NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals – Here are all the NCERT solutions for Class 9 Maths Chapter 8. This solution contains questions, answers, images, explanations of the complete Chapter 8 titled Quadrilaterals of Maths taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Maths, then you must come across Chapter 8 Quadrilaterals. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals in one place.
NCERT Solutions Class 9 Maths Chapter 8 QUADRILATERALS
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Class
9
Subject
Maths
Book
Mathematics
Chapter Number
8
Chapter Name
QUADRILATERALS
NCERT Solutions Class 9 Maths chapter 8 QUADRILATERALS
Class 9, Maths chapter 8, QUADRILATERALS solutions are given below in PDF format. You can view them online or download PDF file for future use.
NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals
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Question & Answer
Q.1:The angles of the quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Ans : 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 \(360^{\circ}\),
Therefore, 3x + 5x + 9x + 13x = \(360^{\circ}\)
30x = \(360^{\circ}\)
x = \(12^{\circ}\)
Hence, the angles are
\(\begin{array}{l}{3 x=3 \times 12=360} \\ {5 x=5 \times 12=60^{\circ}} \\ {9 x=9 \times 12=108^{\circ}} \\ {13 x=13 \times 12=156^{\circ}}\end{array}\)
Q.2:If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Ans :
Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is \(90^{\circ}\).
In \(\triangle \mathrm{A} \mathrm{BC}\) and \(\triangle \mathrm{D} \mathrm{CB}\),
AB = DC (Opposite sides Of a parallelogram are equal)
BC = BC (Common)
AC = DB (Given)
\(\therefore \triangle A B C \cong \triangle D C B\) (By SSS Congruence rule)
\(\Rightarrow \angle A B C=\angle D C B\)
It is known that the sum of the measures of angles on the same side of transversal is \(180^{\circ}\).
\(\begin{array}{l}{\angle A B C+\angle D C B=180^{\circ}(A B \| C D)} \\ {\Rightarrow \angle A B C+\angle A B C=180^{\circ}} \\ {\Rightarrow 2 \angle A B C=180^{\circ}} \\ {\Rightarrow \angle A B C=90^{\circ}}\end{array}\)
Since ABCD is a parallelogram and one of its interior angles is \(90^{\circ}\) , ABCD is a rectangle.
Q.3:Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Ans :
Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each Other at right
angle i.e., OA = OC, OB = OD, and \(\angle A O B=\angle B O C=\angle C O D=\angle A O D=90^{\circ}\). To prove ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are equal.
In \(\triangle \mathrm{AOD} \text { and } \Delta \mathrm{COD}\) ,
OA = OC (Diagonals bisect each other )
\(\angle A O D=\angle C O D\) (Given)
OD = OD (Common)
\(\begin{array}{l}{\therefore \triangle A O D=\Delta C O D(B y \text { SAS congruence rule) }} \\ {\therefore A D=C D(1)}\end{array}\)
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.
Q.4:Show that the diagonals of a square are equal and bisect each other at right angles.
Ans :
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 \(\angle A O B=90^{\circ}\) .
In \(\triangle A B C \text { and } \Delta D C B\),
AB = DC (Sides Of a square are equal to each Other)
\(\angle A B C=\angle D C B\) (All interior angles are of \(90^{\circ}\) )
BC = CB (Common side)
\(\begin{array}{l}{\therefore \triangle \mathrm{ABC}=\Delta \mathrm{DCB}(\mathrm{By} \text { SAS congruency })} \\ {\therefore \mathrm{AC}=\mathrm{DB}(\mathrm{By} \mathrm{CPCT})}\end{array}\)
Hence, the diagonals of a square are equal in length.
In \(\triangle \mathrm{AOB} \text { and } \Delta \mathrm{COD}\) ,
\(\angle A O B=\angle C O D\) (Vertically Opposite Angle )
\(\square A B O=\square C D O\) (Alternate interior angles)
AB = CD (Sides of a square are always equal )
\(\begin{array}{l}{\square \triangle A O B \square \Delta C O D(B y \text { AAS congruence rule) }} \\ {\square A O=C O \text { and } O B=O D(B y C P C T)}\end{array}\)
Hence, the diagonals of a square bisect each other.
