A quadrilateral is a polygon which contains four side whereas, a parallelogram is a type of quadrilateral, whose pairs of opposite sides are parallel. One of the most common properties of all quadrilaterals is that sum of all the angles is 360 degrees. Some of the different types of quadrilaterals are:
Parallelograms have some unique properties such as:
- Opposite sides are congruent
- Opposite angles are congruent
- It has adjacent angles which are supplementary
- It has diagonals which bisect each other.
Download PDF of RS Aggarwal Class 9 Solutions Chapter 9 – Quadrilaterals And Parallelograms
Students must practice these solutions to gain more marks in the exam. We have provided here all the exercise questions of RS Aggarwal Maths solution to help you to revise the complete syllabus. It is an important tool to prepare effectively for the Class 9 Maths examination. All the solutions are explained in a detailed step-by-step manner for the better understanding of the students.
Q.1: Three angles of a quadrilateral measure 56°, 115°, 84°. Find the measure of the fourth angle.
Let, the measure of the fourth angle be X∘
Since the sum of the angle of a quadrilateral is 360∘, we have:
56 + 115 + 84 + X = 360
⇒ 255 + X = 360
⇒ X = 105
Hence, the measure of the fourth angle is 105∘
Q.2: The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find the angles.
Let, ∠A = 2X∘
Since the sum of the angles of a quadrilateral is 360∘, we have:
2X + 4X + 5X + 7X = 360∘
⇒ 18X = 360∘
⇒ X = 20∘
Therefore, Angle A = 40∘
Angle B = 80∘
Angle C = 100∘
Angle D = 140∘
Q.3: In the adjoining figure, ABCD is a trapezium in which AB ∥ DC. If ∠ A = 55°and ∠ B = 70°, find ∠ C and ∠ D.
We have AB∥DC
Angle A and Angle D are the interior angles on the same side of the transversal line AD, whereas angle B and Angle C are the same interior angles on the same side of transversal line BC.
Now, Angle A + Angle D = 180∘
⇒ ∠D = 180∘ – ∠A
Therefore, ∠D = 180∘ – 55∘ = 125∘
Again, ∠B + ∠C = 180∘
⇒ ∠C = 180∘ – ∠B
Therefore, ∠C = 180∘ – 70∘ = 110∘
Q.4: In the adjoining figure, ABCD is a square and ΔEDC is an equilateral triangle. Prove that (i) AB = BE, (ii) ∠ DAE = 15°.
ABCD is a square in which AB = BC = CD = DA. ΔEDC is an equilateral triangle in which ED = EC = DC
∠EDC = ∠DEC = ∠DCE = 60∘.
To prove: AE = BE and ∠DAE = 15∘
Proof: In ΔADE and ΔBCE, we have:
AD = BC [Sides of a square]
DE = EC [Sides of an equilateral triangle]
∠ADE = ∠BCE = 90∘ + 60∘ = 150∘
ΔADE ≅ ΔBCE
i.e., AE = BE
Now, ∠ADE = 150∘
DA = DC [Sides of a square]
DC = DE [Sides of an equilateral triangle]
So, DA = DE
ΔADE and ΔBCE are isosceles triangles.
i.e., ∠DAE = ∠DEA
Q.5: In the adjoining figure, BM ⊥ AC and DN ⊥ AC. If BM = DN, prove that AC bisects BD.
A quadrilateral ABCD, in which BM ⊥ AC and DN ⊥ AC and BM = DN
To prove: AC bisects BD, or DO = BO
Proof: Let, AC and BD intersect at O
Now, in ΔOND and ΔOMB, we have:
∠OND = ∠OMB (90∘ each)
∠DON = ∠BOM (Vertically opposite angles)
Also, DN = BM (Given)
i.e., ΔOND ≅ ΔOMB (AAS congruence rule)
Therefore, OD = OB (CPCT)
Hence, AC bisects BD
Q.6: In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that:
(i) AC bisects ∠ A and ∠ C, (ii) BE = DE, (iii) ∠ ABC = ∠ADC.
Given: ABCD is a quadrilateral in which AB = AD and BC = DC
(i) In ΔABC and ΔADC, we have:
AB = AD (Given)
BC = DC (Given)
AC is common
i.e., ΔABC ≅ ΔADC (SSS Congruence rule)
Therefore, ∠BAC = ∠DAC and
∠BCA = ∠DCA (By CPCT)
Thus, AC bisects ∠A and ∠C
(ii) Now, in ΔABE and ΔADE, we have:
AB = AD (Given)
∠BAE = ∠DAE (Proven above)
AE is common.
