NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

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Written by Team Trustudies
Updated at 2021-02-14


NCERT solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1

Q1 ) The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given: The ratios of the angles of quadrilateral are 3: 5: 9: 13.

Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.

We know that, sum of angles of a quadrilateral = 360?
? 3x + 5x + 9x + 13x = 360?
? 30x = 360?
? x = 12?

Thus, we get,
angles of quadrilateral
=> 3x = 3 × 12? = 36?
=>5x = 5 × 12? = 60?
=> 9x = 9 × 12? = 108?
=> 13x = 13 × 12? = 156?

Q2 ) If the diagonals of a parallelogram are equal, then show that it is a rectangle.



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given: Parallelogram is ABCD whose diagonals AC and BD are equal.

image

? AC = BD

To prove: ABCD is a rectangle.

Proof:
In ?ABC and ?DCB, we have,
AC = BD ...(Given)
AB = CD
(Opposite sides of parallelogram)
BC = CB ...(Common sides)
? ?ABC??DCB
(By SSS rule)
? ?ABC = ?DCB ...(i)(By CPCT)

But from figure, DC || AB and transversal CB intersects them

?ABC + ?DCB = 180?
(interior angles on same side of transversal)
?ABC + ?ABC = 180?
(from(i))
? 2 ?ABC = 180?
? ?ABC = 90?
Also, ?DCB = 90?

Thus, we can say that, ABCD is a parallelogram and some angles are 90?.

Hence it is proved that, ABCD is a rectangle.

Q3 ) Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given: A quadrilateral ABCD whose diagonals AC and BD bisect each other at right angles.

? OA = OC and OB = OD
And also,
?AOD=?AOB=?COD=?BOC=90?.

image

To prove: ABCD is a rhombus.

Proof:
In ?OAB and ?ODC, we have,

OA = OC and OB = OD ...(given)
?AOB = ?COD
(vertically opposite angles)
? ?OAB ? ?OCD
(By SAS rule)
? AB = CD ...(i)(By CPCT)

Similarly, in ?OAD and ?OBC, we have,
OA = OC and OD = OB ...(given)
?AOD = ?BOC
(vertically opposite angles)
? ?OAD ? ?OCB ...(By SAS rule)
? AD = BC ...(ii)(By CPCT)
Similarly, we can prove that,
AB = CD and
CD = BC ...(iii)

Hence, from (i), (ii) and (iii), we get,
AB = BC = AD = CD
Hence, it is proved that ABCD is a rhombus.

Q4 ) Show that the diagonals of a square are equal and bisect each other at right angles.



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given: A square ABCD whose diagonals AC and BD intersect at O.

To prove: Diagonals are equal and bisect each other at right angles.
image
i.e, AC = BD, OD = OB, OA = OC and AC?BD

Proof: In ?ABC and ?BAD, we have,
BC = AD ...(given)
AB = BA ...(Common side)
?ABC = ?BAD = 90?
? ?ABC ? ?BAD ...(By SAS rule)
? AC = BD ...(By CPCT)

Similarly, in ?OAB and ?OCD, we have,
AB = DC ...(given)
?OAB = ?DCO
(? AB || CD and transversal AC intersect)
?OBA = ?BDC
(? AB || CD and transversal BD intersect)
? ?OAB ? ?OCD
(By SAS rule)
? OA = OC and OB = OD
(By CPCT)

Now, in ?AOB and ?AOD, we have,
OA = OD ...(proved earlier)
AB = AD
(sides of square)
AO = OA
(Commom side)
? ?AOB ? ?AOD
(By SSS rule)
? ?AOB = ?AOD ...(By CPCT)

?AOB + ?AOD = 180?
(linear pair axiom)
?AOB = ?AOD = 90?
? AO?BD
? AC?BD.
Also, AC = BD, OA = OC, OB = OD and AC?BD

Hence, it is proved that diagonals are equal and bisect each other at right angles.

Q5 ) Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given: A quadrilateral ABCD in which AC = BD and AC ?BD such that OA = OC and OB = OD. So, ABCD is a parallelogram.
image
To prove: ABCD is a square.

