MANISH KUMAR
MATHEMATICS QUADRILATERALS
1.
INTRODUCTION WE have studies in details about the properties of a triangle. We also know that the triangle is a figure obtained by joining thee non collinear points in pair. In this chapter we shall discuss about four non-collinear points such that no three on them are collinear. QUADRILATERAL We know that the figure obtained on joining three non-collinear points in pairs is a triangle. If we mark four points and join them is some order, then there possibilities for the figure obtained: ● ● ● ● (i) If all the points are collinear (in the same line), we obtain a line segment. (ii) If three out of four points are collinear, we get a triangle.
(iii) If no three points out of four are collinear, we obtain a closed figure with four sides.
Each of the figure obtained by joining four points in order is called a quadrilateral. (quad means four and lateral for sides). 2.
CONSTITUENTS OF A QUADRILATERAL A quadrilateral has four sides, four angles and four vertices.
In quadrilateral ABCD, BC, CD and DA are the four sides; A,B, C and D are the four vertices and A, B, C and D are the four angles formed at the vertices.
3.
QUADRILATERALS IN PRACTICAL LIFE We find so many objects around us which are of the shape of quadrilateral - the floor, walls, ceiling, windows of our classroom, the blackboard, each face of the duster, each page of our mathematics book, the top of our study table, etc. Some of these are given below.
MANISH KUMAR
MATHEMATICS
4.
SOME RELATED TERMS TO QUADRILATERALS3
(i) (ii)
In a quadrilateral ABCD, we have VERTICES :- The points, A, B, C and D are called the vertices of quadrilateral ABCD. SIDES :- The line segments AB, BC, CD and DA are called the sides of quadrilateral ABCD.
(iii)
DIAGONALS:- The line segments AC and BD are called the diagonals of quadrilateral ABCD.
(iv)
ADJACENT SIDES :-The sides of quadrilateral are said to be adjacent sides if they have la common end point. Here, in the above figure, (AB, BC), (BC, CD), (CD, DA) and (DA, AB) are four pairs of adjacent sides or consecutive sides of quadrilateral ABCD. OPPOSITE SIDES:- Two sides of a quadrilateral are said to be opposite sides if they have no common end point. Here, in the above figure, (AB, DC) and (BC, AD) are two pairs of opposite sides of quadrilateral ABCD. CONSECUTIVE ANGLES:- Two angles of a quadrilateral are said to be consecutive angles if they have a common arm. Here, in the above figure, A, B B, C C, D and D, A are four pairs of
(v)
(vi)
(vii)
consecutive angles. OPPOSITE ANGLES :- Two angles of a quadrilateral are said to be opposite angles if they have no common arm. Here in the given figure A, C and A, B are two pairs of opposite angles of
5. (i)
quadrilateral ABCD. VARIOUS TYPES OF QUADRILATERALS PARALLELOGRAM :- A quadrilateral in which both pair of opposite sides are parallel is called a parallelogram.
In figure, ABCD is a quadrilateral in which AB||DC, BC||AD.
quadrilateral ABCD is a parallelogram. (ii)
RHOMBUS :- A parallelogram whose all sides are equal is called rhombus.
In figure, ABCD is a parallelogram in which AB = BC = CD = D, AB||DC and BC||AD.
Parallelogram ABCD is a rhombus. (iii)
RECTANGLE: - A parallelogram, whose each angle is equal to 900, is called a rectangle. In figure, ABCD is a parallelogram in which
A B C D 900 AB||DC and BC||AD.
MANISH KUMAR
MATHEMATICS
Parallelogram ABCD is a rectangle. (iv)
SQUARE :- A rectangle in which a pair of adjacent sides are equal is said to be a square. In figure, ABCD is a rectangle in which A B C D 900 AB = BC, BC = CD, CD = DA, DA = AB. i.e., AB = BC = CD = DA.
rectangle ABCD is a square. (v)
TRAPEZIUM :- A quadrilateral in which exactly one pair of opposite sides is parallel, is called a trapezium.
In figure, ABCD is a quadrilateral in which AB||DC.
ABCD is trapezium. (vi)
ISOSCALES TRAPEZIUM :- A trapezium whose non-parallel sides are equal is called an isosceles trapezium.
In figure ABCD is a trapezium in which AB||DC and BC = AD.
trapezium ABCD is isosceles trapezium. (vi)
KITE :- A quadrilateral in which two pairs of adjacent sides are equal is called a kite. In figure, ABCD is a quadrilateral in which AB = AD and BC = CD.
quadrilateral ABCD is a kite. 6.
ANGLE SUM PROPERTY OF A QUADRILATERAL. THEOREM-I : The sum of the four angles of a quadrilateral is 360. Given : A quadrilateral ABCD. To Probe : A B C D 3600 Construction : Join AC. Proof :
STATEMENT 1.
In ABC 1 4 6 1800
2.
In ADC
2 3 5 1800 3.
(1 2) (3 4) 5 6 1800 1800
4.
A C D B 3600
REASON Sum of the all angles of triangles is equal to 180
0
0
Sum of the all angles of triangle is equal to 180 Adding (1) & (2), we get
MANISH KUMAR 5.
MATHEMATICS
A B C D 3600
Hence, proved Ex.1
There angles of a quadrilateral measure 560, 1000 and 880. Find the measure of the fourth angle.
Sol.
Let the measure of the fourth angle be x0.
560 + 1000 + 880 + x0 = 3600
[Sum of all the angles of quadrilateral is 3600]
244 + x = 360 x = 360 - 244 = 116 Hence, the measure of the fourth angle is 1160.
Ex.2
The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Sol.
Let the four angles of the quadrilateral be 3x, 5x, 9x and 13x.
3x + 5x + 9x + 13x = 360
0
[NCERT]
[Sum of all the angles of quadrilateral is 360 ] 0
30x = 3600 x = 120 Hence, the angles of the quadrilateral are 3 × 120 = 360, 5 × 120 = 600, 9 × 120 = 1080 and 13 × 120 = 1560.
Ex.3
In the given figure, sides AB and CD of the quadrilateral ABCD are produced. Find the value of x.
Sol.
Since, ADE ADC = 1800 [Linear pair]
1000 + ADC = 1800 ADC = 1800 - 1000 = 800 In quadrilateral ABCD
ADC A ABC C 3600
800 + 600 + ABC = 1200 = 3600 ABC + 2600 = 3600
[Sum of all the angles of quadrilateral is 3600]
MANISH KUMAR
MATHEMATICS
ABC = 1000 ABC + x = 1800
But,
100 + x = 180 0
Hence, x = 800.
0
[Linear pair]
MANISH KUMAR Ex.4 Sol.
MATHEMATICS
In the given figure, ABCD is a quadrilateral in which AE and BE are the angle bisectors of A and B. Prove that C D AEB. . Given : ABCD is a quadrilateral in which AE and BE are the angle bisectors of A and B To Prove : A D 2AEB Proof :
1. (i) (ii)
Ex.5 Sol.
STATEMENT In triangle ABE, ABE BAE AEB 1800
ABE BAE 1800 AEB Now, in quadrilateral ABCD, A B C D 3600 1 ( A B C D) 3600 2 1 1 1 A B (C D) 1800 2 2 2 1 BAE ABE (C D) 1800 2 1 1800 AEB (C D) 1800 2 1 (C D) AEB 2 C D 2AEB Hence proves.
