Math Class 9 Question Bank

Question 1 of 42

1. Question
1 point(s)
Three angles of a quadrilateral are equal. It is a parallelogram ?
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Hint

It need not be parallelogram we may have $\angle A=\angle B=\angle C=75^o$ and $\angle D=135^o$
Question 2 of 42

2. Question
1 point(s)
Diagonals of a parallelogram are perpendicular to each other. Is this statemetn true ? Give reason for your answer.
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Hint

No, diagonals of a parallelogram are not perpendicular to each other, they only bisect each other.
Question 3 of 42

3. Question
1 point(s)
Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, find the lengths of AC and BD.
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Hint

$AC=2\times3=6\ cm\\ \\ BD=2\times2=4\ cm$
As diagonals bisect each other in a parallelogram
Question 4 of 42

4. Question
1 point(s)
The sides BA and DC of quadrilateral ABCD are produced as shown in the figure.

Prove that $x + y = a + b $
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Hint

$b+\angle ADB=180^o$ (Linear pair)
$\angle APB=180-b\\ \\ \angle DCB=180-a\\ \\ x+y+180-b+180-a=360\\ \\ x+y-a-b=0\\ \\ x+y=a+b$
Question 5 of 42

5. Question
1 point(s)
In the figure. $ABCD $ is a square. A line segment $DX $ cuts the side $BC $ at $X $ and the diagonal $AC $ at $O $ such that $\angle COD=105^o $. Find the of $x $.
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Hint

$\angle COX=180-105=75\\ \\ \angle OCX=\dfrac{90}{2}=45\\ \\ x=180-(105+45)\\ \\ =180-150=30^o$
Question 6 of 42

6. Question
1 point(s)
In $\triangle $ABC,D and E are mid points of AB and AC. If AD = 3.5 cm ; AE = 4 cm ; DE = 2.5 cm, find the perimeter of $\triangle $ABC
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Hint

$AB=2\times3.5=7\ cm\\ \\ AC=2\times4=8\ cm$
( $\because\ D$ and $E$ are mid points)
$DE=\dfrac{1}{2}\times BC\Rightarrow BC=2\times DE=5\ cm\\ \\ AB+BC+AC=7+8+5=20\ cm$
Question 7 of 42

7. Question
1 point(s)
ABCD is a parallelogram, L and M are points on AB and DC respectively such that AL = MC. Prove that LM and BD bisect each other.
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Hint

In $\triangle OMD$ and $OLB$
$\angle DOM=\angle BOL$ (VOA)
$\angle OMD=\angle OLB$ (alt. $\angle S$ )
$DM=BL\ (\because \ AB-AL=CD-CM)\\ \\ \triangle OMD\ \cong\ \triangle\ OLB$ (AAS)
$OM=OL,OD,OB$ (cpct)
$O$ bisect $LM$ and $BD$
Question 8 of 42

8. Question
1 point(s)
l, m and n are three parallel lines intersected by transversal p and q such that l, m and n cut equal intercepts AB and BC on p. Show that l, m, n cut off equal intercept DE and EF on q also.
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Hint

$ACFD$ is divided into two triangles $AB=AC\Rightarrow B$ is the mid point of $AC$
In $\triangle \ ACF,D$ is the mid point of $AC$
By mid point theorem, $G$ is the mid point of $AF$
In $\triangle AFD,G$ is the mid point of $AF$
$\Rightarrow E$ is the mid point of $DF$ .
$DE=EF$ (by mid point theorem)
Question 9 of 42

9. Question
1 point(s)
In the figure. ABCD is a parallelogram BA is produced to E such that AE = AD. ED is produced to meet BC produced at F. Show that CD = CF.
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Hint

$\angle 2=\angle 4\\ \\ \angle 1=\angle 3$ Corresponding angle
$AE=AD$ (given)
So, $\angle1=\angle2$
$\Rightarrow\angle 3=\angle4\qquad CF=CD$
Question 10 of 42

10. Question
1 point(s)
In the given figure. PQRS is a parallelogram and line segments PA and RB bisect the angles P and R respectively. Show that PA||RB.
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Hint

