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Quadrilaterals
1.
2.
A quadrilateral ABCD is a parallelogram if
(a) AB = CD
(b) AB BC
(c) ∠A = 600 , ∠C = 600 , ∠B = 1200
(d) AB = AD
In figure, ABCD and AEFG are both parallelogram if ∠C = 800 , then∠DGF is
(a) 1000
(b) 600
(c) 800
(d) 1200
3.
4.
5.
In a square ABCD, the diagonals AC and BD bisects at O. Then ∆AOB is
(a) acute angled
(b) obtuse angled
(c) equilateral
(d) right angled
ABCD is a rhombus. If ∠ACB = 300 , then ∠ADB is
(a) 300
(b) 1200
(c) 600
(d) 450
The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.
6.
Show that each angles of a rectangle is a right angle.
7.
A transversal cuts two parallel lines prove that the bisectors of the interior angles enclose a rectangle.
8.
Prove that diagonals of a rectangle are equal in length.
9.
10.
In a parallelogram ABCD, bisectors of adjacent angles A and B intersect each other at P. prove that ∠APB = 900
In figure diagonal AC of parallelogram ABCD bisects ∠A show that (i) if bisects ∠C and (ii) ABCD is a rhombus
11.
12.
In figure ABCD is a parallelogram. AX and CY bisects angles A and C. prove that AYCX is a
parallelogram.
The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
13.
Prove that if the diagonals of a quadrilateral are equal and bisect each other at right angles then it is a square.
ANSWERS
Ans01.
(c)
Ans05.
Suppose angles of quadrilateral ABCD are 3x, 5x, 9x, and 13x
Ans02.
Ans03.
(c)
∠A + ∠B + ∠C + ∠D = 3600
(d)
Ans04.
(c)
[ sum of angles of a quadrilateral is 3600 ]
3 x + 5 x + 9 x + 13 x = 3600
30 x = 3600
x = 120
∴
∠A = 3 x = 3 × 12 = 360
∠B = 5 x = 5 ×12 = 600
∠C = 9 x = 9 × 12 = 1080
∠D = 13 x = 13 × 12 = 1560
Ans06.
We know that rectangle is a parallelogram whose one angle is right angle.
Let ABCD be a rectangle.
∠A = 900
To prove ∠B = ∠C = ∠D = 900
Proof: ∵ AD BC and AB is transversal
∴
∠A + ∠B = 1800
900 + ∠B = 1800
∠B = 1800 − 900
= 900
∠C = ∠A
Ans07.
∴
∠C = 900
∠D = ∠B
∴
∠D = 900
∵ AB CD and EF cuts them at P and R.
∴ ∠APR = ∠PRD
1
1
∴ ∠APR = ∠PRD
2
2
i.e.
∠1 = ∠2
∴
PQ RS
[ Alternate int erior angles ]
[ Alternate]
Ans08.
ABCD is a rectangle AC and BD is diagonals.
To prove AC = BD
Proof: In ∆ DAB and CBA
AD = BC [In a rectangle opposite sides are equal]
∠A = ∠B
[900 each]
AB = AB common [common]
∴
∆DAB ≅ ∆CAB
[ By SAS ]
∴
AC = BD
[ By CPCT ]
Ans09.
Given ABCD is a parallelogram is and bisectors of ∠A and ∠B intersect each
other at P.
To prove ∠APB = 900
Proof:
1
1
∠1 + ∠2 = ∠A + ∠B
2
2
1
(i )
= ( ∠A + ∠B )
→
2
In ∆APB
But ABCD is a parallelogram AD BC
∠1 + ∠2 + ∠APB = 1800
∴
∠A + ∠B = 1800
1
∠1 + ∠2 = ×1800
2
= 900
∴
Ans10.
(i)
AB DC and AC is transversal
∴
and
∠1 = ∠2
∠3 = ∠4
But
∠1 = ∠3
∴
∠ 2 = ∠4
∴
(ii)
∴
∴
∴
( Alternate angles )
( Alternate angles )
AC bi sec sts ∠C
In ∆ABC and ∆ADC
AC = AC
[common]
∠1 = ∠3
[ given]
∠2 = ∠4
[ proved ]
∆ABC ≅ ∆ADC
AB = AD
ABCD is a r hom bus
[ By CPCT ]
900 + ∠APB = 1800
∠APB = 900
Hence proved
[ By angle sum property ]
Ans11.
Given in a parallelogram AX and CY bisects ∠A and ∠C respectively and we have
to show that AYCX in a parallelogram.
