**Q.1: In the adjoining figure, ABCD is a parallelogram in which ∠A = 72°. Calculate ∠B, ∠C and ∠D.**

**Sol: **

ABCD is a parallelogram and ∠ A = 72°

We know that opposite angles of a parallelogram are equal

∴ ∠A = ∠C and ∠B = ∠D

∴ ∠C = 72°

∠A and ∠B are adjacent angles

i.e., ∠A + ∠B = 180°

∠B = ∠D =108°

Hence, ∠B = ∠D = 108° and ∠C = 72°

**Q.2: In the adjoining figure, ABCD is a parallelogram in which ∠ DAB = 80° and ∠ DBC = 60°. Calculate ∠ CDB and ∠ ADB.**

**Sol: **

**Given:** ABCD is a parallelogram and ∠ DAB = 80° and ∠ DBC = 60°

**To find:** Measure of ∠CDB and ∠ ADB

In parallelogram ABCD, AD

∴ ∠ DBC = ∠ ADB = 60° (Alternate interior angles) . . . . . . . (i)

As ∠ DAB and ∠ ADC are adjacent angles, ∠ DAB + ∠ ADC = 180°

∴∠ ADC = 180° – ∠ DAB

∠ ADC = 180° – 80° = 100°

Also, ∠ ADC = ∠ ADB + ∠ CDB

∴∠ ADC = 100°

∠ ADB + ∠ CDB = 100° . . . . . . . (ii)

**From (i) and (ii), we get:**

60° + ∠ CDB = 100°

∠ CDB = 100° – 60° = 40°

Hence, ∠ CDB = 40° and ∠ ADB = 60°

**Q.3: In the adjoining figure, ABCD is a parallelogram in which ∠ A = 60°. If the bisectors of ∠ A and ∠ B meet DC at P, prove that (i) ∠ APB = 90°, (ii) AD = DP and PB = PC = BC, (iii) DC = 2AD.**

**Sol:**

ABCD is a parallelogram

∴ ∠ A = ∠ C and ∠ B = ∠ D (Opposite angles)

And ∠ A + ∠ B = 180° (Adjacent angles are supplementary)

∴ ∠ B = 180° – ∠ A

180° – 60° = 120° (Since, ∠ A = 60°)

∴ ∠ A = ∠ C = 60° and ∠ B = ∠ D = 120°

**(i)** In

∴ ∠ APB = 180°- (30° + 60°) = 90°

**(ii)** In

∴ ∠APB = 180° – (30° + 120°) = 30°

Thus, ∠PAD = ∠APB = 30°

Hence,

In triangle PBC, ∠PBC = 60°. ∠BPC = 180° – (90° – 30°) = 60° (Opposite angle of ∠A)

∠PBC = ∠BPC = ∠BCP

Hence,

**(iii)** DC = DP + PC

From (ii), we have:

Hence,

DC = DP + PC

From (ii) we have:

DC = AD + BC

i.e. DC = AD + AD. Therefore, DC = 2AD

**Q.4: In the adjoining figure, ABCD is a parallelogram in which ∠BAO = 35°, ∠DAO =**

**40° and ∠COD = 105°.**

**Calculate (i) ∠ABO, (ii) ∠ODC, (iii) ∠ACB, (iv) ∠CBD**

**Sol:**

ABCD is a parallelogram.

AB || DC and BC || AD

**(i)** In

∠ABO = 180° – (35° + 105°) = 40°

**(ii)** ∠ODC and ∠ABO are alternate interior angles

∠ODC = ∠ABO = 40°

**(iii)** ∠ACB = ∠CAD = 40° (Alternate interior angles)

**(iv)** ∠CBD = ∠ABC – ∠ABD

∠ABC = 180° – ∠BAD

∠ABC = 180° – 75° = 105°

∠LCBD = 105° – ∠ABD

∠CBD = 105° – 40° = 65° (Adjacent angles are supplementary) and cABD = cABO

**Q.5: In a parallelogram ABCD, if ∠A = (2x+25) ° and ∠B = (3x-5) °, find the value of X and the measure of each angle of the parallelogram.**

**Sol:**

ABCD is a parallelogram

i.e., ∠A = ∠C and ∠B = ∠D (Opposite angles)

Also, ∠A + ∠B = 180° (Adjacent angles are supplementary)

(2x + 25) ° + (3x — 5) ° = 180

5x +20 = 180

5x =160

x = 32°

∠A= 2 x 32 + 25 = 89° and ∠B = 3 x 32 — 5 = 91°

Hence, x = 32°, ∠A = ∠C = 89° and ∠B = ∠D = 91°

**Q.6: If an angle of a parallelogram is four-fifths of its adjacent angle, find the angles of the parallelogram.**

**Sol:**

Let, ABCD be a parallelogram

∠A = ∠C and ∠B = ∠D (Opposite angles)