$\angle P=\angle R,\angle Q=\angle S$ (opposite angles of a parallelogram)
$\angle P+\angle Q=180^o\\ \\ \angle Q+\angle R=180^o\\ \\ \angle R+\angle S=180^o\\ \\ \angle S+\angle P=180^o\\ \\ \angle APS=\angle APB=\dfrac{\angle P}{2}\\ \\ \angle QRB=\angle BRS=\dfrac{\angle R}{2}\\ \\ \angle APB=\angle BRS \ (\because\ \angle P=\angle R)$
opposite angles of a quadrilateral are equal.
So, $ARBP$ is a parallelogram
Hence $PA||BR$
Question 11 of 42

11. Question
1 point(s)
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that :
(i) $\triangle APB\cong\triangle CQD $
(ii) $AP=CQ $
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Hint

In $\triangle APB$ and $\triangle CQD$
$\angle ABP=\angle CDQ$ (alt. $\angle S$ )
$AB=CD$ (opposite sides)
$\angle APB=\angle CQD=90^o$
$\triangle APB\cong\triangle CQD$ (ASA)
$AP=CQ$ (cpct)
Question 12 of 42

12. Question
1 point(s)
ABCD is a quadrilateral in which P, Q, R and S are mid points of AB, BC, CD and DA respectively. Show that PQRS is a parallelogram.
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Hint

In $\triangle ADC$
$SR||AC$
as $S$ and $R$ are the mid point $SR=\dfrac{1}{2}AC$ …(1)
In $\triangle DAC$
$PQ||AC\\ \\ PQ=\dfrac{1}{2}AC$ …(2)
So, $PQ=SR$
$SR||PQ$
So, $PQRS$ is a parallelogram
Question 13 of 42

13. Question
1 point(s)
In the figure. it is given that BDEF and FDCE are parallelograms. Show that BD = CD.
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Hint

$BD=EF\\ \\ CD=EF$ opposite sides of a parallelogram all equal
So, $BD=CD$ (Proved)
Question 14 of 42

14. Question
1 point(s)
In a cycle quadrilateral $PQRS $, if $\angle P-\angle R=50^o $. then find the measure of $\angle P $ and $\angle Q $
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Hint

$\angle P+\angle r=180^o\\ \\ \angle P=180^o-\angle r\\ \\ \angle P-\angle r=50\\ \\ 180-r-r=50\\ \\ 130=2r\ \Rightarrow \ r=65\\ \\ P=180-\angle r=180-65=115\\ \\ 115=180-\angle r\\ \\ \angle r=65^o$
Question 15 of 42

15. Question
1 point(s)
In $\triangle $ABC , AD is the median and DE || AB. Prove that BE is another median.
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Hint

Given $AD$ is a median and $DE||BA$
In $\triangle ABC$
$D$ is mid point of $AB$ and $DE||AB$
By converse of mid point theorem, $BE$ bisects $AC$
So $BE$ is the median
Question 16 of 42

16. Question
1 point(s)
In $\triangle $ABC, AB = 5 cm , BC = 8 cm and AC = 7 cm. If D and E are respectively mid–points of AB and BC, determine the length of DE. Give reasons.
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Hint

$DE||AC$ (By mid point theorem)
$DE=\dfrac{1}{2}AC=\dfrac{1}{2}\times7=3.5\ cm$
Question 17 of 42

17. Question
1 point(s)
In the figure, it is given the BDEF and FDCE are parallelograms. If BD = 4 cm, determine CD.
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Hint

In parallelogram $BDEF$
$BD=EF$ ….(1)
In parallelogram $FDCE$
$FE=CD$ ….(2)
From equation (1) and (2)
$BD=CD$
So $CD=4\ cm$
Question 18 of 42

18. Question
1 point(s)
In the figure, PQRS is a parallelogram. Find the values of x and y.
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Hint

$8x=32$ (alternate angles)
$\Rightarrow x=\dfrac{32}{8}=4^o$
$9y=27^o$ (alternate angles)
$y=3^o$
Question 19 of 42

19. Question
1 point(s)
In the figure, $\angle AOB=90^o,AC=BC,OA=12\ cm $. and $OC=6.5\ cm $ Find the measure of $OB $.
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Hint