REASON
Sum of the all angles of triangles is 180
Sum of the all angles of quadrilateral is 360
0
[ AE and BE are the angle bisector of A and B repectively.] [Form (i)]
In figure, ABCD is a quadrilateral in which AB = AD and BC = CD. Prove that (i) AC bisects A and C (ii) BE = DE. Given : ABCD is a quadrilateral in which AB = AD and BC = CD To Prove : (i) AC bisects A and C (ii) BE = DE. Proof :
STATEMENT In ABC and ADC, , AB = DA BC = CD and AC = AC ABC ADC BAC DAC and ACB ACD Therefore, AC bisect A and C. Hence proved STATEMENT (ii) Now, in BCD and DCE, BC = CD BCE DCE and CE = CE BCE DCE BE DE
0
REASON
1.
[Given] [Given] [Common] [By SSS congruence rule] [CP.CT.] [CP.CT]]
REASON [Given] [ ACB ACD ] [Common] [By SAS congruence rule] [CP.CT.]
MANISH KUMAR
MATHEMATICS
Hence proved
In a quadrilateral ABCD, AO and BO are the bisectors of A and B respectively. Prove that 1 AOB (C D). 2 Sol. Given : In a quadrilateral ABCD, AO and BO are the bisectors of A and B 1 To Prove : (i) AOB (C D) 2 Proof : STATEMENT REASON 0 sum of all the angles of triangle is 180 (i) In AOB 1 1 AOB 1 2 1800 [1 A ] and C B] 0 2 2 AOB 180 (1 2) Ex.6
1 1 A B C D 360 0 AOB 1800 A B A B 360 0 ( C D ) 2 2 1 0 AOB 180 (A B) 2 1 0 AOB 180 (3600 (C D) 2 1 AOB 1800 1800 (C D) 2 1 AOB ( C D) 2 Hence proved Ex.7 In fig. bisectors of B and D of quadrilateral ABCD meet CD and AB produces at P and Q 1 respectively. Prove that P Q (ABC ADC) 2 Sol. Given : bisectors of B and D of quadrilateral ABCD meet CD and AB produced at P and Q 1 To Prove : P Q (ABC ADC ) 2 Proof : STATEMENT REASON sum of all the angles of (1) In PBC, 0 tiriangle is 180 P 4 C 1800 1 Adding (i) and (ii), (i) P B C 1800 2 (2) In QAD, A B C D 3600
Q A 1 1800 1 (ii) Q A D 1800 2 1 1 (iii) P Q A A B D 1800 1800 2 2 1 1 P Q A C B D 3600 2 2 1 P Q A C (B D) A B C D 2 1 P Q (B D) 2
[Sum of the angles of a 0 quadrilateral equal is 360 ]
MANISH KUMAR 1 ( ABC ADC) 2 Hence proved
P Q
MATHEMATICS
MANISH KUMAR Ex.8
MATHEMATICS
In quadrilateral ABCD B 90 0 , C - D 60 0
and A - C - D 10 0. Find A, C and D.
Sol.
A B C D 3600 (Sum of the four angles of a quadrilateral is 3600)
A C D 3600 B A C D 3600 900 A C D 2700
.....(1)
It is given that
A C D 100
......(2)
C D 60 0
.....(3)
Adding (1) and (2), we get
(A C D) (A C D) 2700 100
A C D A C D 2800 2A 2800
A
2800 2
A 1400 From (1), 1400 + A D = 2700
C D 2700 1400 C D 1300 Adding (3) and (4), we get
(C D) C D 600 1300
C D C D 1900 2 C 1900
C
1900 2
C 950 Subtracting (3) from (4), we get
(C D) (C D) 1300 600
C D C D 700 2 D 70 0
C
700 2
…..(4)
MANISH KUMAR
MATHEMATICS
D 3500 Ex.9
In quadrilateral ABCD
A C 140 0 , A : C1 : 3 and B : D 5 : 6. Find the A, B, C and D. Sol.
A C 1400
(Given)
A C 1 : 3
(Given)
sum of ratio = 1 + 3 = 4
A
1 1400 350 4
and C
3 1400 350 3 1050 4
Sum of all the angles of quadrilateral is 3600 We have A B C D 3600
350 B 1050 D 3600 B D 1400 3600 B D 3600 1400 B D 2200 It is given that B : D 5 : 6 sum of ratios = 5 + 6 = 11
B and
5 2200 200 6 1100 11
D
6 2200 200 6 1200 11
Hence, A 350 , B 1000 , C 1050 and D 1200 7.
PROPERTIES OF A PARALLELOGRAM THEOREM-1 A diagonal of a parallelogram divides it into two congruent triangles. Given : ABCD is a parallelogram and AC is a diagonal which form two triangles CAB and ACD. To Prove : ACD CAB Proof :
1. 2. (i) (ii) 3.
STATEMENT AB || DC AND AD || BC AB || DC AND AC is a transversal ACD CAB AD || BC and AC is a transversal CAD ACB In ACD and CAB, ACD CAB AC = AC
REASON ABCD is a parallelogram Alternate angles [From (2)] [Common] [From (3)] [By ASA congruence rule]
MANISH KUMAR
MATHEMATICS
CAD ACB Therefore, ACD CAB Hence, proved. THEOREM-2 In a parallelogram, opposite sides are equal. Given : ABCD is a parallelogram To Prove : AB = CD and BC = DA Construction : Join AC. Proof :
STATEMENT 1.
AB || DC and AD || BC
2.
In ABC and CDA
REASON
Since ABCD is a parallelogram
BAC DCA Alternage angles
AC = AC
ACB CAD
Common
ABC CDA
Alternate angles
AB = CD and BC = DA
By ASA congruence rule C.P.C.T.
Hence, proved. THEOREM-3 In a parallelogram, opposite angles are equal.
Given : ABCD is a parallelogram. To Prove : A C and B D
Proof :
STATEMENT 1.
AB || DC and AD || BC
2.
AB || DC and AD is a transversal
REASON
Since ABCD is a parallelogram
A D 1800 3.
AD || BC and DC is transversal
D C 1800 4.
A D D C A C
Sum of consectuve interior angles is 1800
Sum of consectuve interior angles is 180
From (2) & (3)
0
MANISH KUMAR 5.
Similarly, B D A C and B D
Hence proved.
MATHEMATICS
MANISH KUMAR
MATHEMATICS
THEOREM-4 : The diagonals of a parallelogram bisect each other. Given : ABCD is a parallelogram, diagonals AC and BD intersect at O. To Prove: OA = OC and OB = OD
Proof :
STATEMENT 1.
AB || DC and AD || BC
2.
In AOB and COD
BAO DCO
REASON
ABCD is a parallelogram
Alternate angles
AB = CD
ABO CDO
Opposite sides of a parallelogram
3.
AOB COD
Alternate angles
4.
OA OC and OB OD
By ASA congruence rule [C.P.C.T.]
Hence Proved. CONDITION FOR A QUADRILATERAL TO BE PARALLELOGRAM : THEOREM-5 : If each pair of opposite sides of a quadrilateral is equal, then it is parallelogram. Given : A quadrilateral ABCD in which AB = CD and CB = AD. To Prove : ABCD is a parallelogram. Contruction : Join AC. Proof :
STATEMENT 1.
