In Exercise 17.1 of Chapter 17, we shall discuss problems based on various types of quadrilaterals like – trapezium, isosceles trapezium, parallelogram, rhombus, rectangle, square. In this exercise, we mainly study the properties of a parallelogram. Students who find difficulty in understanding the concepts can refer to RD Sharma Solutions which is prepared by our expert tutors with utmost care, which helps students gain competence on the concept. Ultimately the main aim is to help students boost their confidence level and achieve high marks in their exams. Pdf can be downloaded easily from the links given below.

## Download the Pdf of RD Sharma Solutions for Class 8 Maths Exercise 17.1 Chapter 17 Understanding Shapes- III (Special Types of Quadrilaterals)

### Access answers to Maths RD Sharma Solutions For Class 8 Exercise 17.1 Chapter 17 Understanding Shapes- III (Special Types of Quadrilaterals)

**1. Given below is a parallelogramÂ ABCD.Â Complete each statement along with the definition or property used.
(i)Â ADÂ = **

**(ii)Â âˆ DCB =
(iii)Â OCÂ = **

**(iv)Â âˆ DAB +Â âˆ CDAÂ =**

**Solution:**

(i)Â ADÂ = BC. Because, diagonals bisect each other in a parallelogram.

(ii)Â âˆ DCBÂ =Â âˆ BAD.Â Because, alternate interior angles are equal.

(iii)Â OCÂ = OA. Because, diagonals bisect each other in a parallelogram.

(iv)Â âˆ DAB+Â âˆ CDAÂ = 180Â°. Because sum of adjacent angles in a parallelogram is 180Â°.

**2. The following figures are parallelograms. Find the degree values of the unknownsÂ x, y, z.**

**Solution:**

**(i)** âˆ ABC = âˆ y = 100^{o} (opposite angles are equal in a parallelogram)

âˆ x + âˆ y = 180^{o} (sum of adjacent angles is = 180^{o }in a parallelogram)

âˆ x + 100^{o} = 180^{o}

âˆ x = 180^{o} – 100^{o}

= 80^{o}

âˆ´ âˆ x = 80^{o} âˆ y = 100^{o}âˆ z = 80^{o} (opposite angles are equal in a parallelogram)

**(ii)** âˆ RSP + âˆ y = 180^{o} (sum of adjacent angles is = 180^{o }in a parallelogram)

âˆ y + 50^{o} = 180^{o}

âˆ y = 180^{o} – 50^{o}

= 130^{o}

âˆ´ âˆ x = âˆ y = 130^{o} (opposite angles are equal in a parallelogram)

âˆ RSP = âˆ RQP = 50^{o} (opposite angles are equal in a parallelogram)

âˆ RQP + âˆ z = 180^{o} (linear pair)

50^{o} + âˆ z = 180^{o}

âˆ z = 180^{o} â€“ 50^{o}

= 130^{o}

âˆ´ âˆ x = 130^{o} âˆ y = 130^{o} âˆ z = 130^{o}

**(iii)** In Î”PMN

âˆ NPM +Â âˆ NMP +Â âˆ MNP =Â 180Â° [Sum of all the angles of a triangle is 180Â°]

30Â°Â +Â 90Â°Â +Â âˆ z =Â 180Â°

âˆ z =Â 180Â°-120Â°

=Â 60Â°

âˆ y =Â âˆ z =Â 60Â° [opposite angles are equal in a parallelogram]

âˆ z =Â 180Â°-120Â° [sum of the adjacent angles is equal to 180Â° in a parallelogram]

âˆ z =Â 60Â°

âˆ z +Â âˆ LMN =Â 180Â° [sum of the adjacent angles is equal to 180Â° in a parallelogram]

60Â°Â +Â 90Â°+Â âˆ x =Â 180Â°

âˆ x =Â 180Â°-150Â°

âˆ x =Â 30Â°

âˆ´ âˆ x = 30^{o} âˆ y = 60^{o} âˆ z = 60^{o}

**(iv)** âˆ x =Â 90Â° [vertically opposite angles are equal]

In Î”DOC

âˆ x +Â âˆ y +Â 30Â°Â =Â 180Â° [Sum of all the angles of a triangle is 180Â°]