In \(\triangle A O B \text { and } \Delta C O B\) ,
As we had proved that diagonals bisect each other , therefore ,
AO = CO
AB = CB (Sides of a square are equal)
BO = BO (Common)
\(\begin{array}{l}{\square \triangle A O B \square \Delta C O B(B y \text { SSS congruence rule) }}\end{array}\)
\(\square \square A O B=\square C O B(B y C P C T)\)
However, \(\square A O B+\square C O B=180^{\circ}\) (Linear pair )
\(\begin{array}{l}{2 \square A O B=180^{\circ}} \\ {\square A O B=90^{\circ}}\end{array}\)
Hence, the diagonals of a square bisect each other at right angles.
Q.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 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, 0B = OD, and \(\angle A O B=\angle B O C=\angle C O D\)\(=\angle A O D=90^{\circ}\). 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 \(90^{\circ}\) .
In \(\triangle \mathrm{AOB} \text { and } \Delta \mathrm{COD}\),
AO = CO (Diagonals bisect each other)
OB = OD (Diagonals bisect each other)
\(\angle A O B=\angle C O D(\text { Vertically opposite angles })\)
\(\square \triangle \mathrm{AOB} = \square \Delta \mathrm{COD}(\text { SAS congruence rule) }\)
\(\begin{array}{l}{\square A B=C D(B y C P C T) \ldots(1)} \\ {\text { And, } \square O A B=\square O C D(B y C P C T)}\end{array}\)
However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel.
\(\square A B\|C D \ldots(2)\)
From equations (1) and (2), we obtain
ABCD is a parallelogram.
In \(\triangle \mathrm{AOD} \text { and } \triangle \mathrm{COD}\) ,
AO = CO (diagonals bisect each other)
\(\square A O D=\square C O D\left(\text { Given that each is } 90^{\circ}\right)\)
OD = OD (Common)
\(\begin{array}{l}{\square \triangle A O D \square \Delta C O D(S A S \text { congruence rule) }} \\ {\square A D=D C \ldots(3)}\end{array}\)
However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD)
\(\square A B=B C=C D=D A\)
Therefore, all the sides of quadrilateral ABCD are equal to each other.
In \(\triangle A D C \text { and } \Delta B C D,\) ,
AD = BC (Already proved)
AC = BD (Given)
DC =CD (Common)
\(\begin{array}{l}{\square \triangle \mathrm{ADC} \square \Delta \mathrm{BCD}(\mathrm{SSS} \text { Congruence rule) }} \\ {\square \square \mathrm{ADC}=\square \mathrm{BCD}(\mathrm{By} \mathrm{CPCT})}\end{array}\)
However, \(\square A D C+\square B C D=180^{\circ}(\text { Co-interior angles })\)
\(\begin{array}{l}{\square \square A D C+\square A D C=180^{\circ}} \\ {\square 2 \square A D C=180^{\circ}} \\ {\square \square A D C=90^{\circ}}\end{array}\)
One of the interior angles of quadrilateral ABCD is a right angle.
Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angles is \(90^{\circ}\) .Therefore, ABCD is a square.
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Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is \(90^{\circ}\). In \(\triangle \mathrm{A} \mathrm{BC}\) and \(\triangle \mathrm{D} \mathrm{CB}\), AB = DC (Opposite sides Of a parallelogram are equal) BC = BC (Common) AC = DB (Given) \(\therefore \triangle A B C \cong \triangle D C B\) (By SSS Congruence rule) \(\Rightarrow \angle A B C=\angle D C B\) It is known that the sum of the measures of angles on the same side of transversal is \(180^{\circ}\). \(\begin{array}{l}{\angle A B C+\angle D C B=180^{\circ}(A B \| C D)} \\ {\Rightarrow \angle A B C+\angle A B C=180^{\circ}} \\ {\Rightarrow 2 \angle A B C=180^{\circ}} \\ {\Rightarrow \angle A B C=90^{\circ}}\end{array}\) Since ABCD is a parallelogram and one of its interior angles is \(90^{\circ}\) , ABCD is a rectangle.
Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each Other at right angle i.e., OA = OC, OB = OD, and \(\angle A O B=\angle B O C=\angle C O D=\angle A O D=90^{\circ}\). To prove ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are equal. In \(\triangle \mathrm{AOD} \text { and } \Delta \mathrm{COD}\) , OA = OC (Diagonals bisect each other ) \(\angle A O D=\angle C O D\) (Given) OD = OD (Common) \(\begin{array}{l}{\therefore \triangle A O D=\Delta C O D(B y \text { SAS congruence rule) }} \\ {\therefore A D=C D(1)}\end{array}\) 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.