Proof: Let AC and BD intersect at a point O.
In ?ABO and ?ADO, we have,
BO = OD ...(given)
AO = OA ...(Common side)
?AOB = ?AOD = 90? ...(given)
? ?ABO ? ?ADO
(By SAS rule)
? AB = AD ...(By CPCT)
Also, AB = DC and AD = BC
(opposite sides of parallelogram)
? AB = BC = DC = AD ...(i)

Similarly, in ?ABC and ?BAD, we have,
AC = BD ...(given)
AB = BA ...(Common side)
BC = AD ...(from(i))
? ?ABC ? ?BAD
(By SSS rule)
? ?ABC = ?BAD ...(ii)(By CPCT)

But ?ABC + ?BAD = 180? >br> (linear pair axiom)
?ABC = ?BAD = 90? ...(from (ii))
? AB = BC = CD = DA, and ?A = 90?

Hence, it is proved that, ABCD is a square.

Q6 ) Diagonal AC of a parallelogram ABCD bisects ?A (see figure). Show that
(i) It bisects ?C also,
(ii) ABCD is a rhombus.
image



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given, diagonal AC of a parallelogram ABCD bisects ?A.
image
? ?DAC = ?BAC = 12 ?BAD ...(i)
Here, AB || CD and AC is the transversal.
?DCA = ?CAB ...(ii)(alternate angles)
?BCA = ?DAC ...(iii)(alternate angles)

From eq. (i), (ii) and (iii), we get,
?DAC = ?BAC = ?DCA = ?BCA
Now, ?BCD = ?BCA + ?DCA
? ?BCD= ?DAC + ?CAB
? ?BCD = ?BAD
Thus, we can say that, diagonal AC bisects ?C.

Now, in ?OAD and ?OCD, we have,
OA = OC
(since, diagonals bisects each other)
DO = OD ...(Common side)
?AOD = ?COD = 90?
? ?OAD ? ?OCD
(By SAS rule)
? AD = CD ...(By CPCT)

Now, AB = CD and AD = BC
(sides of parallelogram)
? AB = CD = AD = BC

Hence it is proved that, ABCD is a rhombus.

Q7 ) ABCD is a rhombus. Show that diagonal AC bisects ?A as well as ?C and diagonal BD bisects ?B as well as ?D.



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given: ABCD is a rhombus.
i.e., AD = AB = BC = CD ... (i)
image
To prove: (i) Diagonal AC bisect ?A as well as ?C.
(ii) Diagonal BD bisects ?B as well as ?D.

Proof: Let AC and BD are the diagonals of rhombus ABCD.
In ?ABC and ?ADC, we have,
AD = AB ...(Given)
AC = CA ...(Common side)
CD = BC ...(From eq. (i))
? ?ABC ? ?ADC ...(By SSS rule)
Thus, ?DAC = ?BAC ...(By CPCT)
Also, ?DCA = ?BCA
Also, ?DAC = ?DCA
And ?BAC = ?BCA
This shows that, Diagonal AC bisect ?A as well as ?C.

Now, in ?BDC and ?BDA, we have,
AB = BC ...(Given)
BD = BD ...(Common side)
AD = CD ...(Given)
? ?BDC ? ?BDA ...(By SSS rule)
Thus, ?BDA = ?BDC ...(By CPCT)
Also, ?DBA = ?DBC
Also, ?BDA = ?DBA
And ?BDC = ?DBC
This shows that, Diagonal BD bisects ?B as well as ?D.
Hence, proved.

Q8 ) 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 ?B as well as ?D.



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given: ABCD is a rectangle. AB = CD and BC = AD ...(i)
image

To prove: (i) ABCD is a square. i.e., AB = BC = CD = DA
(ii) Diagonal BD bisects ?B as well as ?D.

Proof: In ?ADC and ?ABC, we have,
?DAC = ?BAC
(Since, AB || DC and AC is transversal that intersects)
Similarly, ?DCA = ?BCA
AC = CA ...(Common side)
? ?ADC ? ?ABC ...(By ASA rule)
?AD = AB ...(By CPCT)
Also, CD = BC ...(ii)
Thus, from eq. (i) and(ii), we get,
AB = BC = AD = CD

Hence, it is proved that ABCD is a square.

