Quadrilaterals - Solutions
CBSE Class 9 Mathematics
NCERT Solutions
CHAPTER 8
Quadrilaterals(Ex. 8.1)
1. The angles of a quadrilateral are in the ratio 3: 5: 9: 13. Find all angles of the quadrilateral.
Ans. Let in quadrilateral ABCD,
Since, sum of all the angles of a quadrilateral = 360o
Now
And
Hence angles of given quadrilateral are and
2. If the diagonals of a parallelogram are equal, show that it is a rectangle.
Ans. Given: ABCD is a parallelogram with diagonal AC = diagonal BD
To prove: ABCD is a rectangle.
Proof: In triangles ABC and ABD,
AB = AB [ Common ]
AC = BD [ Given ]
AD = BC [ Opposite sides of a ]
[ By SSS congruency ]
[ By C.P.C.T.] ……….(i)
But
[ The sum of consecutve angles of a parallelogram is] ……….(ii)
From eq. (i) and (ii),
Simillarly, the other two angles are of each.
Hence ABCD is a rectangle.
3. Show that if diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Ans. Let ABCD is a quadrilateral.
Given: The diagonals AC and BD bisect each other at right angle at point O.
OA = OC, OB = OD
And
To prove: ABCD is a rhombus, i.e.we have to prove that AB = BC = CD = DA
Proof: In and ,
OA = OC [ Given ]
[ Given ]
OB = OD [ Given ]
[ By SAS congruency ]
AD = BC [ By C.P.C.T. ] ……….(i)
Again, In and ,
OA = OC [ Given ]
[ Given ]
OB = OD [ Given ]
[ By SAS congruency ]
AB = CD [ By C.P.C.T. ] ……….(ii)
Now In and ,
OA = OC [ Given ]
[ Given ]
OB = OB [ Common ]
AOB COB [ By SAS congruency ]
AB = BC [ By C.P.C.T. ] ……….(iii)
From eq. (i), (ii) and (iii),
AD = BC = CD = AB
And the diagonals of quadrilateral ABCD bisect each other at right angle.
Therefore, ABCD is a rhombus.
4. Show that the diagonals of a square are equal and bisect each other at right angles.
Ans. Given: ABCD is a square. AC and BD are its diagonals intersect each other at point O.
To prove:(I) AC = BD
(ii) OA = OC and OB = OD
(iii) AC BD at point O.
Proof: In triangles ABC and BAD,
AB = AB [ Common ]
ABC = BAD = [ Because ABCD is a square ]
BC = AD [ Sides of a square ]
ABC BAD [ By SAS congruency ]
AC = BD [ By C.P.C.T. ]
Thus, the diagonals of a square are equal. Hence proved (i).
Again in AOB and COD,
= [ Alternate angles ]
[ Vertically opposite angles ]
AB = CD [ Sides of a square ]
[By AAS congruency ]
OA = OC AND OB = OD [ By C.P.C.T. ]
Thus, diagonals of a square bisect each other. Hence proved (ii).
Now in triangles AOB and AOD,
AO = AO [ Common ]
AB = AD [ Sides of a square ]
OB = OD [ Diagonals of a square bisect each other ]
AOB AOD [ By SSS congruency ]
AOB = AOD [ By C.P.C.T. ]
But AOB + AOD = [ Linear pair ]
AOB = AOD =
OA BD or AC BD. Hence proved (iii).
5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Ans. Let ABCD be a quadrilateral in which equal diagonals AC and BD bisect each other at right angle at point O.
Given: AC = BD ……….(i)
OA = OC ...........(ii)
And OB = OD ……….(iii)
To Prove: ABCD is a square. i.e all the sides are equal and all the angles are of each.
Proof: Now OA + OC = OB + OD
OC + OC = OB + OB [ Using (i), (ii) & (iii) ]
2OC = 2OB
OC = OB ……….(iv)
From eq. (i), (ii), (iii) and (iv), we get, OA = OB = OC = OD ……….(v)
Now in AOB and COD,
OA = OD [ proved ]
AOB = COD [ Vertically opposite angles ]
OB = OC [ proved ]
AOB COD [ By SAS congruency ]
AB = CD [ By C.P.C.T. ] ……….(vi)
Similarly, BOC AOD [ By SAS congruency ]
BC = AD [ By C.P.C.T. ] ……….(vii)
From eq. (vi) and (vii), it is concluded that ABCD is a parallelogram because opposite sides of a quadrilateral are equal.
