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# NCERT Solution - Quadrilaterals Class 9 Notes | EduRev

## Class 9 : NCERT Solution - Quadrilaterals Class 9 Notes | EduRev

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Class IX  Chapter 8 – Quadrilaterals
Maths

Exercise 8.1 Question 1:
The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the
Let the common ratio between the angles be x. Therefore, the angles will be 3x, 5x,
9x, and 13x respectively.
As the sum of all interior angles of a quadrilateral is 360º,
3x + 5x + 9x + 13x = 360º
30x = 360º x
= 12º
Hence, the angles are
3x = 3 × 12 = 36º 5x =
5 × 12 = 60º
9x = 9 × 12 = 108º 13x =
13 × 12 = 156º Question 2:
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Page 2

Cbse-spot.blogspot.com
Cbse-spot.blogspot.com
1

Class IX  Chapter 8 – Quadrilaterals
Maths

Exercise 8.1 Question 1:
The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the
Let the common ratio between the angles be x. Therefore, the angles will be 3x, 5x,
9x, and 13x respectively.
As the sum of all interior angles of a quadrilateral is 360º,
3x + 5x + 9x + 13x = 360º
30x = 360º x
= 12º
Hence, the angles are
3x = 3 × 12 = 36º 5x =
5 × 12 = 60º
9x = 9 × 12 = 108º 13x =
13 × 12 = 156º Question 2:
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Cbse-spot.blogspot.com
Cbse-spot.blogspot.com
2

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that
one of its interior angles is 90º.
In ?ABC and ?DCB,
AB = DC (Opposite sides of a parallelogram are equal)
BC = BC (Common)
AC = DB (Given)
?ABC  ?DCB (By SSS Congruence rule)

ABC = DCB
It is known that the sum of the measures of angles on the same side of transversal is
180º.

ABC + DCB = 180º (AB || CD)
ABC +  ABC = 180º
2 ABC = 180º
ABC = 90º
Since ABCD is a parallelogram and one of its interior angles is 90º, ABCD is a rectangle.
Question 3:
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it
is a rhombus.
Page 3

Cbse-spot.blogspot.com
Cbse-spot.blogspot.com
1

Class IX  Chapter 8 – Quadrilaterals
Maths

Exercise 8.1 Question 1:
The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the
Let the common ratio between the angles be x. Therefore, the angles will be 3x, 5x,
9x, and 13x respectively.
As the sum of all interior angles of a quadrilateral is 360º,
3x + 5x + 9x + 13x = 360º
30x = 360º x
= 12º
Hence, the angles are
3x = 3 × 12 = 36º 5x =
5 × 12 = 60º
9x = 9 × 12 = 108º 13x =
13 × 12 = 156º Question 2:
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Cbse-spot.blogspot.com
Cbse-spot.blogspot.com
2

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that
one of its interior angles is 90º.
In ?ABC and ?DCB,
AB = DC (Opposite sides of a parallelogram are equal)
BC = BC (Common)
AC = DB (Given)
?ABC  ?DCB (By SSS Congruence rule)

ABC = DCB
It is known that the sum of the measures of angles on the same side of transversal is
180º.

ABC + DCB = 180º (AB || CD)
ABC +  ABC = 180º
2 ABC = 180º
ABC = 90º
Since ABCD is a parallelogram and one of its interior angles is 90º, ABCD is a rectangle.
Question 3:
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it
is a rhombus.
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3

Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right
angle i.e., OA = OC, OB = OD, and  AOB =  BOC = COD =  AOD = 90º. To prove
ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD
are equal.
In ?AOD and ?COD,
OA = OC (Diagonals bisect each other)
AOD =  COD (Given)
OD = OD (Common)
?AOD  ?COD (By SAS congruence rule)
Similarly, it can be proved that
AD = AB and CD = BC (2)
From equations (1) and (2),
AB = BC = CD = AD
Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a
parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that
ABCD is a rhombus.
Question 4:
Show that the diagonals of a square are equal and bisect each other at right angles.
Page 4

Cbse-spot.blogspot.com
Cbse-spot.blogspot.com
1

Class IX  Chapter 8 – Quadrilaterals
Maths

Exercise 8.1 Question 1:
The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the
Let the common ratio between the angles be x. Therefore, the angles will be 3x, 5x,
9x, and 13x respectively.
As the sum of all interior angles of a quadrilateral is 360º,
3x + 5x + 9x + 13x = 360º
30x = 360º x
= 12º
Hence, the angles are
3x = 3 × 12 = 36º 5x =
5 × 12 = 60º
9x = 9 × 12 = 108º 13x =
13 × 12 = 156º Question 2:
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Cbse-spot.blogspot.com
Cbse-spot.blogspot.com
2

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that
one of its interior angles is 90º.
In ?ABC and ?DCB,
AB = DC (Opposite sides of a parallelogram are equal)
BC = BC (Common)
AC = DB (Given)
?ABC  ?DCB (By SSS Congruence rule)

ABC = DCB
It is known that the sum of the measures of angles on the same side of transversal is
180º.