REASON
In ABC and CDA AB = CD
[Given]
CB = AD
[Given]
AC = AC
[Common]
ABC CDA
[By SSS congruence rule]
BAC DCA
[C.P.C.T.]
2.
Similarly BCA DAC
Alternate interior angles
BC || AD 3.
AB || DC and BC || AD
4.
ABCD is parallelogram
Hence Proved:
[C.P.C.T.] alternate interior angles [From (1) & (2)]
MANISH KUMAR
MATHEMATICS
THEOREM-6 If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram : Given : A quadrilateral ABCD in which c and B D To Prove : ABCD is parallelogram Construction : Join AC & BD Proof : STATEMENT 1.
REASON
A C
Given
1 1 A C 2 2
Halves of equal are equal
BAC DCA (i) AB || DC
Alternage angles
B D
Given
1 1 B D 2 2
DBC ADB
Halves of equal are equal
(ii) AB || BC 2.
AB || DC, AB || BC
Alternate angles
3.
ABCD is a paallelogam
From (i) & (ii)
Hence proved THEOREM-7: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. Given : ABCD is a quadrilateral whose diagonals AC and BD intersect at point O such that OA = OC and OB = OD. To prove : ABCD is a parallelogram. Proof : STATEMENT 1.
REASON
In AOB and COD, OA = OC
[Given]
AOB COD
[Vertically opposite angles]
OB = OD
[Given]
AOB COD
[By SAS congruence rule]
BAO DCO
CPCT
AB || DC
2.
Similarly, AD || BC
3.
ABCD is a parallelogram.
Alternate interior angles
MANISH KUMAR
MATHEMATICS
Hence proved. THEOREM-8 : - A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel. Given : ABCD is a quadrilateral in which AB - CD and AB||DC To Prove : ABCD is a parallelogram. Construction : Join AC. Proof :
STATEMENT 1.
REASON
In ABC and CDA [Given] [Common]
AB = CD AC = AC
Alternate interior angles [ AB || DC] and AC intersects
BAC DCA
ABC CDA
BCA DAC
AD || BC
2.
AB || CD and AD||BC
3.
ABCD is parallelogram.
them [BY SAS] CPCT Alternate interior angles.
Hence, Proved. Ex.10
In figure, ABCD is a parallelogram. Compute the values of x and y.
Sol.
Since ABCD is a parallelogram. Therefore AB||DC and AD||BC. Now, AB||DC and transversal BD intersect them
ABD BDC
[ Alternate angles are equal]
12x = 60 X
60 12
x=5 and, AD || BC and transversal BD intersects them.
DBC ADB
7y = 28
MANISH KUMAR
MATHEMATICS
y=4 Hence, x = 5 and y = 4 Ex.11 Sol.
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. [NCERT] Given : ABCD is a quadrilateral where diagonals AC and BD intersect each other at right angles at O. To Prove: Quadrilateral ABCD is a rhombus. Proof : In AOB and AOD ,
Ex.12 Sol.
AO = AO
[Common]
OB = OD
[Given]
AOB AOD
[Each = 900
AOB AOD
[SAS Rule]
AB = AD Similarly, we can prove that AB = BC ...... (i) BC = CD ...... (ii)
[C.P.C.T.]
CD = AD ......(iii) (i), (ii), (iii) and (iv), we obtain AB = BC = CD = DA Show that the diagonals of a square are equal and bisect each other at right angles. Given : ABCD is a square. To Prove: (i) AC = BD (ii) AC and BD bisect each other at right angles. Proof : (i) In ABC and BAD AB = BA BC = AD
[Common] [Opp. sides of square ABCD]
ABC BAD
[Each = 900 ( ABCD is a square]
ABC BAD
[SAS Rule]
AC = BD
[C.P.C.T.]
(ii) In OAD and OCB AD = CD
[Opp. sides of square ABCD]
OAB OCB
[ AD||BC and transversal AC intersect them]
ODA OBA
[ AD||BC and transversal BD intersects them]
OAD OCB
OA = OC Similarly, we can prove that OB = OD
[ASA Rule] ......(i) ....(ii)
From (i) and (ii), AC and BD bisect each other. Again in OBA and ODA OB = OD BA = DA OA = OA
[SSS Rule]
[From (ii)] [Opp. sides of square ABCD [Common]
[NCERT]
MANISH KUMAR
MATHEMATICS
[C.P.C.T.]
But
AOB AOD = 1800 [Linear pair] AOB AOD = 900
AC and BD bisect each other at right angles. Ex.13
Show that if the diagonals of quadrilateral are equal and bisect each other at right angles. then it is a square. [NCERT]
Sol.
Given : the diagonals AC and BD of a quadrilateral ABCD are equal and bisect each other at right angles. To Prove : Quadrilateral ABCD is a square. Proof : In OAD and OCB ,
OA - OC
[Given]
OD = OB
[Given]
AOD COB
[Vertically Opposite Angles]
OAD OCB
[SAS Rule]
AD = CB
[C.P.C.T.]
ODA OBC
[C.P.C.T.]
AD||BD Now, AD = CB and AD||CB
Quadrilateral ABCD is a || gm. In AOB and AOD , AO = AO
[Common]
OB = OD
[Given]
AOB AOD
[Each = 900 (Given)]
[SAS Rule]
AOB AOD
AB = AD Now,
ABCD is a parallelogram and AB = AD
ABCD is a rhombus. Again, in ABC and BAD , AC = BD
[Given]
BC = AD
[ ABCD is a Rhombus]
AB = BA
[Common]
ABC BAD
ABC BAD
[C.P.C.T.]
AD||BC and transversal AB intersect them.
[Opposite sides of || gm ABCD]
ABC BAD =180 [Sum of consecutive interior angles on the same side of the transversal is 180 0\ 0
MANISH KUMAR Similarly, BCD ADC =900
ABCD is a square.
MATHEMATICS
MANISH KUMAR Ex.14
MATHEMATICS
ABCD is a rhombus. Show that diagonal AC bisect A as well as C and diagonal BD bisect B as well as D .
Sol.
Given : ABCD is a rhombus. To prove : (i) Diagonal AC bisects A as well as C . (ii) Diagonal BD bisect B as well as D . Proof : ABCD is a rhombus
AD = CD DAC DCA
....(1)
[l Angles opposite to equal sides of a triangle are equal.]
Also, CD||AB and transversal AC intersects them
DAC BCA ...(2)
Alt. Int. S
From (1) and (2) DCA BCA
AC bisect C Similarly AC bisect A (ii) Proceeding similarly as in (i) above, we can prove that BD bisect B as well as D . Ex.15
ABCD is a rectangle in which diagonal AC bisects A as well as C . Show that (i) ABCD is a square (ii) diagonal BD bisect B as well as D .
Sol.
Given : ABCD is a rectangle in which diagonal AC bisect A as well as C . To prove : (i) ABC is a square (ii) diagonal BD bisects B as well as D . Proof : (i) AB || DC and transversal AC intersects them.
ACD CAB CAB CAD
But
[Alt. Int. S ]
ACD CAD
[Sides opposite to equal angles of a triangle are equal]
AD = CD
ABCD is a square (ii) In BDA and DBA BD = DB
[Common]
DA = BC
[Sides of a square ABCD]
AB = DC
[Sides of a square ABCD]
BDA DBC
[SSS Rule]
ABD CBD
[C.P.C.T.]
