Courses

## Mathematics (Maths) Class 9

Created by: Full Circle

## Class 9 : Short Answer Type Questions- Quadrilaterals Class 9 Notes | EduRev

The document Short Answer Type Questions- Quadrilaterals Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Question 1. In the adjoining figure, if âˆ  B = 68Â°, then find âˆ  A, âˆ  C and âˆ  D.
Solution:
âˆµ Opposite angles of a parallelogram are equal.
âˆ´ âˆ  B= âˆ  D
â‡’ âˆ  D = 68Â°                           [âˆµ âˆ B = 68Â°,    given]

âˆµ âˆ  B and âˆ  C are supplementary.
âˆ´ âˆ  B + âˆ  C = 180Â°
â‡’ âˆ  C = 180Â° - âˆ  B = 180Â° - 68Â° = 112Â°
Since âˆ A and âˆ C are opposite angles.
âˆ´ âˆ  A= âˆ  C
â‡’ âˆ  A = 112Â°                                 [âˆµ âˆ  C = 112Â°]
Thus, âˆ  A = 112Â°, âˆ  D = 68Â° and âˆ  C = 112Â°

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:
âˆµ Opposite sides of a parallelogram are equal.
âˆ´ AB = CD = 4.5 cm, and BC = AD

Now, AB + CD + BC + AD = 21 cm
â‡’ AB + AB + BC + BC = 21 cm
â‡’2[AB + BC] = 21 cm
â‡’ 2[4.5 cm + BC] = 21 cm
â‡’ [4.5 cm + BC] = (21/2)
= 11.5 cm â‡’ BC = 11.5 âˆ  4.5 = 7 cm
Thus, BC = 7 cm, CD = 4.5 cm and AD = 7 cm.

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

âˆ´ âˆ  B + âˆ  C = 180Â°
â‡’ (3x - 10)Â° + (2x + 10)Â° = 180Â°
â‡’ 3x + 2x - 10Â° + 10Â° = 180Â°
â‡’ 5x = 180Â°
â‡’ x= (1800/5)= 36Â°
Thus, the required value of x is 36Â°.

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

âˆ´ âˆ  ABC = 90Â°
â‡’ âˆ  ABO + âˆ  CBO = 90Â°
â‡’ âˆ  OBA + âˆ  OBC = 90Â°
â‡’ 27Â° + âˆ  OBC = 90Â°
â‡’ âˆ  OBC = 90Â° - 27Â° = 63Â°

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

âˆ´ If âˆ  A = x, then âˆ  B = 2x, âˆ  C = 3x and âˆ  D = 4x. âˆ´ âˆ  A + âˆ  B + âˆ  C + âˆ  D = 360Â°
â‡’ x + 2x + 3x + 4x = 360Â° â‡’ 10x = 36Â°
â‡’ x= (3600/10)= 36Â°
âˆ´ âˆ  A = x = 36Â° âˆ  B = 2x = 2 x 36Â° = 72Â° âˆ  C = 3x = 3 x 36Â° = 108Â° âˆ  D = 4x = 4 x 36Â° = 144Â°

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.
âˆ´ E must be the mid-point of AC.
âˆ´ AE = EC â‡’ x = 5 cm

Also, DE || BC â‡’ DE = (1/2)BC
âˆ´ 2DE = 2(((1/2))BC)
â‡’ 2DE = BC
â‡’ 2 x 6 cm = BC or BC = 12 cm
â‡’ 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 âˆ  A = 36Â° and âˆ  B = 81Â°, then find âˆ  C and âˆ  D.
Solution:
âˆµ AB || CD and AD is a transversal.            [âˆµ ABCD is a trapezium in which AB || CD]
âˆ´ âˆ  A + âˆ  D = 180Â°
â‡’ âˆ  D = 180Â° -  âˆ  A = 180Â° - 36Â° = 144Â°
Again, AB || CD and BC is a transversal.