Therefore, ΔABE ≅ ΔADE (SAS congruence rule)
⇒ BE = DE (By CPCT)
(iii) ΔABC ≅ ΔADC (Proven above)
Therefore, Angle ABC = Angle ADC (By CPCT)
Q.7: In the given figure, ABCD is a square and ∠PQR = 90°. If PB = QC = DR, prove that (i) QB = RC, (ii) PQ = QR (iii) ∠ QPR = 45°
Let, ABCD be a parallelogram
∠ A = ∠ C and ∠ B = ∠ D (Opposite angles)
Let ∠ A be the smallest angle whose measure is x°
∴ ∠B = (2x – 30)°
Now, ∠ A + ∠ B = 180° (Adjacent angles are supplementary)
⇒ x + 2x – 30° = 180°
⇒ 3x = 210°
⇒ x = 70°
∴ ∠B = 2 x 70°- 30° = 110°
Hence, ∠ A = ∠ C = 70°; ∠ B = ∠ D = 110°
Q.8: If O is a point within a quadrilateral ABCD, show that OA + OB + OC + OD > AC + BD.
Let, ABCD be a quadrilateral whose diagonals are AC and BD and O is any point within the quadrilateral
Join O with A, B, C, and D
We know that the sum of any sides of a triangle is greater than the third side
So, in ΔAOC, OA + OC > AC
Also, ΔBOD, OB + OD > BD
Adding these inequalities, we get:
(OA + OC) + (OB + OD) > (AC + BD)
⇒ OA + OB + OC + OD > AC + BD
Q.9: In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that: (i) AB + BC + CD + DA > 2AC (ii) AB + BC + CD > DA (iii) AB + BC + CD + DA > AC + BD
Given: ABCD is a quadrilateral and AC is one of its diagonals
(i) In ΔABC, AB + BC > AC . . . . . . . . (1)
In ΔACD, CD + DA > AC . . . . . . . . . . (2)
Adding inequalities (1) and (2), we get:
AB + BC + CD + DA > 2AC
(ii) In ΔABC, we have :
AB + BC > AC . . . . . . . . (1)
We also know that the length of each side of a triangle is greater than the positive difference of the the other two sides.
In ΔACD, we have:
AC > |DA–CD| . . . . . . . (2)
From (1) and (2), we have:
AB + BC > |DA–CD|
⇒ AB + BC + CD > DA
(iii) In ΔABC, AB + BC > AC
In ΔACD, CD + DA > AC
ΔBCD, BC + CD > BD
ΔABD, DA + AB > BD
Adding these inequalities, we get:
2 (AB + BC + CD + DA) > 2 (AC + BD)
⇒ (AB + BC + CD + DA) > (AC + BD)
Q.10: Prove that the sum of all angles of a quadrilateral is 360°.
Sol: Let ABCD be a quadrilateral and ∠1, ∠2, ∠3 and ∠4 are its four angles as shown in the figure.
Join BD which divides ABCD in two triangles, ΔABD and ΔBCD.
In ΔABD, we have:
∠ 1 + ∠ 2 + ∠ A = 180° . . . . . . . . (i)
In ΔBCD, we have:
∠ 3 + ∠ 4 + ∠ C = 180° . . . . . . . (ii)
On adding (i) and (ii), we get:
(∠1 + ∠3) + ∠A + ∠C + (∠ 4 + ∠2) = 360°
⇒ ∠ A + ∠ C + ∠ B + ∠ D = 360° ( Since,∠ 1 + ∠ 3 = ∠ B; ∠ 4 + ∠ 2 = ∠ D)
∴ ∠ A + ∠ C + ∠ B + ∠ D = 360°
RS Aggarwal Class 9 Solutions Chapter 9 – Quadrilaterals And Parallelograms
The RS Aggarwal Class 9 Solutions provides all the answers in a proper stepwise method which makes you well versed with all the solutions including the tougher ones. If you want to score good marks in your Class 9 examination, then, practicing RS Aggarwal Class 9 Maths solutions is an utmost necessity. It is one of the best study material when it comes a providing a question bank to practice from and also test the students understanding of concepts.