But
CBD CBD
[ CB = CD (Sides of square ABCD)]
ABD CBD
BD bisect B . Now, ABD CBD
ABD ADB
[ AB = AD]
CBD CDB
[ CB = CD]
ADB CBD
MANISH KUMAR Ex.16
Sol.
MATHEMATICS
BD Bisect D .
ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD respectively. Show that : Δ APB Δ (i) CQD (ii) AP = CQ Given : ABCD is a parallelogram and AP and CQ are perpendiculars form vertices A and C on diagonal BD respectively. To prove : (i) APB CQD (ii0 AP = CQ Proof : (i) APB and CQD AB = CD [Opp. side of || gm ABCD] ABP CDQ
[ AB || DC and transversal BD intersect them]
APB CQD
[Each = 900]
APB CQD
[AAS Rule]
Ex.17
(ii) APB ACQD [Proved above in (i)] AP = CQ [C.P.C.T.] ABCD is a trapezium in which AB||CD and AD = BC. Show that (i) A B (ii) C D
Sol.
(iii) ABC BAD (iv) diagonal AC = diagonal BD. Given : ABCD is a trapezium in which AB || CD and AD = BC. To prove : (i) A B (ii) C D (iii) ABC BAD (iv) diagonal AC = diagonal BD. Construction : Extend AB and draw a line through C parallel to DA intersecting AB produced at E. Proof : (i) AB || CD [Given] and AD || EC [By construction] [A quadrilateral is a parallelogram if a pair of opposite AECD is a parallelogram sides is parallel and is of equal length] [Opp. sides of a || gm are equal] AD = EC But AD = BC [Given] EC = BC .... (1) [Angles of opposite to equal of a triangle are equal] CBE CEB
B CBE 1800 = 1800 .....(2) [Linear Pair]
AD || EC and transversal AE intersect them [By construction]
A CEB 1800
….(3)
[The sum of consecutive interior angles on the same] side of the transversal is 1800]
From (2) and (3) B CBE A CEB But CBE CEB
[From (1)]
MANISH KUMAR
MATHEMATICS
B A or A B (ii) AB || CD
A D 1800
and B C 180
Ex.18
Sol.
[The sum of consecutive interior angles of the same sides
0
of the transversal is 1800]
A D B C But A B [Prove in (i)] D C or C D (iii) In ABC and BAD AB = BA [Common] BC = AD [Given] [From (i)] ABC BAD [SAS Rule] ABC BAD (iv) ABC BAD [From (iii) above] [C.P.C.T.] AC = BD In figure ABCD is a parallelogram in which D = 720. Find A, B and C .
We have But
D 720 B D
[Opposite angles of the parallelogram]
B 72 Now, AB || CD and AD and BC are two transversals. 0
So,
[Interior angles on the same side of the transversal AD] A D 1800 0 0 A + 72 = 180 A = 1800 - 720 = 1080 [Opposite angles of the parallelogram] C
Hence, A 1800 , B 720 and C 1800 Ex.19. Prove that the sum of any two consecutive angles of a parallelogram is 180 0.
Sol.
Since ABCD is a parallelogram, therefore, AB||CD and AD||BC. AB||CD and AD and BC are two transversal.
A D 1800
[Interior angles on the same side of the transversal is 1800]
and B C 180 Similarly AD || BC and AB and CD are two transversal. 0
and
A B 1800 C D 180
...(1) ....(2) ....(3)
0
....(4)
MANISH KUMAR
MATHEMATICS
Hence A B B C C D D A 1800 [Using (1), (2), (3) and (4)]
MANISH KUMAR Ex.20
MATHEMATICS
In figure ABCD is a parallelogram and AP and CQ are bisectors of A and C . Prove that AP||CQ.
Sol.
We have
A C
gm
[Opposite angles of a|| ]
1 1 A C 2 2
PAQ PCQ
...(1)
Now, AB||CD and CQ is a transversal, Therefore, PCQ PCQ
PAQ BQC
[From (1)]
Ex.21
But, these are corresponding angles formed when AP and CQ are intersected by transversal AB. AP||CQ Hence proved. Show that each angles of rectangle is a right angle.
Sol.
Let ABCD be a rectangle. WE know that a rectangle is a parallelogram in which one angle is right angle In rectangle ABCD and A 900 We know that a rectangle is a particular type of parallelogram. Also, the opposite angles in a parallelogram are equal. Therefore,
A C
and
B D
C 90
Now,
A B C D 3600
90 0 B 90 0 D 3600
B D 1800 and B D
B D
Hence,
A B C D 900
[By theorem]
0
1 1800 900 2
Ex.22
Show that the diagonals of a rectangle are equal.
Sol.
In rectangle ABCD, AC and BD are diagonals. We have A B C D 900 In ABC and DCB , we have AB = DC
(Pair of opposite side of rectangle ABCD)
ABC DCB and BC = CB
ABC DCB
AC = BD
(Each = 900) (Common side) (SAS congruence criteria) (By CPCT)
MANISH KUMAR Ex.23 Sol.
Hence Proved Show that the diagonals of a rhombus are perpendicular to each other. In a rhombus ABCD, Diagonals AC and BD intersect each other at O. Then OA = OC and OB = OD ( Diagonals of parallelogram bisect each other) Also AB = BC = CD = DA ( All four sides of Rhombus are equal) In AOB and COB OA = OC, AB = BC and OB = OB [Common side] [By SSS congruence] AOB COB
AOB COB
Also,
[Sum of linear pair of angles] AOB COB 180 1 AOB COB 1800 900 2
Sol.
[By CPCT] 0
Ex.24
MATHEMATICS
Similarly, AOD COD 900 Therefore, diagonals AC and BD and perpendicular to each other. AB and CD are two parallel lines and a transversal intersects AB at X and CD at Y. Prove that the bisector of the interior angles form a rectangle. Given : AB and CD are two parallel lines are transversal intersect AB at X and CD at Y. To Prove : The bisectors of the interior angles form a rectangle. Proof : AB || CD and EF intersect them [Alternate S ] BXY CYX 1 1 BXY CXY [Halves of equal are equal] 2 2 1 3 But these angles from a pair of equal alternate angles of lines XQ and SY and transversal XY. ...(1) XQ || SY Similarly, we can prove that SX || YQ ....(2) From (1) & (2) SYQX is a parallelogram A quadrilateral is a parallelogram if both pairs of its opposite sides are parallel. Now,
BXY DYX 1800 | consecutive interior s | 1 1 1 BXY DYX 180 2 2 2 1 2 900 1 2 XQY 1800
90 XQY 180 0
[Angles sum property of a ]
0
XQY 1800 90 0
XQY 900 YSX 900
[ Opposite s of a ||gm are equal]
and
SXQ 90
0
[ Consecutive interior angles are supplementary]
Now,
SXQ 90
0
[Opposite s of a a || gm are equal] SYQ 90 0 Thus each angle of the parallelogram SYQX is 900
MANISH KUMAR
MATHEMATICS
Hence proved. Hence parallelogram SYXQ is a rectangle.
Ex.25
If the diagonals of a parallelogram are equal, them show that it is a rectangle.