$AC=\sqrt{12^2+6.5^2}\\ \\ =\sqrt{144+42.25}\\ \\ =\sqrt{186.25}\\ \\ =13.6$
$AC=BC$ (given)
$OB=\sqrt{(13.6)^2-(6.5)^2}=\sqrt{184.96-42.25}\\ \\ =\sqrt{142.71}=11.9\ cm$
Question 20 of 42

20. Question
1 point(s)
ABCD is a rhombus in which altitude from D on side AB bisects AB. Find the angles of the rhombus.
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Hint

In $\triangle AED$ and $\triangle BED$
$DE=DE$ (common)
$\angle AED=\angle BED$ (each $90^o$ )
$AE=BE$ (given)
$\triangle AED\cong\triangle BED$ (SAS)
$AD=BD$ (cpct)
But $AD=AB$ (sides of rhombus are equal)
So, $AD=BP=AD$
$\triangle ABD$ is an equilateral $\triangle$
$\angle A=60^o\\ \\ \angle A=\angle C=60^o$
$\angle ABC+\angle BCD=180^o$ (linear pair)
$\angle ABC=120$
$\angle ABC=\angle ADC=120^o$ (opposite angles of rhombus are equal)
Question 21 of 42

21. Question
1 point(s)
In a triangle ABC, median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.
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Hint

$AD$ is a median
So, $BD=CD$ and also $AD=DX$ (given)
$AS$ the diagonals of quadrilateral $ABCD$ bisect each other
$ABCD$ is a parallelogram
Question 22 of 42

22. Question
1 point(s)
Show that the quadrilateral formed by joining the mid–points of sides of a square is also a square.
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Hint

$MN=\dfrac{1}{2}\times DB\\ \\ MN=\dfrac{1}{2}\times AC\ (\because\ AC=BD)$ …(1)
$FO=\dfrac{1}{2}\times DB=\dfrac{1}{2}\times AC$ ….(2)
$ON=\dfrac{1}{2}\times AC$ …(3)
$FM=\dfrac{1}{2}\times AC$ ….(4)
From equation (1), (2), (3) and (4)
$OF=ON=MN=FM$
So FONM is a square
Question 23 of 42

23. Question
1 point(s)
In the figure, P is the mid–point of side BC of a parallelogram ABCD such that $\angle $BAP = $ \angle$DAP. Prove that AD = 2CD.
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Hint

$\angle BAP=\angle DAP\\ \\ \angle BAP=\dfrac{1}{2}\angle A$ …(1)
$ABCD$ is a parallelogram
$AD||BC\\ \\ \angle A+\angle B=180^o$ (Linear pair)
$\angle B=180^o-\angle A$ …(2)
In $\triangle ABP$
$\angle BAP+\angle APB+\angle B=180^o$ (angle sum property)
$\dfrac{1}{2}\angle A+\angle APB+180-\angle A=180^o\\ \\ \angle A-\dfrac{1}{2}\angle A=\angle APB$
From (1) and (2)
$\angle BAP=\angle BPA\\ \\ BP=AB\\ \\ \dfrac{1}{2}BC=AB\\ \\ \Rightarrow BC=2AB$
$\Rightarrow AD=2CD$ (opposite sides of a parallelogram) (Proved)
Question 24 of 42

24. Question
1 point(s)
Two parallel lines l and m are intersected by transversal p. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.
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Hint

$\angle QAC=\angle ACS$ (alt $\angle S$ )
$\dfrac{1}{2}\angle QAC=\dfrac{1}{2}\angle ACS\\ \\ \angle BAC=\angle ACD$
$AB||DC$ (as alt $\angle S$ are equal)
Similarly $BC||AD$
$ABCD$ is a parallelogram
$\angle QAC+\angle CAR=180$ (Linear pair)
$\dfrac{1}{2}\angle QAC+\dfrac{1}{2}\angle CAR=90^o\\ \\ \angle BAC+\angle CAD=90^o\\ \\ \angle BAD=90^o$
$ABCD$ is a rectangle
Question 25 of 42

