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**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= (180^{0}/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= (360^{0}/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= (195

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, AD^{2} = AO^{2} + DO^{2}

â‡’ AD^{2} = (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:** ABCD is a quadrilateral

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 = (252^{0}/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)

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