Sol.
Given : In parallelogram ABCD, AC = BD.
[NCERT]
To Prove : || gm ABCD is a rectangle. Proof : In ACB and BDA AC = BD
| Given
AB = BA
| Common
BC = AD
| Opposite sides of || gm ABCD
ACB BDA
| SSS Rule
ABC BDA
....(1) c.p.c.t.
Again AD || BC | Opp. side of || gm ABCD and transversal AB intersect them.
BAD ABC 1800
….(2)
Sum of consecutive interior angles on the same side of the transversal is 1800
From (1) and (2),
BAD ABC 900
A 900
|| gm ABCD is a rectangle. Ex.26
ABCD is a parallelogram and line segments AX, CY bisect the angles A and C respectively. Show that AX || CY.
Sol.
Given : ABCD is a parallelogram and line segments AX, CY bisect the angles A and C respectively. To prove : AX | CY. Proof : ABCD is a parallelogram.
A C
1 1 A A 2 2
| Opposite s | Halves of equals are equal
1 2
...(1) | AX is the bisector of A and CY is the bisector of C
Now,
AB || DC and CY intersect them
2 3
From (1) and (2), we get
1 3
...(2)
| Alternate interior s
MANISH KUMAR But these are corresponding angles
AX || CY.
MATHEMATICS
MANISH KUMAR
MATHEMATICS
MID - POINT THEOREM THEOREM 1 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side. GIVEN :
E and F are the mid-points of the sides AB and AC respectively of the ABC .
TO PROVE : EF || BC. CONSTRUCTION : Through the vertex C, CG is drawn parallel to AB and it meets EF (produced) in G. PROOF :
STATEMENT In AEF and CGF AF = CF AFE CFG EAF GCF ….(i) AEF CGF AE = CG …..(ii) BE = CG BE || CG Therefore, BCGE is a parallelogram. EG || BC EF || BC.
REASON ( F is mid-point of AC) (Pair of vertically opposite angles) (Pair of alternate angles) (By SAS congruence rule) ( E is mid-point of AB) From (i) and (ii) by construction
Hence proved. Converse of Theorem 1 : The line drawn through the mid-point of one side of a triangle and parallel to another side of the triangle. GIVEN : ABC is a triangle in which D is mid-point of AB and DE || BC. TO PROVE : E is mid-point of AC. CONSTRUCTION :Let E is not the mid-point of AC. If possible, let F is the mid-point of AC. Join DF. PROOF :
STATEMENT D is the mid-point of AB and F is the mid-point of AC. DF || BC But, it is given that DE || BC
E and F concide. Hence, E is the mid-point of AC.
REASON [Given] [By construction] [By mid-point theorem] This is not possible that two lines parallel to the same line intersect each other. So, our supposition is wrong.
MANISH KUMAR
MATHEMATICS
Hence proved. THEOREM 2 : Length of the line segment joining the mid points of two sides of a triangle is equal to half the length of the third side.
GIVEN : In ABC, , EF is the-line segment joining the mid-points of the sides AB and AC of ABC
1 BC. 2 CONSTRUCTION : Through C, draw CG||BA, CG meets EF (produced) at G. PROOF : TO PROVE : EF
STATEMENT In AEF and CGF , we have AF = CF 1 2 and 3 4 aef cgf (i) AE = CG and (ii) EF = FG Also, (iii) AE = BE we have BE = CG BE || CG BCGE is a parallelogram BC = EG = EF + FG = EF + EF = 2 EF 2EF = BC 1 EF BC 2 Hence proved. Ex.27 Sol.
REASON ( F is mid-point of AC) (Vertically opposite angles) (pair of interior alternate angles) (By COCT) (By CPCT) ( E is mid-pont of AB) Then from (i) and (iii) By construction, we have ( CG || BA) (By (ii)
In the following figure, D, E, and F are respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. Prove that DEF is also an equilateral triangle. Given : D, E and F are respectively the mid-points of sides BC, CA and AB of a equilateral triangle ABC. To prove : DEF is also an equilateral triangle. Proof : since the segment joining the mid points of two sides of a triangle is half of the third side. Therefore D and E are the mid point of BC and AC respectively. 1 ...(i) DE AB 2 E and F are the mid point of AC and AB respectively. 1 ...(ii) EF BC 2 F and D are the mid point of AB and BC respectively 1 ....(iii) FD AC 2 is an equilateral triangle AB = BC = CA 1 1 1 AB BC CA 2 2 2
MANISH KUMAR
MATHEMATICS
DE = EF = FD
Ex.28 Sol.
Ex.29
using (i), (ii) & (ii) Hence, DEF is an equilateral triangle. Hence Proved In figure, D and E are the mid-point of the sides AB and AC respectively of ABC . If BC = 5.6 cm, find DE. D is mid-point of AB and E is mid-point of AC. 1 DE BC 2 1 5.6 cm 2.8 cm. 2 In figure, E and F mid-points of the sides AB and AC respectively of the ABC , G and H are midpoints of the sides AE and AF respectively of the AEF . If GH = 1.8 cm, find BC.
1 BC ....(1) ( E and F are mid-points of sides AB and AC of ABC ) 2 1 ....(2) ( G and H are mid-point of sides AE and AF of AEF ) GH EF 2 From (1) and (2), We have 1 1 1 GH BC BC 2 2 4 BC = 4 × GH = 4 1.8 cm = 7.2 cm Hence, BC = 7.2 cm. Ex.30 In ABC, D,E and F are the mid- point of BC, BA and AC: Sol.
Sol.
EF
(i) Prove that AEF EBD (ii) Prove that CDEF is a parallelogram. It is given that D, E and F are the mid-points of the sides BC, BA and AC respectively, in ABC . 1 1 EF BC, DE AC 2 2 1 Now, BD = DC BC 2 1 and AF = CF = BC 2 BD = DC = EF ...(1) and AF = CF = DE ...(2) If E is the mid-point of AB. Therefore , AE = BE …(3) (i) In AEF and EBD EF = BD (By 1) AF = ED (By 2) and AE = BE (By 3)
MANISH KUMAR Therefore, AEF EBD (ii) In quadrilateral CDEF, DC = EF and DE = CF Therefore, CDEF is a parallelogram.
MATHEMATICS (By SSS congruence) (By 1) (By 2)
MANISH KUMAR Ex.31 Sol.
MATHEMATICS
Prove that the line segments joining the mid-points of the sides of quadrilateral forms a parallelogram. GIVEN : Point E, F G and H are the mid-points of the sides AB, BC, CD and DA respectively, of the quadrilateral ABCD. TO PROVE : EFCG is a parallelogram. CONSTRUCTION : Join the diagonal AC of the quadrilateral ABCD. PROOF : In ABC , E and F are the mid-points of BA and BC.
EF||AC
1 AC 2 Similarly, In ADC , we have GH||AC and
EF
...(1)
1 AC ...(2) 2 From (1) and (2), we get EF || GH and EF = GH EFGH is a parallelogram. Hence Proved. Ex.32 ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that : and
GH
1 AC (ii) PQ = SR 2 (iii) PQRS is a parallelogram. [NCERT] GIVEN : ABCD is a quadrilateral in which P, Q, R, and S are mid-points of AB, BC, CD and DA. AC is a diagonal. (i) SR || AC and SR =
Sol.