Let ∠A = x° and ∠B = (4x/5)°

Now, ∠A + ∠B = 180° (Adjacent angles are supplementary)

x + (4x/5) = 180° => 9x/5 = 180° i.e. x = 100°

Now, ∠A = 100° and ∠B = (4/5) x 100° = 80°

Hence, ∠A = ∠C = 100°; ∠B = ∠D = 80°

**Q.7: Find the measure of each angle of a parallelogram, if one of its angles is 30° less than twice the smallest angle.**

**Sol:**

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: ABCD is a parallelogram in which AB = 9.5cm and its perimeter is 30cm. Find the length of each side of a parallelogram.**

**Sol:**

ABCD is a parallelogram

The Opposite sides of a parallelogram are parallel and equal

∴ AB = DC = 9.5 cm

Let, BC = AD = x

∴ Perimeter of ABCD = AB + BC + CD + DA = 30cm

Hence, AB = DC = 9.5 cm and BC = DA = 5.5 cm

**Q.9: In each of the figures given below, ABCD is a rhombus. Find the value of X and Y in each case.**

**Sol:**

ABCD is a rhombus and a rhombus is also a parallelogram. A rhombus has four equal sides.

**(i)** In

i.e., x = 35°

Now, ∠B + ∠ C = 180°(Adjacent angles are supplementary)

But, ∠C = x + y = 70°

Hence, x = 35°and y = 35°

**(ii)** The diagonals of a rhombus are perpendicular bisector of each other.

So, in

∴ x = 50 °

In

So, ∠ ABD = ∠ ADB = 50°

Hence, x = 50°and y = 50°

**(iii)** ∠ BAC = ∠ DCA (Alternate interior angles)

i.e., x = 62°

In

Also, ∠BOC = 90°

∴ ∠ OBC = 180° – (90° + 62°) = 28°

Hence, x = 62°and y = 28°

**Q.10: The lengths of the diagonals of a rhombus are 24cm and 18cm respectively. Find the length of each side of the rhombus.**

**Sol:**

Let, ABCD be a rhombus

∴ AB = BC = CD = DA

Here, AC and BD are the diagonals of ABCD, where AC = 24 cm and BD = 18 cm

Let, the diagonals intersect each other at O

We know that the diagonals of a rhombus are perpendicular bisectors of each other

∴

And OB = BD / 2 = 18 / 2 = 9 cm.

Now, AB^{2 }= OA^{2 }+ OB^{2 }(Pythagoras theorem)

^{2} = (12)^{2 }+ (9)^{2}

^{2 }=144 + 81 = 225

Hence, the side of the rhombus is 15 cm

**Q.11: Each side of a rhombus is 10cm long and one of its diagonals measures 16cm.Find the length of the other diagonal and hence find the area of the rhombus.**

**Sol:**

Let, ABCD be a rhombus

AB = BC = CD = DA = 10cm

Let, AC and BD be the diagonals of ABCD. Let AC = x and BD = 16 cm and O be the intersection point of the diagonals

We know that the diagonals of a rhombus are perpendicular bisectors of each other:

Now,

=>36 x4 =

Hence, the other diagonal of the rhombus is 12 cm.

Therefore, Area of the rhombus = 12 x 12 x 16 = 96 cm^{2}

**Q.12: In each of the figures given below, ABCD is a rectangle. Find the values of x and y in each case.**

**Solution:**

**(i)** ABCD is a rectangle

The diagonals of a rectangle are congruent and bisect each other. Therefore, in

OA = OB

∠OAB = ∠OBA = 35°

x = 90° – 35° = 55° and ∠AOB = 180° – (35° + 35°) = 110°

y = ∠AOB = 110° [Vertically opposite angles]

Hence, x = 55° and y = 110°

**(ii)** In

OA = OB

Now, ∠OAB = ∠OBA = 12 x 180° – 110° = 35°

y = ∠BAC = 35° [Interior alternate angles]

Also, x = 90° – y

x = 90° – 35° = 55°

Hence, x = 55° and y = 35°

**Q.13: In the adjoining figure, ABCD is a square. S line segment CX cuts AB at X and the diagonal BD at O such that ∠COD =80° and ∠OXA = x°. Find the value of x°.**

**Sol:**

The angles of a square are bisected by the diagonals.