25. Question
1 point(s)
In a parallelogram ABCD, bisector of $\angle $A. also, bisects BC at X. Prove that AD = 2 AB.
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Hint

$\angle 1=\angle2$ (given)
$BX=XC$ ( $X$ is the mid point)
$\angle3=\angle2$ (alt. $\angle S$ )
$\angle3=\angle1$ (as $\angle1=\angle2$ )
$AB=BX$ (side opposite to equal angles are equal)
$AB=\dfrac{BC}{2}\\ \\ 2AB=BC$
$AD=BC$ (opposite sides of parallelogram)
$AD=2AB$ (Proved)
Question 26 of 42

26. Question
1 point(s)
AD is the median of $\triangle $ABC. E is the midpoint of AD. BE produced meets AC at F. Show that $\mathrm{AF=\dfrac{1}{3}AC} $.
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Hint

$DG$ is parallel to $EF$ and $E$ is the mid point of $AD$
By converse of mid point theorem, $F$ is the mid point of $AG$ and $AF=FG$ …(1)
Since $D$ is the mid point and $DG$ parallel to $BF$ , by the converse of mid point theorem, $G$ is the mid point of $CF$ and $FG=GC$ …(2)
From (1) and (2)
$AF=FG=GC$
$AF=\dfrac{1}{3}AC$ (Proved)
Question 27 of 42

27. Question
1 point(s)
In a quadrilateral ABCD. AO and BO are the bisector of $\angle $A and $\angle $B respectively. Prove that $\mathrm{\angle AOB=\dfrac{1}{2}(\angle C+\angle D)} $
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Hint

$\angle A+\angle B=180^o\\ \\ \dfrac{1}{2}\angle A+\dfrac{1}{2}\angle B=90^o$
In $\triangle AOB$
$\angle OAB+\angle OBA+\angle AOB=180^o$
$\angle AOB=180-90=90$ …(1)
$\angle d+\angle c=180$
$\dfrac{1}{2}(\angle d+\angle c)=90$ …(2)
From equation (1) and (2)
So, $\angle AOB=\dfrac{1}{2}\ (\angle d+\angle c)$
Question 28 of 42

28. Question
1 point(s)
In the figure. $ABCD $ is a square. If $\angle PQR=90^o $ and $PB=QC=DR $, prove that $ QB = RC. PQ = QR,\angle QPR=45^o$.
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Hint

$CD=BC$ (sides of square)
$CD=CR+DR$ …(2)
$BC=BQ+QC$ ….(3)
$CR+DR=BQ+QC$
But $DR=QC$ (given)
So, $CR=BQ$
In $\triangle PBQ$ and $\triangle RCQ$
$PB=QC$ (given)
$\angle PBQ=\angle RCO$ (each $90$ )
$BQ=RC$ (Proved above)
$\triangle PBQ\cong\triangle\ RCQ$ (SAS)
$PQ=RQ$ (cpct)
$\Rightarrow \angle QRP=\angle QPR\\ \\ \angle PQR+\angle QRP+\angle QPR=180^o\\ \\ 90+2\angle QPR=180^o\\ \\ \angle QPR=\dfrac{90}{2}=45^o$
Question 29 of 42

29. Question
1 point(s)
In the figure, points A and B are on the same side of a line m. AD $\bot $ m and meet m at D and E respectively. If C is the mid point of AB. prove that CD = CE.
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Hint

$AD\bot m,BE\bot m$
$C\times \bot m$ (by construction)
$AD||BE||CX$
$AC=CB$ (given)
$DX=XE \ (\because\ AC=CB)$
In $\triangle DCX$ and $\triangle ECX$
$CX=CX$ (common)
$\angle D\times C=\angle E\times C\ (\because\ DX=XE)\\ \\ \triangle DCX\cong\triangle\ ECX$ (SAS)
$CD=CE$ (cpct)
Question 30 of 42

30. Question
1 point(s)
ABCD is parallelogram. On diagonal BD are points P and Q such that DP = BQ. Show that APCQ is a parallelogram
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Hint