TO PROVE :
(i) SR || AC and SR =
1 AC 2
(ii) PQ = SR (iii) PQRS is a parallelogram. PROOF : (i) In DAC , S is the mid-point of DA and R is the mid-point of DC
SR || AC and SR =
1 AC 2
[Mid-point theorem]
(ii) In BAC,
P is the mid-point of AB and Q is the mid-point of BC PQ || AC and PQ = But from (i), SR =
1 AC 2
[Mid-point theorem]
1 AC 2
PQ = SR (iii) PQ || AC SR || AC PQ || SR Also, PQ = SR PQRS is a parallelogram
[From (ii)] [From (i)] [Two lines parallel to the same line are parallel to each other] [From (ii)] A quadrilateral is a parallelogram if a pair of opposite
MANISH KUMAR
MATHEMATICS sides is parallel and is of equal length]
Ex.33
ABCD is a rhombus and P, Q, R and S are the mid-point of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Sol.
Given : ABCD is a rhombus, P, Q, R, S are the mid-points of AB, BC, CD, DA respectively. PQ, QR, RS and SP are joined. To Prove : PQRS is a rectangle. Construction : Join AC and BD. Proof : In triangles RDS and PBQ.
DS = QB
[Halves of opposite side of || gm ABCD which are equal]
DR = PB
[Halves of opposite of || gm ABCD which are equal]
SDR QBP
[Opposite s of || gm ABCD which are equal]
RDS PBQ
[SAS Axiom]
SR = PQ
[C.P.C.T.]
In RCQ and PAS. RC = AP
[Halves of opposite side of || gm ABCD which are equal]
CQ = AS
[Halves of opposite side of || gm ABCD which are equal]
RCQ PAS
[Opposite s of || gm ABCD which are equal]
RCQ PAS
[SAS Axiom]
RQ = SP
[C.P.C.T.]
In PQRS, SR = PQ and RQ = SP PQRSis a parallelogram In CDB,
R and Q are the mid-point of DC and CB respectively. RQ || DB
RF || EO.
Similarly, RE || FO
OFRE is a || gm
R EOF 90 0 [ Opposite s of a || gm are equal and diagonals of rhombus intersect at 900]
MANISH KUMAR Thus PQRS is rectangle.
MATHEMATICS
MANISH KUMAR Ex.34
MATHEMATICS
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is rhombus.
Sol.
Given : ABCD is rectangle. P,Q,R and S are the mid-points of AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined. To prove : Quadrilateral PQRS is a rhombus Construction : Join AC. Proof : In ABC ,
P and Q are the mid-point of AB and BC respectively. PQ || AC
and
PQ
1 AC 2
...(1)
In ADC,
S are R are the mid-point of AD and DC respectively. SR || AC
and
SR
1 AC 2
..(2)
From (1) and (2), PQ || SR and
PQ = SR
Quadrilateral PQRS is a parallelogram
....(3)
In rectangle ABCD,
AD = BC
[Opposite sides]
1 1 AD BC 2 2
[Halves of equal are equal]
As = BQ In APS and BPQ ,
AP = BP
[ P is the mid-point of AB]
AS = BQ
[Proved above]
PAS PBQ
[Each = 900]
APS BPQ
[SAS Axiom]
PS = PQ From (3) and (4) PQRS is a rhombus.
...(4)
[C.P.C.T.]
MANISH KUMAR Ex.35 Sol.
MATHEMATICS
Show that the quadrilateral formed by joining the mid-point of the consecutive sides of a square is also a square is also a square. Given : ABCD is a square. P, Q, R and S are the mid-points of the consecutive sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined. To prove : PQRS is a square. Construction : Join AC and BD Proof :
RQ || AC and RQ =
1 AC 2
1 AC 2 RQ = SP and RQ || SP Similarly, SR = PQ and RQ || PQ PQRS is a parallelogram RQ || AC RE|| HO SR || PQ HR || OE OERH is a parallelogram. R HOE SP || AC and SP =
But HOE = 90
[Opposite s of a || gm]
0
R 900 PQRS is a rectangle. But AC = BD PQ = QR = RS = SP PQRS is a square. Hence Proved. Ex.36 Sol.
Prove that the figure formed by joining the mid-points of the consecutive sides of a quadrilateral is a parallelogram. Given : ABCD is a quadrilateral. P, Q, R and S are the mid-points of the consecutive sides AB, BC, CD and DA respectively. To prove : PQRS is a parallelogram. Construction : Joing BD. Proof : In CBD,
Q is the mid-point of BC dna R is the mid-pind o CD. QR || BD and QR
1 BD 2
...(1)
In ABD,
P is the mid-point of AB and S is the mid-point of AD. PS || BD and PS
1 2
...(2)
From (1) and (2), QR = PS and QR || PS. Thus a pair of opposite sides of PQRS are parallel and equal.
PQRS is a parallelogram. Hence Proved.
MANISH KUMAR Ex.37
MATHEMATICS
In triangle ABC, points M and N on des AB and AC respectively are taken so that AM =
1 1 AC. Prove that MN = BC. 4 4 Given : In triangle ABC, points M and N on the sides AB and AC 1 1 respectively are taken so that AM AB and AN = AC. 4 4 1 To prove : MN BC. 4 Construction : Join EF where E and F are the middle points of AB and AC respectively. Proof : E is the mid-point of AB and F is the mid-point of AC. 1 ...(1) EF||BC and EF = BC 2 1 1 Now, AE = AB and AM AB 2 4 1 AM = AE 2 1 Similarly, An = AF 2 M and N are the mid-points of AE and AF respectively. AN =
Sol.
MN||EF and MN
1 11 EF = BC 2 22
[From (1)]
1 BC. 4 Ex.38 The diagonals of a rectangle ABCD meet at O. If BOC = 440, find OAD . Sol. We have, MN =
BOC BOA 1800
[Linear pairs]
440 + BOA = 1800 BQA = 1360 Since diagonals of a rectangle are equal and they bisect each other. In OAB OA = OB [ Angles opp. to equal sides are equal] 1 2 Now, In OAB , where
1 2 BOA 1800
21 1360 1800
21 = 1800 - 1360 21 = 440 1 = 220 Since each angle of a rectangle is a right angle. Therefore, BAD = 900 1 3 = 900
220 + 3 = 900 3 = 900 - 22 = 680 Hence Proved.
1 AB and 4
MANISH KUMAR Ex.39 Sol.
MATHEMATICS
PQRS is a square. Determine SRP . Since PQRS in a square. PS = SR and PSR = 900 In PSR PS = SR 1 2 [ Angles opp. to equal sides are equal] But, 1 2 PSR = 1800 [ PSR = 900] 21 + 900 = 1800
21 = 900 1 = 450 Ex.40 Sol.
ABCD is rectangle with BAC 32 0 . Determine DBC . Suppose the diagonals AC and BD intersect at O. In OAB OA = OB [ Diagonals of a rectangle are equal and they bisect each other] OAB OBA
BAC DBA DBA 320 Now,
[ BAC = 320 (given)]
ABC 900
DBA DBA 900 320 + DBC = 900 DBC = 900 - 320 = 580 Ex.41 Sol.