90Â°Â +Â 30Â°Â +Â âˆ y =Â 180Â°

âˆ y =Â 180Â°-120Â°

âˆ y =Â 60Â°

âˆ y =Â âˆ z =Â 60Â° [alternate interior angles are equal]

âˆ´ âˆ x = 90^{o} âˆ y = 60^{o} âˆ z = 60^{o}

**(v)** âˆ x +Â âˆ POR =Â 180Â° [sum of the adjacent angles is equal to 180Â° in a parallelogram]

âˆ x +Â 80Â°Â =Â 180Â°

âˆ x =Â 180Â°-80Â°

âˆ x =Â 100Â°

âˆ y =Â 80Â° [opposite angles are equal in a parallelogram]

âˆ SRQ =âˆ x =Â 100Â°

âˆ SRQ +Â âˆ z =Â 180Â° [Linear pair]

100Â° +Â âˆ z =Â 180Â°

âˆ z =Â 180Â°-100Â°

âˆ z =Â 80Â°

âˆ´ âˆ x = 100^{o} âˆ y = 80^{o} âˆ z = 80^{o}

**(vi)** âˆ y =Â 112Â° [In a parallelogram opposite angles are equal]

âˆ y +Â âˆ VUT =Â 180Â° [In a parallelogram sum of the adjacent angles is equal to 180Â°]

âˆ z + 40Â° +Â 112Â°Â =Â 180Â°

âˆ z =Â 180Â°-152Â°

âˆ z =Â 28Â°

âˆ z =âˆ x =Â 28Â° [alternate interior angles are equal]

âˆ´ âˆ x = 28^{o} âˆ y = 112^{o} âˆ z = 28^{o}

**3. Can the following figures be parallelograms? Justify your answer.**

**Solution:**

**(i)** No, opposite angles are equal in a parallelogram.

**(ii)** Yes, opposite sides are equal and parallel in a parallelogram.

**(iii)** No, diagonals bisect each other in a parallelogram.

**4. In the adjacent figureÂ HOPEÂ is a parallelogram. Find the angle measuresÂ x, yÂ andÂ z. State the geometrical truths you use to find them.
**

**Solution:**

We know that

âˆ POH +Â 70Â°Â =Â 180Â° [Linear pair]

âˆ POH =Â 180Â°-70Â°

âˆ POH =Â 110Â°

âˆ POH =Â âˆ x =Â 110Â° [opposite angles are equal in a parallelogram]

âˆ x +Â âˆ z + 40Â° =Â 180Â° [sum of the adjacent angles is equal to 180Â° in a parallelogram]

110Â°Â +Â âˆ z + 40Â° =Â 180Â°

âˆ z =Â 180Â° – 150Â°

âˆ z =Â 30Â°

âˆ z +âˆ y = 70Â°

âˆ y + 30Â° = 70Â°

âˆ y = 70Â°- 30Â°

âˆ y = 40Â°

**5.** **In the following figuresÂ GUNSÂ andÂ RUNSÂ are parallelograms. FindÂ xÂ andÂ y.**

**Solution:**

**(i)** 3y â€“ 1 = 26 [opposite sides are of equal length in a parallelogram]

3y = 26 + 1

y = 27/3

y = 9

3x = 18 [opposite sides are of equal length in a parallelogram]

x = 18/3

x = 6

âˆ´ x = 6 and y = 9

**(ii)** y â€“ 7 = 20 [diagonals bisect each other in a parallelogram]

y = 20 + 7

y =Â 27

x – y = 16 [diagonals bisect each other in a parallelogram]

x -27 = 16

x = 16 + 27

= 43

âˆ´ x = 43 and y = 27

**6. In the following figureÂ RISKÂ andÂ CLUEÂ are parallelograms. Find the measure ofÂ x.**

**Solution:**

In parallelogram RISK

âˆ RKS +Â âˆ KSI =Â 180Â° [sum of the adjacent angles is equal to 180Â° in a parallelogram]

120Â°Â +Â âˆ KSI =Â 180Â°

âˆ KSI =Â 180Â° – 120Â°

âˆ KSI =Â 60Â°

In parallelogram CLUE

âˆ CEU =Â âˆ CLU =Â 70Â° [opposite angles are equal in a parallelogram]

In Î”EOS

70Â°Â +Â âˆ x + 60Â° =Â 180Â° [Sum of angles of a triangles is 180Â°]

âˆ x =Â 180Â° – 130Â°

âˆ x =Â 50Â°

âˆ´ x = 50Â°

**7. Two opposite angles of a parallelogram are (3 xÂ – 2)^{o} and (50 –Â x)^{o}. Find the measure of each angle of the parallelogram.**

**Solution:**

We know that opposite angles of a parallelogram are equal.