In $\triangle DPC$ and $\triangle QAB$
$DP=BQ$ (given)
$DC=AB$ (opposite sides)
$\angle DCP=\angle BAQ$ (alt. $\angle S$ )
$\triangle DPC\cong\triangle\ QAB$ (SAS)
$PC=AQ$ (cpct)
Similarly we can prove that
$AP=CQ$
So $APCQ$ is a parallelogram
Question 31 of 42

31. Question
1 point(s)
In the figure, diagonal BD of parallelogram ABCD bisects $\angle $B. Show that it bisects $\angle $D also.
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Hint

In $\triangle\ ADB$ and $\triangle\ BDC$
$\angle ABD=\angle DBC$ (given)
$AD=BC$ (opposite sides)
$BD=BD$ (common)
$\triangle\ ADB\cong\triangle\ BDC$ (SAS)
$\angle ADB=\angle BDC$ (cpct)
Question 32 of 42

32. Question
1 point(s)
Prove that quadrilateral formed by bisectors of the angles of a parallelogram is a rectangle.
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Hint

Since $PQRS$ is a parallelogram $PQ||RS$
$\angle P+\angle S=180$ (Linear pair)
$\dfrac{1}{2}\angle P+\dfrac{1}{2}\angle S=90^o\\ \\ \angle APS+\angle PSA=90^o\\ \\ \angle PAS=90^o$
$\angle BAD=90^o$ (V.O.A)
$\angle ADC=90^o$
$\angle DCB=90^o$ and $\angle CBA=90^o$
$ABCD$ is a quadrilatral whose each angle is $90^o$
Hence $ABCD$ is a rectangle
Question 33 of 42

33. Question
1 point(s)
In the figure, $PS $ and $RT $ are medians of $\triangle PQR $ and $ SM||RT$. Prove that $QM=\dfrac{1}{4}PQ $.
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Hint

$RT$ is the median of $\triangle\ PQR$
$TQ=TP$
In $\triangle\ QRT$
$S$ is the mid point of $QR$
$SM||TR$
So, $M$ is the mid point of $TQ$ (by converse of mid point theorem)
$QM=\dfrac{1}{2}TQ\\ \\ =\dfrac{1}{2}\times\dfrac{1}{2}\times PQ=\dfrac{1}{4}PQ$
Question 34 of 42

34. Question
1 point(s)
In the figure, PQRS is a square. M is the midpoint of PQ and AB $\bot $ RM . Prove that RA = RB.
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Hint

In $\triangle\ APM$ and $\triangle\ BQM$
$\angle APM=\angle BQM=90^o$
$\angle AMP=\angle BMQ$ (V.O.A)
$PM=QM$ ( $M$ is the mid point)
$\triangle \ APM\cong\triangle \ BQM$ (ASA)
$AM=BM$ (cpct)
In $\triangle\ AMR$ and $\triangle\ BMR$
$AM=BM$
$\angle AMR=\angle BMR$ (each $90^o$ )
$RM=RM$ (common)
$\triangle AMR\cong\triangle\ BMR$ (SAS)
$RA=RB$ (cpct)
Question 35 of 42

35. Question
1 point(s)
In the figure $ABC $ is an isosceles triangle in which $AB = AC. CD || AB $ and $ AD$ is bisectors of exterior $\angle CAE $ of $\triangle ABC $. Prove that $ \angle CAD=\angle BCA$ and $ABCD $ is a parallelogram.
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Hint

In $\triangle ABC$
$AB=AC$
$\Rightarrow \angle B=\angle C$ …(1)
$\angle EAC=\angle B+\angle C$ (ext. angle is equal to the sum of interior opposite angle)
$\angle EAC=2\angle C\ (\because\ \angle B=\angle C)\\ \\ 2\angle CAD=2\angle C\\ \\ \angle CAD=\angle C\\ \\ \Rightarrow \angle DAC=\angle BCA$
But they form alternate $\angle S$
$AD||BC$
Also $CD||AB$
$ABCD$ is a parallelogram
Question 36 of 42

36. Question
1 point(s)
P, Q, R, and S are respectively the mid–points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AD = BC. Prove that PQRS is a rhombus.
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Hint