ABCD is a rhombus with ABC 560 . Determine ACD . ABCD is a rhombus ABCD is a parallelogram
ABC ADC ADC 560
[ ABC = 560 (given)]
CDA = 280
[ ODC
1 ADC ] 2
In OCD OCD ODC COD 1800
OCD + 280 + 900 = 1800 OCD =620 ACD = 620 Ex. 42 In figure , m and n are three parallel lines intersected by transversals p and q such that , m and n cut-off equal intercepts AB and BC op p. Show that , m and n cut off equal intercepts De and EF on q also. Sol. Join AF. Suppose it interests line m at G. In ACF , B is the mid-point of AC ( AB = BC and BG || CF. Therefore, G is the mid-points of AF. In AFD , G is the mid-point of AF and GD || AD. E is the mid-point of DE DE = EF
MANISH KUMAR Hence , m and n cut of equal intercepts DE and EF on q.
MATHEMATICS
MANISH KUMAR
MATHEMATICS
EXERCISE SUBJECTIVE TYPE QUESTIONS. (A)
VERY SHORT ANSWRS TYPE QUESTIONS :
1.
Three angles of quadrilateral are respectively equal to 1100, 500 and 400 . Find its fourth angles.
2.
In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 2 : 4 : 5. Find the measure of each angles of the quadrilateral.
3.
In a quadrilateral ABCD, A,B 1500 and C = 1300 . Find D .
4.
In a quadrilateral ABCD, A B 2100 and C : D = 2 : 3 Find C and D
5.
What is a trapezium.
6.
State the conditions for quadrilateral ABCD to be a parallelogram.
7.
State the condition for a parallelogram to be a rectangle.
8.
State the type of quadrilateral ABCD if AB = BC = CD = DA, A C and B D .
9.
“In a parallelogram, the diagonals are of equal length.” Is the statement true or false ?
10.
“The diagonals of a parallelogram bisect each other.” In the statement true/False ?
11.
In ABC , LM is the line segment joining the mid-points of the sides BC and CA. If LM = 3.4 cm, find the length of AB.
12.
IF ABC , LM is a line segment parallel to side BC of ABC where L and M divide AB and AC respectively in the ratio 1: 3 If. If BC = 8.4 cm, find LM.
(B)
SHORT ANSWERS TYPE QUESTIONS :
1.
Two opposite angles of a parallelogram are (3x -2)0 and (50 -x)0. Fin the measure of each angle of the parallelogram.
2.
If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.
3.
Find the measure of all the angles of a parallelogram, if one angle is 24 0 less than twice the smallest angle.
4.
The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side /
5.
In a parallelogram ABCD, D = 1350, determine the measure of A and B .
6.
ABCD is a parallelogram in which A 700 . Compute B, C and D .
7.
In figure ABCD is a parallelogram in which DAB 750 and BCD 600 . Compute CDB and ADB .
8.
In figure ABCD is a parallelogram and E is the mid-point of side BC. If De and AB when produced meet at F, prove that AF = 2AB.
MANISH KUMAR
MATHEMATICS
9.
In a parallelogram ABCD, determine the sum of angles C and D .
10.
In a parallelogram ABCD, if B = 1350 determine the measure of its other angles.
11.
ABCD is a square. AC and BD intersect at O. State the measure of AOB .
12.
ABCD is a rectangle with ABD = 400. Determine DBC .
13.
The sides AB and CD of a parallelogram ABCD are bisected t E and F. Prove that EBFD is a parallelogram.
(C)
LONG ANSWER TYPE QUESTIONS :
1.
Given ABC , lines are drawn through A, B and C parallel respectively to the sides BC, CA and AB forming PQR , Show that BC =
2.
1 QR. 2
In the given figure, ABCD is a parallelogram and X, Y are the mid-points of the sides AB and DC respectively. Show that quadrilateral AXCY is a parallelogram.
3.
In figure, three parallel lines , m and n are intersected by a transversal p at points A, B and C respectively and transversal q at D, E and F respectively. If AB : BC = 1 : 2, prove that DE : EF = 1 : 2.
4.
ABC is a triangle right angles at B; and P is the mid-point of AC. Prove that (i) PQ AB (ii) Q is the mid point of AB (iii) PB = PA =
5.
1 AC 2
In figure AD and BE are medians of ABC and BE||DE. Prove that CF =
1 AC 4
MANISH KUMAR 6.
MATHEMATICS
In ABC , AD is the median through A and E is the mid-point of AD. BE produced meets AC in F figure. Prove that AF
7.
1 AC. . 3
P is the mid-point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R. Prove that (i) AR = 2BC (ii) BR = 2BQ.
8.
In figure, ABCD is a trapezium in which side AB is parallel to side DC and E is the mid-points of side AD. If F is a point of the side BC such that the segments EF is parallel to side DC. Prove that F is the mid-point of BC and EF
1 ( AB DC) 2
9.
In the given figure, ABCD is a parallelogram such that AEB CFD . Prove that DAE BCF .
10.
In the figure, ABCD is a square and PAB is a triangle such that AQ = BR. Prove that PQR is an isosceles triangle.
11.
In a ABC D, E, and F are, respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of DEF .
12.
In a triangle ABC, A 500 , V 600 and C 700 . Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.
13.
In a triangle, P, Q, and R are the mid-points of sides BC, CA and AB respectively. If AC = 21 cm, BC = 29 cm and AB = 30 c m, find the perimeter of the quadrilateral ARPQ.
MANISH KUMAR
MATHEMATICS
14.
In a ABC median AD in produced to X such that AD = DX. Prove that ABXC is a parallelogram.
15.
In figure, triangle ABC is right angles at B. Given that AB = 9cm, AC = 15cm and D, E are the mid-points of the sides AB and AC respectively, calculate (i) The length of BC
16.
(ii) The area of ADE
In figure, M, N and P are the mid- points of AB, AC and BC respectively. If MN = 2c, NP = 3.5 cm and MP = 2.5cm, calculate BC, AB and AC.
17.
ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of PQR is double the perimeter of ABC .
18.
In a parallelogram, show that the angle bisector of two adjacent angles intersect at right angles.
19.
In figure ABC and DEF are such that AB = DE, BC = EF, AB || DE and BC || EF. Show that (i) ABED is a parallelogram. (ii) BEFC is a parallelogram. (iii) ADFC is a parallelogram.
20.
In figure, ABCD is a parallelogram. E and F are mid-points of the sides AB and CD respectively. AF and DE intersects at P ; BF and CE intersects at Q. Prove that (i) AECF is a parallelogram (ii) BEDF is a parallelogram. (iii) PEQF is a parallelogram.
21.
In figure, ABCD is a parallelogram. E and F are two point on the diagonal AC such that AE = CF. Show that (i) AEB CFD (ii) AED CFB (iii) BEF DFE
MANISH KUMAR
MATHEMATICS
(iv) BEDF is parallelogram.
22.
In figure, ABCD is a trapezium such that AB || CD and AD = BC. Line drawn through the vertex B and parallel to AD meets DC (produced) at E show that (i) ABED is a parallelogram. (ii) A C B D =1800
(D)
NCERT Questions :
1.
In figure, ABCD is a parallelogram. Diagonal AC bisect A . Show that (i) it bisect C also (ii) ABCD is a rhombus.
2.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that : (i) APD CQB (ii) AP = CQ (iii) AQB CPD (iv) AQ = CP (v) APCQ is a parallelogram
3.