Now in ABC and BAD,
AB = BA [ Common ]
BC = AD [ proved above ]
AC = BD [ Given ]
ABC BAD [ By SSS congruency ]
ABC = BAD [ By C.P.C.T. ] ……….(viii)
But ABC + BAD = [ ABCD is a parallelogram ] ……….(ix)
ABC + ABC = [ Using eq. (viii) and (ix) ]
2ABC = ABC =
……….(x)
But ABC = ADC [ Opposite angles of a parallelogram are equal. ]
ABC = ADC = ……….(xi)
BAD = BDC = ……….(xii)
From eq. (xi) and (xii), we get
ABC = ADC = BAD = BDC = ……….(xiii)
Now in AOB and BOC,
OA = OC [ Given ]
AOB = BOC = [ Given ]
OB = OB [ Common ]
AOB COB [ By SAS congruency ]
AB = BC ……….(xiv)
From eq. (vi), (vii) and (xiv), we get,
AB = BC = CD = AD ……….(xv)
Now, from eq. (xiii) and (xv), we have a quadrilateral whose sides are equal make an angle of with each other.
ABCD is a square. Hence Proved
6. Diagonal AC of a parallelogram ABCD bisects A (See figure). Show that:
(i) It bisects C also.
(ii) ABCD is a rhombus.
Ans. Diagonal AC bisects A of the parallelogram ABCD.
(i) Since AB DC and AC intersects them.
1 = 3 [ Alternate angles ] ……….(i)
Similarly 2 = 4 ……….(ii)
But 1 = 2 [ Given ] ……….(iii)
3 = 4 [ Using eq. (i), (ii) and (iii) ]
Thus AC bisects C.
(ii) 2 = 3 = 4 = 1 [ Proved above ]
Thus, 2 = 3
AD = CD [ Sides opposite to equal angles ] ..............(iv)
But as ABCD is a parallelogram, therfore AB = CD and BC = AD
So, we get AB = CD = AD = BC [ Using (iv) ]
Hence ABCD is a rhombus.
7. ABCD is a rhombus. Show that the diagonal AC bisects A as well as C and diagonal BD bisects B as well as D.
Ans. ABCD is a rhombus. Therefore, AB = BC = CD = AD
Let O be the point of intersection of diagonals.
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[ Angles opposite to equal sides are equal ]
[ Alternate interior angles ]
Thus, diagonal AC bisects
Also, [ Alternate interior angles ]
So, we get ,
Thus AC bisects also.
Similarly,
[ Angles opposite to equal sides are equal ]
[ Alternate interior angles ]
Thus, diagonal BD bisects .
Again, [ Alternate interior angles ]
So, we get ,
Thus, BD bisects also.
Hence Proved.
8. ABCD is a rectangle in which diagonal AC bisects A as well as C. Show that:
(i) ABCD is a square.
(ii) Diagonal BD bisects both B as well as D.
Ans. ABCD is a rectangle.
Therefore AB = DC ……….(i)
And BC = AD
Also A = B = C = D =
(i) In ABC and ADC
1 = 2 and 3 = 4
[AC bisects A and C (given)]
AC = AC [Common]
ABC ADC [ By ASA congruency ]
AB = AD ……….(ii)
From eq. (i) and (ii), AB = BC = CD = AD
Hence ABCD is a square.
(ii) In ABD and CBD
AB = BC [ Since ABCD is a square ]
AD = DC [ Since ABCD is a square ]
BD = BD [ Common ]
ABD CBD [ By SSS congruency ]
ABD = CBD [ By C.P.C.T. ] ……….(iii)
And ADB = CDB [ By C.P.C.T. ] ……….(iv)
From eq. (iii) and (iv), it is clear that diagonal BD bisects both B and D.
Hence, Proved.
9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (See figure). Show that:
(i) APD CQB
(ii) AP = CQ
(iii) AQB CPD
(iv) AQ = CP
(v) APCQ is a parallelogram.