$SR|| AC\\ \\ \Rightarrow SR=\dfrac{1}{2}AC$ …(1)
(By mid point theorem)
$PQ||AC\Rightarrow PQ=\dfrac{1}{2}AC$ ….(2)
$PQ=SR=\dfrac{1}{2}AC$
Similarly in $\triangle BCD,\ RQ||BD$
$\Rightarrow RQ=\dfrac{1}{2}BD$
In $\triangle BAD,SP||BD,SP=\dfrac{1}{2}BD$
$RQ=SP=\dfrac{1}{2}BD=\dfrac{1}{2}AC\ (\because \ AC=BD)\\ \\ SR=PQ=SP=RQ$
All sides are equal so, $PQRS$ is a rhombus
Question 37 of 42

37. Question
1 point(s)
In a parallelogram ABCD, E and F are the mid– points of sides AB and CD respectively. Show that the line segment AF and EC trisects the diagonal BD.
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Hint

$ABCD$ is a parallelogram
$AB||CD\\ \\ \Rightarrow AE|| CF$
and $AB=CD$
$\dfrac{1}{2}AB=\dfrac{1}{2}CD\\ \\ AE=CF$
In $AECF$
$AE||CF$ and $AE=CF$
$AECF$ is a parallelogram
$AF||CE$
$PF||CQ$ and $AP||EQ$
In $\triangle\ DQC,PF||CQ$ and $F$ is the mid point of $DC$
So, $P$ is the mid point of $DQ$
$PQ=DP$ …(1)
In $\triangle\ ABP$
$E$ is the mid point of $AB$
$AP||EQ$
$Q$ is the mid point of $BP$
$PQ=QB$ ….(2)
From (1) and (2)
$DP=PQ=BQ$
This is the line segment trisects the diagonal $BD$
Question 38 of 42

38. Question
1 point(s)
ABC is a triangle right angled at C. A line through the mid–point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) $\mathrm{MD\bot AC} $
(ii) $\text D $ is mid–point of $\text{AC} $
(iii) $\mathrm{MC=MA=\dfrac{1}{2}AB} $
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Hint

(i) In $\triangle\ ABC$
$M$ is the mid point of $AB$
$MD|| BC$
$\therefore\ D$ is the mid point of $AC$ (Line drawn through mid point of one sides of a triangle, parallel to another side, bisect the third side)
(ii) As $MD||BC$ and $AC$ is transversal
$\angle MDC+\angle BCD=180^o$ (Interior angles on the same side of transversal are supplementary)
$MD\bot AC$ (Proved)
(iii) In $\triangle\ AMD$ and $\triangle\ CMD$
$AD=CD$ ( $D$ is the mid point of $AC$ )
$\angle ADM=\angle CDM$ (each $90^o$ )
$DM=DM$ (common)
$\triangle\ AMD\cong\triangle\ CMD$ (SAS)
$AM=CM$ (cpct) ….(1)
$AM=\dfrac{1}{2}AB\\ \\ CM=AM=\dfrac{1}{2}AB$
Question 39 of 42

39. Question
1 point(s)
In the figure, ABCD is a parallelogram. E is the mid–point of BC. DE and AB when produced meet at F. Prove that AF = 2AB.
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Hint

In $\triangle\ DEC$ and $\triangle\ BEF$
$\angle DEC=\angle BEF$ (VOA)
$EC=EB$ (given)
$\angle DCE=\angle EBF$ (alternate angles)
$\triangle DEC\cong\triangle BEF$ (ASA)
$DC=FB$ (cpct)…(1)
$DC=AB$ (opposite sides of a parallelogram) …(2)
From (1) and (2)
$BF=AB\\ \\ AF=AB+BF\\ \\ =AD+AB=2AB\\ \\ AF=2AB$
Question 40 of 42

40. Question
1 point(s)
ABCD is a a trapezium in which AB || CD and AD = BC Show that
(i) $\angle A=\angle B $
(ii) $\angle C=\angle D $
(iii) $\triangle ABC\equiv\ \triangle BAD $
(iv) Diagonal $AC= $ diagonal $BD $
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Hint