So, (3x – 2)Â° = (50 – x)Â°

3x^{o} – 2Â° = 50Â° – x^{o}

3xÂ° + x^{o} = 50Â° + 2Â°

4x^{o} = 52Â°

x^{o} = 52^{o}/4

= 13^{o}

Measure of opposite angles are,

(3x â€“ 2)^{o} = 3Ã—13 – 2 = 37Â°

(50 – x)^{o} = 50 – 13 = 37Â°

We know that Sum of adjacent angles = 180Â°

Other two angles are 180Â° – 37Â° = 143Â°

âˆ´ Measure of each angle is 37^{o}, 143^{o}, 37^{o}, 143^{o}

**8. If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.**

**Solution:**

Let us consider one of the adjacent angle as xÂ°

Other adjacent angle is = 2x^{o}/3

We know that sum of adjacent angles = 180Â°

So,

x^{o} + 2x^{o}/3 = 180^{o}

(3x^{o} + 2x^{o})/3 = 180^{o}

5x^{o}/3 = 180^{o}

x^{o} = 180^{o}Ã—3/5

= 108^{o}

Other angle is = 180^{o} â€“ 108^{o} = 72^{o}

âˆ´ Angles of a parallelogram are 72^{o}, 72^{o}, 108^{o}, 108^{o}

**9. The measure of one angle of a parallelogram is 70Â°. What are the measures of the remaining angles?**

**Solution:**

Let us consider one of the adjacent angle as xÂ°

Other adjacent angle = 70Â°

We know that sum of adjacent angles = 180Â°

So,

x^{o} + 70^{o} = 180^{o}

x^{o} = 180^{o} â€“ 70^{o}

= 110^{o}

âˆ´ Measures of the remaining angles are 70Â°, 70Â°, 110Â° and 110Â°

**10. Two adjacent angles of a parallelogram are as 1 : 2. Find the measures of all the angles of the parallelogram.**

**Solution:**

Let us consider one of the adjacent angle as xÂ°

Other adjacent angle = 2xÂ°

We know that sum of adjacent angles = 180Â°

So,

x^{o} + 2x^{o} = 180^{o}

3x^{o} = 180^{o}

x^{o} = 180^{o}/3

= 60^{o}

So other angle is 2x = 2Ã—60 = 120^{o}

âˆ´ Measures of the remaining angles are 60Â°, 60Â°, 120Â° and 120Â°

**11. In a parallelogramÂ ABCD,Â âˆ D= 135Â°, determine the measure ofÂ âˆ AÂ and âˆ B.**

**Solution:**

Given, one of the adjacent angleÂ âˆ D =Â 135Â°

Let other adjacent angleÂ âˆ A beÂ = xÂ°

We know that sum of adjacent angles = 180Â°

x^{o} + 135^{o} = 180^{o}

x^{o} = 180^{o} â€“ 135^{o}

= 45^{o}

âˆ A = x^{o} = 45^{o}

We know that measure of opposite angles are equal in a parallelogram.

So, âˆ A = âˆ C = 45^{o}

And âˆ D = âˆ B = 135^{o}

**12. ABCDÂ is a parallelogram in whichÂ âˆ AÂ = 70Â°. ComputeÂ âˆ B,Â âˆ CÂ andÂ âˆ D.**

**Solution:**

Given, one of the adjacent angle âˆ A = 70^{o}

Other adjacent angle âˆ B be = x^{o}

We know that sum of adjacent angles = 180Â°

x^{o} + 70^{o} = 180^{o}

x^{o} = 180^{o} â€“ 70^{o}

= 110^{o}

âˆ B = x^{o} = 110^{o}

We know that measure of opposite angles are equal in a parallelogram.