âˆ´ âˆ  B + âˆ  C = 180Â°
â‡’ âˆ  C = 180Â° - âˆ  B = 180Â° - 81Â° = 99Â°
âˆ´ The required measures of âˆ  D and âˆ  C are 144Â° and 99Â° 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.
âˆ´ E is the mid-point of AC, and DE = (1/2) BC.
Since, perimeter of DABC = 27 cm
âˆ´ AB + BC + CA = 27 cm
â‡’ 2(AD) + BC + 2(AE) = 27 cm
â‡’ 2(4.5 cm) + BC + 2(4 cm) = 27 cm
â‡’ 9 cm + BC + 8 cm = 27 cm
âˆ´ BC = 27 cm - 9 cm - 8 cm = 10 cm

âˆ´  (1/2)BC =(10/2) = 5 cm
â‡’ DE = 5 cm

Question 3. In the adjoining figure, DE || BC and D is the mid-point of AB. Find the perimeter of Î”ABC when AE = 4.5 cm.
Solution:
âˆµ D is the mid-point of AB and DE || BC.
âˆ´ E is the mid-point of AC and DE = (1/2)BC.

â‡’ 2DE = BC
â‡’ 2 x 5 cm = BC
â‡’ BC = 10 cm
Now DB = 3.5 cm
âˆ´ 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 Î”ABC = 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 âˆ  B = x

âˆ´ âˆ  A= (4/5)x
Since, the adjacent angles of a parallelogram are supplementary.
âˆ´ âˆ  A + âˆ  B = 180Â°
â‡’ (4/5)x + x = 180Â°
â‡’ 4x + 5x = 180Â° x 5
â‡’ 9x = 180Â° x 5
â‡’
âˆ´ âˆ  B = 100Â°
Since âˆ  B= âˆ  D            [Opposite angles of parallelogram]
âˆ´ âˆ  D = 100Â°
Now, âˆ  A= (4/5)x =(4/5) x 100Â° = 80Â°
Also âˆ  A= âˆ  C             [Opposite angles of parallelogram]
âˆ´ âˆ  C = 80Â°
The required measures of the angles of the parallelogram are:  âˆ  A = 80Â°, âˆ  B = 100Â° âˆ  C = 80Â° and âˆ  D = 100Â°

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

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.
âˆ´ O is the mid-point of AC and BD
â‡’ AO =(1/2)AC and DO =(1/2)BD
Also âˆ  AOD = 90Â°.
Now, Î”AOD is a right triangle, in which

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
â‡’ AD2 = (12)2 + (9)2
= 144 + 81 = 225 = 152
â‡’ AD = âˆš(15)2 = 15
â‡’ 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 108Â° and the remaining three angles are equal. Find each of the three equal angles.
Solution:
âˆ´ âˆ A + âˆ B + âˆ C + âˆ D = 360Â°
â‡’ 108Â° + [âˆ B + âˆ C + âˆ D] = 360Â°
â‡’ [âˆ B + âˆ C + âˆ D]
= 360Â° - 108Â° = 252Â°

Since,
âˆ D = âˆ B = âˆ C
âˆ´ âˆ B + âˆ C + âˆ D = 252Â°
â‡’ âˆ B + âˆ B + âˆ B = 252Â°
â‡’  3âˆ B = 252Â°
â‡’ âˆ B = (2520/3) = 84Â°
âˆ´ âˆ B = âˆ C = âˆ D = 84Â°
Thus, the measure of each of the remaining angles is 84Â°.

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.
âˆµ ABCD is a ||gm
âˆ´ Its opposite angles are equal.
â‡’ âˆ A = âˆ C
â‡’ (1/2)âˆ A =(1/2) âˆ C

i.e., âˆ YAX = âˆ YCX             ...(1)
Again DA || BC â‡’ YA || CX                [opposite sides of ||gm]
Also âˆ AYC + âˆ YCX = 180Â°            ...(2)
â‡’ âˆ AYC + âˆ YAX = 180            [From (1) and (2)]
â‡’ 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.

Let us join BE and extend it meet CD produced at P.
In Î”AEB and Î”DEP, we get AB || PC and BP is a transversal,
âˆ´ âˆ ABE = âˆ EPD             [Alternate angles]
AE = ED            [âˆµ E is midpoint of AB]
âˆ AEB = âˆ PED             [Vertically opp. angles]
â‡’ Î”AEB â‰Œ Î”DEP
â‡’ BE = PE and AB = DP [SAS]
â‡’ BE = PE and AB = DP
Now, in Î”EPC, E is a mid point of BP and F is mid point of BC
âˆ´ 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)

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;