∠OBX = 45° [Therefore, LABC = 90° and BD bisects LABC]

And ∠BOX= ∠COD = 80° [Vertically opposite angles]

In

∠AXO = ∠OBX+ ∠BOX [Exterior angle of

= 45° + 80° = 125°

Therefore, x =125°

**Q.14: In the adjoining figure, AL and CM are perpendiculars to the diagonal BD of a parallelogram ABCD. Prove that,**

**(i) **

**(ii) **AL = CM

**Solution:**

ABCD is a parallelogram.

AD || BC and AD = BC

**(i)** In

AD = BC, ∠ALD = ∠CMB and ∠ADL = ∠CMB

**(ii)** As

Therefore, AL = CM

**Q.15: In the adjoining figure, ABCD is a parallelogram in which the bisectors of ∠A and ∠B intersect at a point P. Prove that ∠APB = 90°.**

**Sol:**

ABCD is a parallelogram and ∠A and LB are adjacent angles.

∠A + ∠B= 180°.

∠A2 + ∠B2 = 1802 = 90**°**

In

∠PAB = ∠A/2 and ∠PBA = ∠B/2

∠APB = 180 – (∠PAB + ∠PBA) [Angle sum property of triangle]

∠APB = 180 – ∠A2 + ∠B2

∠APB = 180 – 90 = 90° (Hence proved)

**Q.16: In the adjoining figure, ABCD is a parallelogram. If P and Q are points on AD and BC respectively such that AP = 13AD and CQ =13ADBC, prove that AQCP is a parallelogram.**

**Sol:**

We have:

∠B= ∠D

AD= BC and AB = DC

Also, AD || BC and AB || DC

It is given that AP = 13AD and CQ = 13BC

AP = CQ

In

AB =CD, ∠B= ∠D and DP =QB

i.e.,

PC = QA

Thus, in quadrilateral AQCP, we have:

AP = CQ . . . . . . . (i)

PC =QA . . . . . . . (ii)

Therefore, AQCP is a parallelogram

**Q.17: In the adjoining figure, ABCD is a parallelogram whose diagonals intersect each other at O. A line segment EOF is drawn to meet AB at E and DC at F. Prove that OE = OF.**

**Sol:**

In

OD = 0B, ∠DOF = ∠BOE and ∠FDO = ∠OBE

i.e.,

OF = OE (Hence Proved)

**Q.18: In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC.**

**Sol:**

In

DC = BE, ∠COD = ∠BOE and ∠OCD = ∠OBE

i.e.

We know that, BC = OC + 0B.

Therefore, ED bisects BC.

**Q.19: In the adjoining figure, ABCD is a parallelogram and E is the midpoint of side BC. If DE and AB when produced meet at F, Prove that AF = 2 AB.**

**Sol:**

**Given: **ABCD is a parallelogram.

BE = CE (E is the midpoint of BC)

DE and AB when produced meet at F

**To prove:** AF = 2AB

**Proof:**

In parallelogram ABCD, we have:

AB || DC

∠DCE = ∠EBF (Alternate interior angles)

In ADCE and ABFE, we have:

∠DCE = ∠EBF (Proved above)

∠DEC = ∠BEF (Vertically opposite angles)

Also, BE = CE (Given)

DC = BF (CPCT)

But DC = AB, as ABCD is a parallelogram

DC = AB = BF . . . . . . . (i)

Now, AF = AB + BF . . . . . . . (ii)

From (i), we get:

AF = AB + AB = 2AB (Hence, proved)

**Q.20: A ΔABC is given. If lines are drawn through A, B, C parallel respectively to the sides BC, CA and AB, forming ΔPQR, as shown in the adjoining figure, show that BC = 12QR.**

**Sol:**

BC || QA and CA || QB

i.e., BCQA is a parallelogram.

BC = QA . . . . . . (i)

Similarly, BC || AR and AB || CR

i.e., BCRA is a parallelogram.

BC = AR . . . . . . . . . (ii)

But QR = QA + AR

From (i) and (ii), we get:

QR = BC + BC, QR = 2BC and BC = 12QR

**Q.21: In the adjoining figure, ΔABC is a triangle and through A, B, C lines are drawn, parallel respectively to BC, CA and AB, intersecting at P,Q and R. Prove that the perimeter of ΔPQR is double the perimeter of ΔABC.**

**Sol:**

Perimeter of

Perimeter of

BC || QA and CA || QB

i.e., BCQA is a parallelogram.

BC = QA . . . . . . (iii)

Similarly, BC || AR and AB || CR

i.e., BCRA is a parallelogram.

BC =AR . . . . . . (iv)

But, QR = QA + AR

From (iii) and (iv), we get:

QR = BC + BC

QR = 2BC

BC = 12QR

Similarly, CA = 12PQ and AB = 12PR

**From (i) and (ii), we have: **

Perimeter of

i.e., Perimeter of

Perimeter of