ABC + DCB = 180º (AB || CD)
ABC +  ABC = 180º
2 ABC = 180º
ABC = 90º
Since ABCD is a parallelogram and one of its interior angles is 90º, ABCD is a rectangle.
Question 3:
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it
is a rhombus.
Cbse-spot.blogspot.com
Cbse-spot.blogspot.com
3

Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right
angle i.e., OA = OC, OB = OD, and  AOB =  BOC = COD =  AOD = 90º. To prove
ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD
are equal.
In ?AOD and ?COD,
OA = OC (Diagonals bisect each other)
AOD =  COD (Given)
OD = OD (Common)
?AOD  ?COD (By SAS congruence rule)
Similarly, it can be proved that
AD = AB and CD = BC (2)
From equations (1) and (2),
AB = BC = CD = AD
Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a
parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that
ABCD is a rhombus.
Question 4:
Show that the diagonals of a square are equal and bisect each other at right angles.
Cbse-spot.blogspot.com
Cbse-spot.blogspot.com
4

Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O.
To prove that the diagonals of a square are equal and bisect each other at right angles,
we have to prove AC = BD, OA = OC, OB = OD, and AOB = 90º.
In ?ABC and ?DCB,
AB = DC (Sides of a square are equal to each other)
ABC = DCB (All interior angles are of 90 )
BC = CB (Common side)
?ABC  ?DCB (By SAS congruency)
AC = DB (By CPCT)
Hence, the diagonals of a square are equal in length.
In ?AOB and ?COD,
AOB =  COD (Vertically opposite angles)
ABO = CDO (Alternate interior angles)
AB = CD (Sides of a square are always equal)
?AOB  ?COD (By AAS congruence rule)
AO = CO and OB = OD (By CPCT)
Hence, the diagonals of a square bisect each other.
In ?AOB and ?COB,
As we had proved that diagonals bisect each other, therefore,
AO = CO
Page 5

Cbse-spot.blogspot.com
Cbse-spot.blogspot.com
1

Class IX  Chapter 8 – Quadrilaterals
Maths

Exercise 8.1 Question 1:
The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the
Let the common ratio between the angles be x. Therefore, the angles will be 3x, 5x,
9x, and 13x respectively.
As the sum of all interior angles of a quadrilateral is 360º,
3x + 5x + 9x + 13x = 360º
30x = 360º x
= 12º
Hence, the angles are
3x = 3 × 12 = 36º 5x =
5 × 12 = 60º
9x = 9 × 12 = 108º 13x =
13 × 12 = 156º Question 2:
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Cbse-spot.blogspot.com
Cbse-spot.blogspot.com
2

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that
one of its interior angles is 90º.
In ?ABC and ?DCB,
AB = DC (Opposite sides of a parallelogram are equal)
BC = BC (Common)
AC = DB (Given)
?ABC  ?DCB (By SSS Congruence rule)

ABC = DCB
It is known that the sum of the measures of angles on the same side of transversal is
180º.

ABC + DCB = 180º (AB || CD)
ABC +  ABC = 180º
2 ABC = 180º
ABC = 90º
Since ABCD is a parallelogram and one of its interior angles is 90º, ABCD is a rectangle.
Question 3:
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it
is a rhombus.
Cbse-spot.blogspot.com
Cbse-spot.blogspot.com
3

Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right
angle i.e., OA = OC, OB = OD, and  AOB =  BOC = COD =  AOD = 90º. To prove
ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD
are equal.
In ?AOD and ?COD,
OA = OC (Diagonals bisect each other)
AOD =  COD (Given)
OD = OD (Common)
?AOD  ?COD (By SAS congruence rule)
Similarly, it can be proved that
AD = AB and CD = BC (2)
From equations (1) and (2),
AB = BC = CD = AD
Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a
parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that
ABCD is a rhombus.
Question 4:
Show that the diagonals of a square are equal and bisect each other at right angles.
Cbse-spot.blogspot.com
Cbse-spot.blogspot.com
4

Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O.
To prove that the diagonals of a square are equal and bisect each other at right angles,
we have to prove AC = BD, OA = OC, OB = OD, and AOB = 90º.
In ?ABC and ?DCB,
AB = DC (Sides of a square are equal to each other)
ABC = DCB (All interior angles are of 90 )
BC = CB (Common side)
?ABC  ?DCB (By SAS congruency)
AC = DB (By CPCT)
Hence, the diagonals of a square are equal in length.
In ?AOB and ?COD,
AOB =  COD (Vertically opposite angles)
ABO = CDO (Alternate interior angles)
AB = CD (Sides of a square are always equal)
?AOB  ?COD (By AAS congruence rule)
AO = CO and OB = OD (By CPCT)
Hence, the diagonals of a square bisect each other.
In ?AOB and ?COB,
As we had proved that diagonals bisect each other, therefore,
AO = CO
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5

AB = CB (Sides of a square are equal)
BO = BO (Common)
?AOB  ?COB (By SSS congruency)
AOB = COB (By CPCT)
However, AOB + COB = 180º (Linear pair)
AOB = 180º 2
AOB = 90º
Hence, the diagonals of a square bisect each other at right angles.
Question 5:
Show that if the diagonals of a quadrilateral are equal and bisect each other at right
angles, then it is a square.

Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each
other at O. It is given that the diagonals of ABCD are equal and bisect each other at
right angles. Therefore, AC = BD, OA = OC, OB = OD, and AOB = BOC = COD
AOD = = 90º. To prove ABCD is a square, we have to prove that ABCD is a
parallelogram, AB = BC = CD = AD, and one of its interior angles is 90º.
In ?AOB and ?COD,
AO = CO (Diagonals bisect each other)
OB = OD (Diagonals bisect each other)
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