In ABC and DEF AB = DF, || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively. Show that : (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFC is a parallelogram (v) AC = DF (vi) ABC DEF
4.
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.
MANISH KUMAR 5. 6.
MATHEMATICS
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. ABC is a triangle right angled at C. A. line through the mid-point M of hypotenuse AB and parallel to BD intersects AC and D. Show that : (i) D is the mid - point of AC
(ii) MD AC
(iii) CM = MA =
1 AB 2
7.
Show that the bisectors of angles of a parallelogram form a rectangle.
8.
ABC is a isosceles triangle in which AB = AC. AD bisects exterior angle PAC and CD || AB. Show that (i)
9.
DAC BCA and
(ii) ABCD is a parallelogram.
Two parallel lines and m are intersected by a transversal p. Show that the quadrilateral formed by bisectors of interiors angles is a rectangle.
(E)
WHICH OF THE FOLLOWING STATEMENTS ARE TRUE (T) AND WHICH ARE FALSE (F) : (1) In a parallelogram, the diagonals are equal. (2) In a parallelogram, the diagonals bisect each other. (3) In a parallelogram, the diagonals intersect at right angles. (4) In any quadrilateral, if a pair of opposite sides is equal, it is a parallelogram. (5) If all angles of a quadrilateral are equal, it is a parallelogram. (6) If all sides of quadrilateral are equal, it is parallelogram.
(F)
(7) If three sides of quadrilateral are equal, it is a parallelogram. (8) If three angles of a quadrilateral are equal, it is parallelogram. FILL IN THE BLANKS : (1) The triangle formed by joining the mid-points of the sides of an isosceles triangle is____ (2) The triangle formed by joining the mid-points of the sides of right-triangle is_____ (3) The figure formed by joining the mid-points of the mid-points of the consecutive sides of a quadrilateral is _____
MANISH KUMAR
MATHEMATICS
(4) If a line is divided by three parallel line into two-segments of lengths in the ratio 1 : 3 another line will the divided by these parallel lines into two-segments of lengths in the ratio___
OBJECTIVE TYPE QUESTIONS : 1.
If is quadrilateral ABCD A 900 and AB = BC = CD = DA, then ABCD is a :(A) Parallelogram
2.
(D) None of these
(B) Rectangle
(C) Rhombus
(D) None of these
If the diagonals AC and BD of a quadrilateral ABCD bisect each other, then ABCD is a :(A) Parallelogram
4.
(C) Square
If in quadrilateral ABCD, AB || CD, then ABCD is a :(A) Parallelogram
3.
(B) Rectangle
(B) Rectangle
(C) Rhombus
(D) None of these
In figure EF = (A) 3 cm (B) 2.5 cm (C) 4 cm (D) None of these
5.
In quadrilateral ABCD, if A 600 and B : C : D 2 : 3 : 7 , then D (A) 1750
6.
(B) Parallelogram
(C) Trapezium
(D) Rectangle
(B) Parallelogram
(C) Rectangle
(D) Rhombus
The length of the diagonals of a rhombus are 16 cm and 12 cm. The side of the rhombus is :(A) 10 cm
9.
(D) None of these
If the diagonals of a quadrilateral bisect each other at right angles, then it is a :(A) Trapezium
8.
(C) 1500
In which of the following is the lengths of diagonals equal ? (A) Rhombus
7.
(B) 1350
(B) 12 cm
(C) 9 cm
(D) 8 cm
The length of a side of rhombus is 5m and one of its diagonals is of lengths 8m. The length of the other diagonal is :(A) 5m
10.
(D) 8m
(B) Trapezium
(C) Rectangle
(D) None
The bisectors of any two adjacent angles of a parallelogram intersect at :(A) 300
12.
(C) 6m
IF ABCD is parallelogram with two adjacent angles A and B equal to each other, then the parallelogram is a (A) Rhombus
11.
(B) 7m
(B) 450
The bisectors of the angle of a || gm enclose a :-
(C) 600
(D) 900
MANISH KUMAR (A) Parallelogram 13.
MATHEMATICS (B) Rhombus
(C) Rectangle
(D) Square
The figure formed by joining the mid points of the adjacent sides of a quadrilateral is a :(A) Parallelogram
(B) Rectangle
(C) Square
(D) Rhombus
MANISH KUMAR 14.
The figure formed by joining the mid points of the adjacent sides of a rectangle is a :(A) Square
15.
(C) Trapezium
(D) None of these
(B) Rectangle
(C) Trapezium
(D) None of these
The figure formed by joining the mid points of the adjacent sides of a square is a :(A) Rhombus
17.
(B) Rhombus
The figure formed by joining the mid points of the adjacent sides of rhombus is a : (A) Square
16.
MATHEMATICS
(B) Square
(C) Rectangle
(D) Parallelogram
0
If one angle of a parallelogram is 24 less than twice the smallest angle, then the largest angle of the parallelogram is :(A) 1760
18.
(C) 1120
(D) 1020
If an angle of parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is :(A) 1080
19.
(B) 600 (B) 540
(C) 720
(D) 810
In the given figure, ABCD is a parallelogram in which DAB 750 and DBC 60 0 then, BDC is equal to :
(A) 750 20.
(B) 600
(C) 450
(D) 550
Three angles of a quadrilateral are of magnitudes 800, 950 and 1200. The magnitude of the fourth angle is :(A) 800
(B) 650
(C) 750
(D) 700
MANISH KUMAR
MATHEMATICS
(A) VERY SHORT ANSWER TYPE QUESTIONS : 0 1. 160
0
(F)
4. C 60 D 90 0,
0
6.
Pairs of opposite sides of the quadrilateral be equal and parallel.
7.
One angle must be a right angle
8. Rhombus
9.
False
10. True
0
12. 2.1cm
LONG ANSWERS TYPE QUESTIONS : 0
0
0
0
0
0
0
2. 108 ,72 ,108 ,72
0
1.
37 ,143 ,37 ,143
3.
680,1120,680,1120
4. 4.5 cm
5.
A 45 0, B 135 0
7. CDB 45 ABD 60 0,
0
AOB 90 0 ,
0
10.
A C 45 0 , B D 135 0 ,
12.
BC 50 0 ,
LONG ANSWER TYPE QUESTIONS: 0
0
0
1.
12 cm
2. 50 ,60 ,70
3.
51 cm
5. 12 CM, 13.5 cm
6.
6 cm , 7cm, 5cm
2
TRUE OR FALSE TYPE QUESTIONS: 1. False
2. True
3. False
6. True
7. False
8. False
4. False
5. True
FILL IN THE BLANKS TYPE QUESTIONS: 1. an isosceles triangle
2. right triangle
3. Parallelogram
5. 1 : 3
OBJECTIVE
Que. Ans. Que. Ans.
0
80
11.
(E)
0
3.
9.. C D 180
(C)
0
2. 30 ,60 ,120 ,150
11. 6.8 cm (B)
EXERCISE
ANSWER KEY
QUADRILATERALS
1 C 11 D
EXERCISE-1
ANSANSWER WER KEY ANSWER KEY KEY 2 D 12 C
3 A 13 A
4 B 14 B
5 A 15 B
6 D 16 B
7 D 17 C
8 A 18 C
9 C 19 C
10 C 20 B