So, âˆ A = âˆ C = 70^{o}

And âˆ D = âˆ B = 110^{o}

**13. The sum of two opposite angles of a parallelogram is 130Â°. Find all the angles of the parallelogram.**

**Solution:**

Consider ABCD as a parallelogram

âˆ A +Â âˆ C = 130^{0}

HereÂ âˆ A andÂ âˆ C are opposite angles

SoÂ âˆ C = 130/2 = 65^{0}

We know that sum of adjacent angles is 180^{0}

âˆ B +Â âˆ D = 180^{0}

65^{0} +Â âˆ D = 180^{0}

âˆ D = 180^{0} – 65^{0} = 115^{0}

âˆ D =Â âˆ B = 115^{0} ( Opposite angles)

Therefore,Â âˆ A = 65^{0},Â âˆ B = 115^{0},Â âˆ C = 65^{0} andÂ âˆ D = 115^{0}.

**14. All the angles of a quadrilateral are equal to each other. Find the measure of each. Is the quadrilateral a parallelogram? What special type of parallelogram is it?**

**Solution:**

Let us consider each angle of a parallelogram as x^{o}

We know that sum of angles = 360^{o}

x^{o} + x^{o} + x^{o} + x^{o} = 360^{o}

4 x^{o} = 360^{o}

x^{o} = 360^{o}/4

= 90^{o}

âˆ´ Measure of each angle is 90^{o}

Yes, this quadrilateral is a parallelogram.

Since each angle of a parallelogram is equal to 90Â°, so it is a rectangle.

**15. Two adjacent sides of a parallelogram are 4 cm and 3 cm respectively. Find its perimeter.**

**Solution:**

We know that opposite sides of a parallelogram are parallel and equal.

So, Perimeter = Sum of all sides (there are 4 sides)

Perimeter = 4 + 3 + 4 + 3

= 14 cm

âˆ´ Perimeter is 14cm.

**16. The perimeter of a parallelogram is 150 cm. One of its sides is greater than the other by 25 cm. Find the length of the sides of the parallelogram.**

**Solution:**

Given, Perimeter of the parallelogram = 150 cm

Let us consider one of the sides as = â€˜xâ€™ cm

Other side as = (x + 25) cm

We know that opposite sides of a parallelogram are parallel and equal.

So, Perimeter = Sum of all sides

x + x + 25 + x + x + 25 = 150

4x + 50 = 150

4x = 150 â€“ 50

x = 100/4

= 25

âˆ´ Sides of the parallelogram are (x) = 25 cm and (x+25) = 50 cm.

**17. The shorter side of a parallelogram is 4.8 cm and the longer side is half as much again as the shorter side. Find the perimeter of the parallelogram.**

**Solution:**

Given, Shorter side of the parallelogram = 4.8 cm

Longer side of the parallelogram = 4.8 + 4.8/2

= 4.8 + 2.4

= 7.2cm

We know that opposite sides of a parallelogram are parallel and equal.

So, Perimeter = Sum of all sides

Perimeter of the parallelogram = 4.8 + 7.2 + 4.8 + 7.2

= 24cm

âˆ´ Perimeter of the parallelogram is 24 cm.

**18. Two adjacent angles of a parallelogram are (3 x-4)^{o} and (3x+10)^{o}. Find the angles of the parallelogram.**

**Solution:**

We know that adjacent angles of a parallelogram are equal.

So, (3x – 4)Â° + (3x + 10)Â° = 180^{o}

3xÂ° + 3x^{o} â€“ 4 + 10 = 180Â°

6x = 180Â° – 6^{o}

x = 174^{o}/6

= 29^{o}

Measure of adjacent angles are,

(3x â€“ 4)^{o} = 3Ã—29 – 4 = 83Â°

(3x + 10)^{o} = 3Ã—29 + 10 = 97Â°

We know that Sum of adjacent angles = 180Â°

âˆ´ Measure of each angle is 83^{o}, 97^{o}, 83^{o}, 97^{o}

**19. In a parallelogramÂ ABCD,Â the diagonals bisect each other atÂ O.Â IfÂ âˆ ABCÂ =30Â°,Â âˆ BDC= 10Â° andÂ âˆ CABÂ =70Â°. Find:
âˆ DAB,Â âˆ ADC,Â âˆ BCD,Â âˆ AOD,Â âˆ DOC,Â âˆ BOC,Â âˆ AOB,Â âˆ ACD,Â âˆ CAB,Â âˆ ADB,Â âˆ ACB,Â âˆ DBC, andÂ âˆ DBA.**

