Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  Short Answer Type Questions: Quadrilaterals

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

Question 1. In the adjoining figure, if ∠ B = 68°, then find ∠ A, ∠ C and ∠ D.
 Solution: 
∵ Opposite angles of a parallelogram are equal.
∴ ∠ B= ∠ D
⇒ ∠ D = 68°                           [∵ ∠B = 68°,    given]

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

∵ ∠ B and ∠ C are supplementary.
∴ ∠ B + ∠ C = 180°
⇒ ∠ C = 180° - ∠ B = 180° - 68° = 112°
Since ∠A and ∠C are opposite angles.
∴ ∠ A= ∠ C
⇒ ∠ A = 112°                                 [∵ ∠ C = 112°]
Thus, ∠ A = 112°, ∠ D = 68° and ∠ C = 112°


Question 2. In the figure, ABCD is a parallelogram. If AB = 4.5 cm, then find other sides of the parallelogram when its perimeter is 21 cm.
 Solution: 
∵ Opposite sides of a parallelogram are equal.
∴ AB = CD = 4.5 cm, and BC = AD

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

Now, AB + CD + BC + AD = 21 cm
⇒ AB + AB + BC + BC = 21 cm
⇒2[AB + BC] = 21 cm
⇒ 2[4.5 cm + BC] = 21 cm
⇒ [4.5 cm + BC] = (21/2)
= 11.5 cm ⇒ BC = 11.5 ∠ 4.5 = 7 cm
Thus, BC = 7 cm, CD = 4.5 cm and AD = 7 cm.
 

Question 3. In a parallelogram ABCD,if (3x ∠ 10)° = ∠ B and (2x + 10)° = ∠ C, then find the value of x.
 Solution:
Since, the adjacent angles of a parallelogram are supplementary.

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

∴ ∠ B + ∠ C = 180°
⇒ (3x - 10)° + (2x + 10)° = 180°
⇒ 3x + 2x - 10° + 10° = 180°
⇒ 5x = 180°
⇒ x= (1800/5)= 36°
Thus, the required value of x is 36°.


Question 4. The adjoining figure is a rectangle whose diagonals AC and BD intersect at O. If ∠ OAB = 27°, then find ∠ OBC.
 Solution:
Since, the diagonals of a rectangle are equal and bisect each other.
∴ OA = OB
⇒ ∠ OBA = ∠ OAB = 27°
Also, each angle of a rectangle measures 90°.

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

∴ ∠ ABC = 90°
⇒ ∠ ABO + ∠ CBO = 90°
⇒ ∠ OBA + ∠ OBC = 90°
⇒ 27° + ∠ OBC = 90°
⇒ ∠ OBC = 90° - 27° = 63°


Question 5. In a quadrilateral, ∠ A : ∠ B : ∠ C : ∠ D = 1 : 2 : 3 : 4, then find the measure of each angle of the quadrilateral.
 Solution: 
Since ∠ A : ∠ B : ∠ C : ∠ D = 1 : 2 : 3 : 4

∴ If ∠ A = x, then ∠ B = 2x, ∠ C = 3x and ∠ D = 4x. ∴ ∠ A + ∠ B + ∠ C + ∠ D = 360°
⇒ x + 2x + 3x + 4x = 360° ⇒ 10x = 36°
⇒ x= (3600/10)= 36°
∴ ∠ A = x = 36° ∠ B = 2x = 2 x 36° = 72° ∠ C = 3x = 3 x 36° = 108° ∠ D = 4x = 4 x 36° = 144°


Question 6. In the figure, D is the mid-point of AB and DE || BC. Find x and y.
 Solution:
Since DE || BC and D is the mid-point of AB.
∴ E must be the mid-point of AC.
∴ AE = EC ⇒ x = 5 cm

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

Also, DE || BC ⇒ DE = (1/2)BC
∴ 2DE = 2(((1/2))BC)
⇒ 2DE = BC
⇒ 2 x 6 cm = BC or BC = 12 cm
⇒ y = 12 cm
Thus, x = 5 cm and y = 12 cm

 

Question 1. In the adjoining figure, ABCD is a trapezium in which AB || CD. If ∠ A = 36° and ∠ B = 81°, then find ∠ C and ∠ D.
 Solution: 
∵ AB || CD and AD is a transversal.            [∵ ABCD is a trapezium in which AB || CD]
∴ ∠ A + ∠ D = 180°
⇒ ∠ D = 180° -  ∠ A = 180° - 36° = 144°
Again, AB || CD and BC is a transversal.