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 \(\angle A O B=90^{\circ}\) . In \(\triangle A B C \text { and } \Delta D C B\), AB = DC (Sides Of a square are equal to each Other) \(\angle A B C=\angle D C B\) (All interior angles are of \(90^{\circ}\) ) BC = CB (Common side) \(\begin{array}{l}{\therefore \triangle \mathrm{ABC}=\Delta \mathrm{DCB}(\mathrm{By} \text { SAS congruency })} \\ {\therefore \mathrm{AC}=\mathrm{DB}(\mathrm{By} \mathrm{CPCT})}\end{array}\) Hence, the diagonals of a square are equal in length. In \(\triangle \mathrm{AOB} \text { and } \Delta \mathrm{COD}\) , \(\angle A O B=\angle C O D\) (Vertically Opposite Angle ) \(\square A B O=\square C D O\) (Alternate interior angles) AB = CD (Sides of a square are always equal ) \(\begin{array}{l}{\square \triangle A O B \square \Delta C O D(B y \text { AAS congruence rule) }} \\ {\square A O=C O \text { and } O B=O D(B y C P C T)}\end{array}\) Hence, the diagonals of a square bisect each other. In \(\triangle A O B \text { and } \Delta C O B\) , As we had proved that diagonals bisect each other , therefore , AO = CO AB = CB (Sides of a square are equal) BO = BO (Common) \(\begin{array}{l}{\square \triangle A O B \square \Delta C O B(B y \text { SSS congruence rule) }}\end{array}\) \(\square \square A O B=\square C O B(B y C P C T)\) However, \(\square A O B+\square C O B=180^{\circ}\) (Linear pair ) \(\begin{array}{l}{2 \square A O B=180^{\circ}} \\ {\square A O B=90^{\circ}}\end{array}\) Hence, the diagonals of a square bisect each other at right angles.
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, 0B = OD, and \(\angle A O B=\angle B O C=\angle C O D\)\(=\angle A O D=90^{\circ}\). 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 \(90^{\circ}\) . In \(\triangle \mathrm{AOB} \text { and } \Delta \mathrm{COD}\), AO = CO (Diagonals bisect each other) OB = OD (Diagonals bisect each other) \(\angle A O B=\angle C O D(\text { Vertically opposite angles })\) \(\square \triangle \mathrm{AOB} = \square \Delta \mathrm{COD}(\text { SAS congruence rule) }\) \(\begin{array}{l}{\square A B=C D(B y C P C T) \ldots(1)} \\ {\text { And, } \square O A B=\square O C D(B y C P C T)}\end{array}\) However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel. \(\square A B\|C D \ldots(2)\) From equations (1) and (2), we obtain ABCD is a parallelogram. In \(\triangle \mathrm{AOD} \text { and } \triangle \mathrm{COD}\) , AO = CO (diagonals bisect each other) \(\square A O D=\square C O D\left(\text { Given that each is } 90^{\circ}\right)\) OD = OD (Common) \(\begin{array}{l}{\square \triangle A O D \square \Delta C O D(S A S \text { congruence rule) }} \\ {\square A D=D C \ldots(3)}\end{array}\) However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD) \(\square A B=B C=C D=D A\) Therefore, all the sides of quadrilateral ABCD are equal to each other. In \(\triangle A D C \text { and } \Delta B C D,\) , AD = BC (Already proved) AC = BD (Given) DC =CD (Common) \(\begin{array}{l}{\square \triangle \mathrm{ADC} \square \Delta \mathrm{BCD}(\mathrm{SSS} \text { Congruence rule) }} \\ {\square \square \mathrm{ADC}=\square \mathrm{BCD}(\mathrm{By} \mathrm{CPCT})}\end{array}\) However, \(\square A D C+\square B C D=180^{\circ}(\text { Co-interior angles })\) \(\begin{array}{l}{\square \square A D C+\square A D C=180^{\circ}} \\ {\square 2 \square A D C=180^{\circ}} \\ {\square \square A D C=90^{\circ}}\end{array}\) One of the interior angles of quadrilateral ABCD is a right angle. Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angles is \(90^{\circ}\) .Therefore, ABCD is a square.