Now, in ?AOB and ?COB, we have,
AB = BC ...(Given)
BO = OB ...(Common side)
OA = OC
(Since,diagonal bisectS each other)
? ?AOB ? ?COB ...(By SSS rule)
? ?OBA = ?OBC ...(By CPCT)
This shows that, Diagonal BD bisects ?B.

Similarly, Now, in ?AOD and ?COD, we have,
AD = CD ...(Given)
OD = DO ...(Common side)
OA = OC
(Since,diagonal bisectS each other)
? ?AOD ? ?COD ...(By SSS rule)
? ?ADO = ?CDO ...(By CPCT)

Hence, it is proved that, Diagonal BD bisects ?D, too.

Q9 ) 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.
image



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given, ABCD is a parallelogram and P and Q are lie on BD such that DP = BQ ...(i)

i) Now,in ?APD and ?CQB, we have,
DP = BQ ...(Given)
AD = BC
(Opposite sides are equal in parallelogram)
?ADP = ?QBC
(Since, AD || BC and BD is a transversal)
? ?APD ? ?CQB ...(By SAS rule)

ii)
? ?APD ? ?CQB
? AP = CQ ...(By CPCT)

iii) Here, Now, in ?AQB and ?CPD, we have,
DP = BQ ...(Given)
AB = CD
(Opposite sides are equal in parallelogram)
?ABQ = ?CDP
(Since, AB || CD and BD is a transversal)
? ?AQB ? ?CPD ...(By SAS rule)

iv)
? ?AQB ? ?CPD
? AQ = CP...(By CPCT)

v) Now,in ?APQ and ?PCQ, we have,
AQ = CP (From part iv))
AP = CQ (From part ii))
PQ = QP ...(Common side)
? ?APQ ? ?PCQ ...(By SSS rule)
? ?APQ = ?PQC
And ?AQP = ?CPQ
(Vertically Opposite angles)

Now, these equal angles form a pair of alternate angle when line segment AP and QC are intersected by a transversal PQ.

? AP || CQ and AQ || CP

Now, both pairs of opposite sides of quadrilateral APCQ are parallel.

Hence, it is proved that APCQ is a parallelogram.

Q10 ) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that
(i) ?APB ? ?CQD
(ii) AP = CQ
image



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given: ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD.

? AB || CD and BD is a transversal, we get,

?CDB = ?DBA ...(i)
Now, in ?APB and ?CQD, we have,
CD = AB ...(Sides of parallelogram)
?CQD = ?APB = 90? ...(Given)
?CDQ = ?ABP ...(From Eq. (i))
? ?APB ? ?CQD ...(By ASA rule)
? AP = CQ ...(By CPCT)
Hence, it is proved.

Q11 ) 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).
image
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



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given: In ?ABC and ?DEF, AB = DE, AB || DE, BC = EF and BC || EF

(i) Now, in quadrilateral ABED,
AB = DE and AB || DE ...(Given)
Since, a pair of opposite sides is equal and parallel
Therefore, ABED is a parallelogram.

(ii) In quadrilateral BEFC,
BC = EF and BC || EF ...(Given)
Since, a pair of opposite sides is equal and parallel
Therefore, BEFC is a parallelogram.

(iii)Since, ABED is a parallelogram,
AD || BE and AD = BE ...(i)
Also, BEFC in a parallelogram,
CF || BE and CF = BE ...(ii)
Thus, from Eq. (i) and (ii), we get,
AD || CF and AD = CF ...(From part (iii))
Therefore, ACFD is a parallelogram.

(v)Since, ACFD is a parallelogram.
we get, AC = DF and AC || DF

(vi)Now, in ?ABC and ?DEF,
AB = DE ...(Given)
BC = EF ...(Given)
and AC = DF ...(From part (v))
Therefore, ?ABC ? ?DEF ...(By SSS test)

Q12 ) ABCD is a trapezium in which AB || CD and AD = BC (see figure).
image
Show that
i) ?A = ?B
ii) ?C = ?D
iii) ?ABC ? ?BAD
iv) Diagonal AC = Diagonal BD
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E].