In ∆ADX and ∆CBY
∠D = ∠B
→
(i ) [opposite angles of a paralle log ram]
1
∠A
2
1
and ∠BCY= ∠C
2
But
∠A=∠C
∴ By (2) and (3), we get
∠DAX =
→
[ given]
[given]
(ii)
→ (iii)
∠DAX = ∠BCY → (iv)
Also
∴
AD = BC
[opposite sides of paralle log ram]
→
(v )
From (i ), (iv) and (v), we get
∴
But
∆ADX ≅ ∆CBY
[ By ASA]
DX = BY
[CPCT ]
AB = CD
AB − BY = CD − DX
[oppsite sides of paralle log ram]
or
AY = CX
But
AY XC
[∵ ABCD is a gm]
∴ AYCX is a paralle log ram
Ans12.
Given ∆ ABC in which E and F are mid points of side AB and AC respectively.
To prove: EF||BC
Construction: Produce EF to D such that EF = FD. Join CD
Proof: In ∆AEF and ∆CDF
AF = FC
[∵ F is mid − po int of AC ]
∠1 = ∠2
[vertically opposite angles ]
EF = FD
[ BY construction]
∴ ∆AEF ≅ ∆CDF
[ By SAS ]
∴
[ By CPCT ]
AE = CD
and AE = BE
∴
[∵ E is the mid − po int]
BE = cd
and AB CD
[∴∠BAC = ∠ACD]
∴ BCDE is a paralle log ram
EF BC
Henceproved
Ans13.
Given in a quadrilateral ABCD, AC = BD, AO = OC and BO = OD and ∠AOB = 900
To prove: ABCD is a square.
Proof: In ∆AOB and ∆COD
OA = OC
[ given]
OB = OD
[ given]
and ∠AOB = ∠COD
[vertically opposite angles ]
∴
∆AOB ≅ ∆COD
[ By SAS ]
∴
AB = CD
[ By CPCT ]
∠1 = ∠2
[ By CPCT ]
But there are alternate angles ∴ AB CD
ABCD is a parallelogram whose diagonals bisects each other at right angles
∴ ABCD is a rhombus
Again in ∆ABD and ∆BCA
AB = BC
[ Sides of a r hom bus ]
AD = AB
[ Sides of a r hom bus ]
and BD = CA
[Given]
∴
∆ABD ≅ ∆BCA
∴
∠BAD = ∠CBA
[ By CPCT ]
There are alternate angles of these same side of transversal
∴
∠BAD + ∠CBA = 1800 or ∠BAD = ∠CBA = 900
Hence ABCD is a square
Quadrilateral
1.
In fig ABCD is a parallelogram. It
( A ) 80°
( B ) 60°
( C ) 20°
( D ) 40°
2.
∠DAB = 60° and ∠DBC = 80° then ∠CDB is
D
C
0
0
80
60
A
B
If the diagonals of a quadrilateral bisect each other, then the quadrilateral must
be.
(a) Square
(b) Parallelogram
(c) Rhombus
(d) Rut angle
3.
The diagonal AC and BD of quadrilateral ABCD are equal and are perpendicular
bisector of each other then quadrilateral ABCD is a
(a) Kite
(b) Square
(c) Trapezium
(d) Rut angle
4.
The quadrilateral formed by joining the mid points of the sides of a quadrilateral
ABCD taken in order, is a rectangle if
(a) ABCD is a parallelogram
(b) ABCD is a rut angle
(c) Diagonals AC and BD are perpendicular
(d) AC=BD
5.
If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
D
6.
C
In fig ABCD is a parallelogram and X,Y are the points on the diagonal BD such that DX<BY show
X
that AYCX is a parallelogram.
O
Y
B
A
G
C
D
7.
Show that the line segments joining the mid points of opposite sides of a quadrilateral
bisect each other.
H
F
A
E
B
8.
ABCD is a rhombus show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D
9.
Prove that a quadrilateral is a rhombus if its diagonals bisect each other at right angles.
10.
11.
Prove that the straight line joining the mid points of the diagonals of a trapezium is parallel to the parallel sides.
In fig ∠B is a right angle in ∆ABC.D is the mid-point of AC.DE AB intersects BC at E. show that
A
(i) E is the mid-point of BC
D
(ii) DE ⊥ BC
(iii) BD = AD
C
B
E
12.
ABC is a triangle and through vertices A, B and C lines are drawn parallel to BC, AC and AB respectively intersecting
at D, E and F. prove that perimeter of ∆DEF is double the perimeter of ∆ABC .
13.
Prove that in a triangle, the line segment joining the mid points of any two sides is parallel to the third side.
ANSWERS
Ans1.
(D)
Ans5.