$AB||CD$ (given)
$AD||CE$ (by construction)
$AECD$ is a parallelogram
$AD=CE$ (opposite sides)
But $AD=BC$ (given)
$BC=EC$
$\angle CBE=\angle CEB$ (angle opposite to equal sides are equal)
$\angle B+\angle CBE=180^o$ (Linear pair)
$AD||CE$ (construction)
$\angle A+\angle CEB=180^o$ (Sum of interior angles)
$\angle B+\angle CBE=\angle A+\angle CEB\\ \\ \angle A=\angle B$
(ii) $AB||DC\\ \\ \angle A+\angle D=180^o\\ \\ \angle B+\angle C=180^o\\ \\ \angle A+\angle D=\angle B+\angle C$
But $\angle A=\angle B$
So, $\angle C=\angle D$
In $\triangle ABC$ and $\triangle BAD$
$AB=AB$ (common)
$BC=AD$ (given)
$\angle ABC=\angle BAD$ (Proved above)
$\triangle\ ABC\cong\triangle\ BAD$ (SAS)
$AC=BD$ (cpct)
Question 41 of 42

41. Question
1 point(s)
$ABCD $ is a parallelogram in which $\angle A=60^o $. If bisectors of $\angle A $ and $\angle B $ meet at $P $, prove that $AD=DP,PC=BC,DC=2AD $.
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Hint

(i) $ABCD$ is a parallelogram
$\angle PAB=\angle PAD=30^o$ ( $AP$ bisects $\angle A$ )
$\angle PAB=\angle APD=30^o$ (alternate $\angle S$ )
In $\triangle APD$
$\angle PAD=\angle APD\\ \\ \Rightarrow PD=AD$
(ii) $\angle A+\angle B=180^o$ (sum of adjacent interior angles)
$\angle B=120^o$
$\angle PBA=\angle PBC=60^o$ ( $PB$ is bisector of $\angle B$ )
$\angle PBA=\angle BPC=60^o$ (alternate angles)
$\angle BPC=\angle PBC\\ \\ \Rightarrow BC=PC$
(iii) $CD=DP+PC$
$=AD+BC\\ \\ =AD+AD\ (\because\ AD=BC)\\ \\ CD=2AD$
Question 42 of 42

42. Question
1 point(s)
Prove that in a triangle, the line segment joining the mid–points of any two sides is parallel to third side and is half of it.
Using the above, if P, Q, R are the mid–points of sides BC, AC and AB of $\triangle $ABC respectively and if PQ = 2.5 cm, QR = 3 cm, PR=3.5 cm, find the lengths of AB, BC and CA.
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Hint

Const:- Draw $CR||BA$ to meet $PQ$ produced at $R$
To prove:- $PQ||BC\\ \\ PQ=\dfrac{1}{2}BC$
Proof:- $\angle QAP=\angle QCP$ (alternate interior)… (1)
$AQ=QC$ ( $\because\ Q$ is mid point of $AC$ ) … (2)
$\angle AQP=\angle CQR$ (vertically opposite) …(3)
$\triangle\ APQ\cong\triangle\ CRQ$ (ASA congruence rule)
$PQ=QR$ (cpct) or $PQ=\dfrac{1}{2}PR$ ….(4)
$AP=CR$ (cpct) ….(5)
But $AP=BP$ ( $P$ is mid point of $AB$ )
$BP=CR$
Also $BP||CR$ (construction)
In quad $BCRP$
$BP=CR$ & $BP||CR$
Therefore quadrilateral $BCRP$ is a $||gm$
$BC||PR$ or $BC||PQ$
$\therefore\ PR=BC\ (\therefore\ BCPR\ \text{is}\ ||gm)\\ \\ \dfrac{1}{2}PR=\dfrac{1}{2}BC$
$PQ=\dfrac{1}{2}BC$ (From equation (4))
$PQ=\dfrac{1}{2}AB\ \Rightarrow AB=2\times PQ=2\times2.5=5\ cm\\ \\ QR=\dfrac{1}{2}BC=BC=2QR=2\times3=6\ cm\\ \\ PR=\dfrac{1}{2}AC\Rightarrow AC=2\times PR=2\times3.5=7\ cm$