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

∴ ∠ B + ∠ C = 180°
⇒ ∠ C = 180° - ∠ B = 180° - 81° = 99°
∴ The required measures of ∠ D and ∠ C are 144° and 99° respectively.


Question 2. In the figure, the perimeter of D ABC is 27 cm. If D is the mid-point of AB and DE || BC, then find the length of DE.
 Solution: 
Since, D is the mid-point of AB and DE || BC.
∴ E is the mid-point of AC, and DE = (1/2) BC.
Since, perimeter of DABC = 27 cm
∴ AB + BC + CA = 27 cm
⇒ 2(AD) + BC + 2(AE) = 27 cm
⇒ 2(4.5 cm) + BC + 2(4 cm) = 27 cm
⇒ 9 cm + BC + 8 cm = 27 cm
∴ BC = 27 cm - 9 cm - 8 cm = 10 cm

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

∴  (1/2)BC =(10/2) = 5 cm
⇒ DE = 5 cm


Question 3. In the adjoining figure, DE || BC and D is the mid-point of AB. Find the perimeter of ΔABC when AE = 4.5 cm.
 Solution:
∵ D is the mid-point of AB and DE || BC.
∴ E is the mid-point of AC and DE = (1/2)BC.

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

⇒ 2DE = BC
⇒ 2 x 5 cm = BC
⇒ BC = 10 cm
Now DB = 3.5 cm
∴ AB = 2(DB) = 2 x 3.5 cm = 7 cm            [D is the mid-point of AB]
Similarly, AC = 2(AE) = 2 x 4.5 cm = 9 cm
Now, perimeter of ΔABC = AB + BC + CA = 7 cm + 10 cm + 9 cm = 26 cm


Question 4. If an angle of a parallelogram is (4/5) of its adjacent angle, then find the measures of all the angles of the parallelogram.
 Solution:
Let ABCD is a parallelogram in which ∠ B = x

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

∴ ∠ A= (4/5)x
Since, the adjacent angles of a parallelogram are supplementary.
∴ ∠ A + ∠ B = 180°
⇒ (4/5)x + x = 180°
⇒ 4x + 5x = 180° x 5
⇒ 9x = 180° x 5
⇒  Class 9 Maths Chapter 8 Question Answers - Quadrilaterals
∴ ∠ B = 100°
Since ∠ B= ∠ D            [Opposite angles of parallelogram]
∴ ∠ D = 100°
Now, ∠ A= (4/5)x =(4/5) x 100° = 80°
Also ∠ A= ∠ C             [Opposite angles of parallelogram]
∴ ∠ C = 80°
The required measures of the angles of the parallelogram are:  ∠ A = 80°, ∠ B = 100° ∠ C = 80° and ∠ D = 100°


Question 5. Find the measure of each angle of a parallelogram, if one of its angles is 15° less than twice the smallest angle.
 Solution:
Let the smallest angle = x
Since, the other angle = (2x ∠ 15°)
Thus, (2x ∠ 15°) + x = 180°             [∵ x and (2x ∠ 15°) are the adjacent angles of a parallelogram]
⇒ 2x ∠ 15° + x = 180°
⇒ 3x ∠ 15° = 180°
⇒ 3x = 180° + 15° = 195°
⇒  x= (1950/3)= 65°
∴ The smallest angle = 65°
∴ The other angle = 2x - 15° = 2(65°) - 15° = 130° -15° = 115°
Thus, the measures of all the angles of parallelogram are: 65°, 115°, 65° and 115°.