Ans. (i) In APD and CQB,
DP = BQ [ Given ]
ADP = QBC [ Alternate angles (ADBC and BD is transversal) ]
AD = CB [ Opposite sides of parallelogram ]
APD CQB [ By SAS congruency ]
(ii) Since APD CQB
AP = CQ [ By C.P.C.T. ]
(iii) In AQB and CPD,
BQ = DP [ Given]
ABQ = PDC [ Alternate angles (ABCD and BD is transversal)]
AB = CD [ Opposite sides of parallelogram]
AQB CPD [ By SAS congruency]
(iv) Since AQB CPD
AQ = CP [ By C.P.C.T.]
(v) In quadrilateral APCQ,
AP = CQ [proved in part (i)]
AQ = CP [proved in part (iv)]
Since opposite sides of quadrilateral APCQ are equal.
Hence APCQ is a parallelogram.
10. ABCD is a parallelogram and AP and CQ are the perpendiculars from vertices A and C on its diagonal BD (See figure). Show that:
(i) APB CQD
(ii) AP = CQ
Ans. Given: ABCD is a parallelogram. AP BD and CQ BD
To prove: (i) APB CQD (ii) AP = CQ
Proof: (i) In APB and CQD,
1 = 2 [ Alternate interior angles ]
AB = CD [ Opposite sides of a parallelogram are equal ]
APB = CQD =
APB CQD [ By AAS Congruency ]
(ii) Since APB CQD
AP = CQ [ By C.P.C.T. ]
11. In ABC and DEF, AB = DE, AB DE, BC = EF and BC EF. Vertices A, B and C are joined to vertices D, E and F respectively (See figure). Show that:
(i) Quadrilateral ABED is a parallelogram.
(ii) Quadrilateral BEFC is a parallelogram.
(iii) AD CF and AD = CF
(iv) Quadrilateral ACFD is a parallelogram.
(v) AC = DF
(vi) ABC DEF
Ans. (i) AB = DE [ Given ]
And AB DE [ Given ]
ABED is a parallelogram.
(ii) BC = EF [ Given ]
And BC EF [ Given ]
BEFC is a parallelogram.
(iii) As ABED is a parallelogram.
AD BE and AD = BE ……….(i)
Also BEFC is a parallelogram.
CF BE and CF = BE ……….(ii)
From (i) and (ii), we get
AD CF and AD = CF
(iv) As AD CF and AD = CF
ACFD is a parallelogram.
(v) As ACFD is a parallelogram.
AC = DF
(vi) In ABC and DEF,
AB = DE [ Given ]
BC = EF [ Given ]
AC = DF [ Proved in (iii) ]
ABC DEF [ By SSS congruency ]
12. ABCD is a trapezium in which AB CD and AD = BC (See figure). Show that:
(i) A = B
(ii) C = D
(iii) ABC BAD
(iv) Diagonal AC = Diagonal BD
Ans. Given: ABCD is a trapezium.
AB CD and AD = BC
To prove: (i) A = B
(ii) C = D
(iii) ABC BAD
(iv) Diagonal AC = Diagonal BD
Construction: Draw CE AD and extend AB to intersect CE at E.
Proof: (i) As AECD is a parallelogram.
[ By construction, CE is parallel to AD and AE is parallel to CD ]
AD = EC
But AD = BC [Given]
BC = EC
3 = 4 [ Angles opposite to equal sides are equal ]
Now 1 + 4 = [ Consecutive Interior angles of a parallelogram ]
And 2 + 3 = [ Linear pair ]
1 + 4 = 2 + 3
1 = 2 [ 3 = 4, so gets cancelled with each other ] ............... (i)
A = B
(ii) 3 = C [ Alternate interior angles ]
And D = 4 [ Opposite angles of a parallelogram ]
But 3 = 4 [ BCE is an isosceles triangle ]
C = D
(iii) In ABC and BAD,
AB = AB [ Common ]
1 = 2 [ Proved , see eqn (i) ]
AD = BC [ Given ]
ABC BAD [ By SAS congruency ]
(iv) We had observed that,
ABC BAD
AC = BD [ By C.P.C.T. ]