle BCF)\\ \\ ar\ (\triangle ADF)=ar\ (ABFC)$
Question 40 of 41

40. Question
1 point(s)
In the figure, M is a point in the interior of a parallelogram PQRS. Show that
(i) $ar\ (\triangle PMQ)+ar\ (\triangle MRS)=\dfrac{1}{2}ar(||gm\ PQRS) $
(ii) $ar\ (\triangle PMS)+ar\ (\triangle MQR)=ar(\triangle PMQ)+ar(\triangle MRS) $
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Hint

(i) $PQ||UV$ (by construction) …(1)
$PS||QR$ (opposite sides)
So $UV||QV$ …(2)
From (1) and (2)
$PQ||UV$ and $PU||QR$
$PUVQ$ is a parallelogram
$ar\ (\triangle PMQ)=\dfrac{1}{2}\ ar(||gm\ PQVU)$ …(3)
$ar\ (\triangle MRS)=\dfrac{1}{2}\ ar\ (||gm\ SRVU)$ …(4)
Adding (3) and (4)
$ar\ (\triangle PMQ)+ar\ (MRS)=\dfrac{1}{2}\ ar\ (PQVU+SRVU)\\ \\ ar\ (\triangle PMQ)+ar\ (\triangle MRS)=\dfrac{1}{2}ar\ (PQRS)$
(ii) $PS||XY$ (by construction) …(1)
$PQ||SR$ (opposite sides)
So $PX||SY$
$PS||XY$ and $PX||SY$
So $PSYX$ is a parallelogram
$ar\ (\triangle PMS)=\dfrac{1}{2}\ ar\ (||gm\ PSYX)\\ \\ ar\ (\triangle MRQ)=\dfrac{1}{2}\ ar(||gm\ XYRQ)$
$ar\ (\triangle PMS)+ar(MRQ)=\dfrac{1}{2}\ ar\ PQRS\\ \\ ar\ (\triangle PMS)+ar\ (\triangle MRQ)=ar\ (\triangle PMQ+ar\ (\triangle MRS))$
(Proved)
Question 41 of 41

41. Question
1 point(s)
In the figure, diagonals AC and BD of quadrilateral ABCD intersect at O, such that OB = OD. If AB = CD, show that
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = (ACB)
(iii) ABCD is a parallelogram.
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Hint

In $\triangle DOE$ and $\triangle BOF$
$\angle DEO=\angle BFO$ (each 90)
$\angle DOE=\angle BOF$ (VOA)
$OD=OB$ (given)
$\triangle DOE\cong\triangle BOF$ (AAS)
$DE=BF$ (cpct) …(1)
Also, $ar\ (\triangle DOE)=ar\ (\triangle BOF)$ …(2)
In $\triangle DEC$ and $\triangle BFA$
$\angle DEC=\angle BFA$ (each 90)
$CD=AB$ (given)
$DE=BF$ (From 1)
$\triangle DEC\cong\triangle BFA$ (RHS)
Also, $ar\ (\triangle DEC=ar\ (\triangle BFA)$ …(3)
Adding (2) and (3) we get
$ar\ (\triangle DOC)=ar\ (\triangle AOB)$
(ii) $ar\ (\triangle DOC)=ar\ (\triangle AOB)$
$\Rightarrow ar\ (\triangle DOC+ar\ \triangle OCB)=ar\ (\triangle AOB)+ar\ (\triangle OCB)\\ \\ \Rightarrow ar\ \triangle DCB=ar\ (\triangle ACB)$
(iii) $DA||BC$
$\angle FBO=\angle EDO$ ….(4) $(\because\ \triangle DOE\cong\triangle BOF)\\ \\ \angle FBA=\angle EDC\ (\because\ \triangle DEC\cong\triangle BFA)$
Adding (4) and (5) we get
$\angle ABD=\angle CDB\\ \\ \therefore DC||AB$
we get $DC||AB$ and $DC||CB$
So, $ABCD$ is a parallelogram