Given A quadrilateral ABCD in which AB = DC and AD = BC
Ans2.
Ans3.
(B)
(B)
To prove: ABCD is a parallelogram
Ans4. (A)
D
C
Construction: Join AC
Prof: In ∆ABC and ∆ADC
AD = BC ( given )
A
AB = DC
B
AC = AC [ common ]
∴∆ABC ≅ ∆ADC [ by sss ]
∴∠BAC = ∠DAC [ By CPCT ]
∴ ABCD Is a parallelogram
Ans6.
ABCD is a parallelogram. The diagonals of a parallelogram bisect bisect each
other
∴OD = OB
But DX = BY [given]
∴ OD − DX = OB − BY
or OX = OY
Now in quad AYCX, the diagonals AC and XY bisect each other
∴ AYCX is a parallelogram.
Ans7.
Given ABCD is quadrilateral E, F, G, H are mid points of the side AB, BC, CD and
DA respectively
To prove: EG and HF bisect each other.
In ∆ABC , E is mid-point of AB and F is mid-point of BC
∴ EF AC And EF =
1
AC....... ( i )
2
Similarly HG AC and GD =
1
AC...... ( ii )
2
From (i) and (ii), EF HG and EF = GH
∴ EFGH is a parallelogram and EG and HF are its diagonals
Diagonals of a parallelogram bisect each other
Thus EG and HF bisect each other.
Ans8.
ABCD is a rhombus
In ∆ABC and ∆ADC
C
D
AB = AD [Sides of a rhombus]
BC = DC [Sides of a rhombus]
AC = AC [Common]
B
A
∴∆ABC ≅ ∆ADC [By SSS Congruency]
∴∠CAB = ∠CAD And ∠ACB = ∠ACD
Hence AC bisects ∠A as well as ∠C
Similarly, by joining B to D, we can prove that ∆ABD ≅ ∆CBD
Hence BD bisects ∠B as well as ∠D
Ans9.
Given ABCD is a quadrilateral diagonals AC and BD bisect each other at O at
right angles
To Prove: ABCD is a rhombus
Proof: ∵ diagonals AC and BD bisect each other at O
D
C
∴ OA = OC , OB = OD And ∠1 = ∠2 = ∠3 = 90°
0
3
Now In ∆BOA And ∆BOC
OA = OC Given
OB = OB [Common]
And ∠1 = ∠2 = 90° (Given)
∴∆BOA = ∆BOC (SAS)
∴ BA = BC (C.P.C.T.)
2
1
A
B
Ans10.
Given a trapezium ABCD in which AB DC and M,N are the mid Points of the
diagonals AC and BD.
We need to prove that MN AB DC
Join CN and let it meet AB at E
C
D
Now in ∆CDN and ∆EBN
M
∠DCN = ∠BEN [Alternate angles]
∠CDN = ∠BEN [Alternate angles]
A
E
N
B
And DN = BN [given]
∴∆CDN ≅ ∆EBN [SAA]
∴ CN = EN [By C.P.C.T]
Now in ∆ACE , M and N are the mid points of the sides AC and CE respectively.
∴ MN AE Or MN AB
Also AB DC
∴ MN AB DC
Ans11.
Proof: ∵ DE AB and D is mid points of AC
Ans12.
∴ BC = AF
In ∆DCE and ∆DBE
CE=BE
DE= DE
And ∠DEC = ∠DEB = 90°
∴∆DCE = ∆DBE
∴∆DCE ≅ ∆DBE
∴ CD = BD
∵ BCAF Is a parallelogram
∵ ABCE Is a parallelogram
∴ BC = AE
AF + AE = 2 BC
A
E
F
Or EF = 2 BC
Similarly ED = 2AB and FD = 2AC
C
B
∴ Perimeter of ∆ABC = AB + BC + AC
Perimeter of ∆DEF = DE + EF + DF
= 2AB+2BC+2AC
= 2[AB+BC+AC]
= 2 Perimeter of ∆ABC
Hence Proved
D
Ans13.
Given: A ∆ABC in which D and E are mid-points of the side AB and AC
respectively
TO Prove: DE BC
A
Construction: Draw CF BA
3
Proof: In ∆ADE and ∆CFE
1
D
E
F
2
∠1 = ∠2
[Vertically opposite
angles]
B
AE=CE
[Given]
And ∠3 = ∠4
[Alternate interior angles]
∴∆ADE ≅ ∆CFE
[By ASA]
∴ DE=FE
[By C.P.C.T]
But DA = DB
∴ DB = FC
Now DB FC
∴ DBCF is a parallelogram
∴ DE BC
Also DE = EF =
4
1
BC
2
C