Question 6. The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively. Find the length of each side of the rhombus.
 Solution
: Since the diagonals of a rhombus bisect each other at right angles.
∴ O is the mid-point of AC and BD
⇒ AO =(1/2)AC and DO =(1/2)BD
Also ∠ AOD = 90°.
Now, ΔAOD is a right triangle, in which

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

AO = (1/2)AC =(1/2)(24 cm) = 12 cm
and DO = (1/2)BD =(1/2)(18 cm) = 9 cm
Since, AD2 = AO2 + DO2
⇒ AD2 = (12)2 + (9)2
= 144 + 81 = 225 = 152
⇒ AD = √(15)2 = 15
⇒ AD = AB = BC = CD = 15 cm (each)
Thus, the length of each side of the rhombus = 15 cm.
 

Question 7. One angle of a quadrilateral is of 108° and the remaining three angles are equal. Find each of the three equal angles.
 Solution:
ABCD is a quadrilateral
∴ ∠A + ∠B + ∠C + ∠D = 360°
⇒ 108° + [∠B + ∠C + ∠D] = 360°
⇒ [∠B + ∠C + ∠D]
= 360° - 108° = 252°

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

Since,
∠D = ∠B = ∠C
∴ ∠B + ∠C + ∠D = 252°
⇒ ∠B + ∠B + ∠B = 252°
⇒  3∠B = 252°
⇒ ∠B = (2520/3) = 84°
∴ ∠B = ∠C = ∠D = 84°
Thus, the measure of each of the remaining angles is 84°.


Question 8. In the figure, AX and CY are respectively the bisectors of opposite angles A and C of a parallelogram ABCD. Show that AX || CY
 Solution. 
∵ ABCD is a ||gm
∴ Its opposite angles are equal.
⇒ ∠A = ∠C
⇒ (1/2)∠A =(1/2) ∠C

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

i.e., ∠YAX = ∠YCX             ...(1)
Again DA || BC ⇒ YA || CX                [opposite sides of ||gm]
Also ∠AYC + ∠YCX = 180°            ...(2)
⇒ ∠AYC + ∠YAX = 180            [From (1) and (2)]
⇒ AX || CY              [As interior angles on the same side of the transversal are supplementary]


Question 9. E and F are respectively the mid points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that: EF || AB and  EF = (1/2)(AB + CD)
 Solution.

Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

Let us join BE and extend it meet CD produced at P.
In ΔAEB and ΔDEP, we get AB || PC and BP is a transversal,
∴ ∠ABE = ∠EPD             [Alternate angles]
AE = ED            [∵ E is midpoint of AB]
∠AEB = ∠PED             [Vertically opp. angles]
⇒ ΔAEB ≌ ΔDEP
⇒ BE = PE and AB = DP [SAS]
⇒ BE = PE and AB = DP
Now, in ΔEPC, E is a mid point of BP and F is mid point of BC
∴ EF || PC and EF =(1/2)PC            [Mid point theorem]
i.e., EF || AB and EF = (1/2) (PD + DC)
= (1/2) (AB + DC)
Thus, EF || AB and EF = (1/2) (AB + DC)

The document Class 9 Maths Chapter 8 Question Answers - Quadrilaterals is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on Class 9 Maths Chapter 8 Question Answers - Quadrilaterals

1. What are quadrilaterals?
Ans. Quadrilaterals are four-sided polygons that have four vertices and four angles. They are classified based on their properties, such as the length of their sides and the measure of their angles.
2. How many types of quadrilaterals are there?
Ans. There are several types of quadrilaterals, including rectangles, squares, parallelograms, rhombuses, trapezoids, and kites. Each type has its own unique properties and characteristics.
3. What is the difference between a square and a rectangle?
Ans. A square is a type of rectangle where all four sides are equal in length and all four angles are right angles. On the other hand, a rectangle has opposite sides that are equal in length and all four angles are right angles, but the adjacent sides may have different lengths.
4. How can we determine if a quadrilateral is a parallelogram?
Ans. A quadrilateral is a parallelogram if both pairs of opposite sides are parallel. Additionally, the opposite sides of a parallelogram are equal in length, and the opposite angles are equal in measure.
5. What is the sum of the measures of the interior angles of a quadrilateral?
Ans. The sum of the measures of the interior angles of a quadrilateral is always 360 degrees. This property holds true for all types of quadrilaterals, regardless of their shape or size.
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