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given: ABCD is trapezium.
AB || CD and AD = BC
Now, extend AB and draw a line through C parallel to DA intersecting AB produced at E.
image
Now, ADCE is a parallelogram.
? AD || CE and AD = CE
But AD = BC
? AD = BC = CE

i) We know that, ?A + ?E = 180?
(Since, interior angles on the same side of the transversal )
? ?E = 180? - ?A
? ?E = 180? - ?A
(Since, BC = EC)

Also, ?ABC = 180? - ?CBE
(Since, ABE is straight line)
?ABC = 180? - 180? + ?A
?B = ?A
Hence, it is proved.

ii)Now, ?A + ?D = 180?
(Since, interior angles on the same side of the transversal)
? ?D = 180? - ?A
? ?D = 180? - ?B
(from eq.(i)) ...(ii)

Also, ?C + ?B = 180? ...180 (Since, interior angles on the same side of the transversal BC)
? ?C = 180? - ?B ...(iii)
from Eq. (ii) and (iii), we get,
?C = ?D
Hence, proved.

iii)Now, in ?ABC and ?BAD, we have,
AD = BC ...(Given)
?A = ?B ...(From Eq.(i))
AB = BA ...(Common side)
? ?ABC ? ?BAD ...(By SAS rule)
Hence, proved.

iv)
? ?ABC ? ?BAD
? AC = BD
Hence, it is proved.

NCERT solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2

Q1 ) ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that
i) SR || AC and SR = (1/2) AC
ii) PQ = SR
iii) PQRS is a parallelogram.
image



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given: P, Q, R, and S are mid-points of the sides.
Therefore, AP = PB, BQ = CQ CR = DR and AS = DS

i) Now, in ?ADC, we have,
S is the midpoint of AD and R is the mid point of CD.
As, we know that,
By midpoint theorem, the line segment joining the mid points of two sides of a triangle is parallel to the third side.
Thus, we can say that, SR || AC ...(i)
and SR = (12 ) AC ...(ii)

ii) Similarly, now, in ?ABC, we have,
PQ || AC ...(iii)
and PQ = (12 ) AC ...(iv)
Now, from (ii) and (iv), we get,
SR = PQ = (12 ) AC ...(v)

Now, from (i), (iii) and (v), we get,
PQ || SR and PQ = SR.
? PQRS is a parallelogram
(Since, a pair of opposite sides of a quadrilateral PQRS is equal and parallel)
Hence, proved.

Q2 ) ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

image
Given: ABCD is a rhombus and P, Q, R and S are mid-points of AB, BC, CD and DA.

By midpoint theorm, Now, in ?ADC, we have,
Thus, SR || AC ...(i)
and SR = (12 ) AC ...(ii)
Similarly, in ?ABC, we have,
PQ || AC ...(iii)
and PQ = (12 ) AC ...(iv)
from (i), (ii), (iii) and (iv), we get,
SR = PQ = (12 ) AC and PQ || SR
? Quadrilateral PQRS is a parallelogram.

Now, we know that diagonals of a rhombus bisect each other at right angles.
? ?EOF = 90?.

Now, By midpoint theorem, we have,
RQ || BD
Thus, RE || OF
As SR || AC ...(from (i))
Thus, FR || OE

? OERF is a parallelogram
So, ?ERF = ?EOF=90°
(since, Opposite angle of a quadrilateral is equal)

Thus, PQRS is a parallelogram with ?R=90°
Hence, it is proved that PQRS is a rectangle.

Q3 ) 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 a rhombus.
image



NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals


Answer :

Given: ABCD is a rectangle.
? ?A = ?B = ?C = ?D = 90? and
AB = CD and BC = AD.

Also, P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.

Therefore, by midpoint theorem,

PQ || BD and PQ = 12 BD and
SR || AC and SR = 12 AC
In rectangle ABCD,
AC = BD