**Solution:**

Firstly let us draw a parallelogram

Given, âˆ ABCÂ = 30^{o},

âˆ ABCÂ =Â âˆ ADCÂ = 30Â° [We know that measure of opposite angles are equal in a parallelogram]

âˆ BDCÂ = 10Â°

âˆ CABÂ =70Â°

âˆ BDA = âˆ ADB = âˆ ADC – âˆ BDC = 30Â° – 10Â° = 20Â°

âˆ DAB = 180Â° – 30Â° = 150Â°

âˆ ADB = âˆ DBC = 20^{o} (alternate angles)

âˆ BCDÂ =Â âˆ DABÂ = 150Â° [we know, opposite angles are equal in a parallelogram]

âˆ DBA =Â âˆ BDC = 10Â°Â [we know, Alternate interior angles are equal]

In Î”ABC

âˆ CAB +Â âˆ ABC +Â âˆ BCA = 180Â°Â [since, sum of all angles of a triangle is 180Â°]

70^{o} + 30^{o} + âˆ BCA = 180Â°

âˆ BCA = 180^{o} â€“ 100^{o}

= 80^{o}

âˆ DAB = âˆ DAC + âˆ CAB = 70^{o} + 80^{o} = 150^{o}

âˆ BCD = 150^{o} (opposite angle of the parallelogram)

âˆ DCA = âˆ CAB = 70^{o}

In Î”DOC

âˆ BDC +Â âˆ ACD +Â âˆ DOC = 180Â°Â [since, sum of all angles of a triangle is 180Â°]

10Â°Â +Â 70Â°Â +Â âˆ DOC = 180Â°

âˆ DOC = 180Â°- 80Â°

âˆ DOC = 100Â°

So, âˆ DOC =Â âˆ AOB = 100Â°Â [Vertically opposite angles are equal]

âˆ DOC +Â âˆ AOD = 180Â°Â [Linear pair]

100Â°Â +Â âˆ AOD = 180Â°

âˆ AOD = 180Â°- 100Â°

âˆ AOD = 80Â°

So, âˆ AOD =Â âˆ BOC = 80Â°Â [Vertically opposite angles are equal]

âˆ CAB = 70^{o}

âˆ ABC +Â âˆ BCDÂ = 180Â° [In a parallelogram sum of adjacent angles is 180Â°]

30Â°Â +Â âˆ ACB +Â âˆ ACDÂ = 180Â°

30Â°Â +Â âˆ ACB +Â 70Â° = 180Â°

âˆ ACBÂ = 180Â° – 100Â°

âˆ ACBÂ = 80Â°

âˆ´ âˆ DAB = 150^{o}, âˆ ADC = 30,Â âˆ BCD = 150^{o},Â âˆ AOD = 80^{o},Â âˆ DOC = 100^{o},Â âˆ BOC = 80^{o}, âˆ AOB = 100^{o}, âˆ ACD = 70^{o},Â âˆ CAB = 70^{o},Â âˆ ADB = 20^{o},Â âˆ ACB = 80^{o},Â âˆ DBC = 20^{o}, andÂ âˆ DBA = 10^{o}.

**20. Find the angles marked with a question mark shown in Figure.**

**Solution:**

In Î”BEC

âˆ BEC +Â âˆ ECB +âˆ CBE = 180Â° [Sum of angles of a triangle is 180Â°]

90Â° + 40Â° +Â âˆ CBE = 180Â°

âˆ CBE = 180Â°-130Â°

âˆ CBE = 50Â°

âˆ CBE = âˆ ADC = 50Â° (Opposite angles of a parallelogram are equal)

âˆ B =Â âˆ D = 50Â° [Opposite angles of a parallelogram are equal]

âˆ A +Â âˆ B = 180Â° [Sum of adjacent angles of a triangle is 180Â°]

âˆ A +Â 50Â°Â = 180Â°

âˆ A = 180Â°-50Â°

So, âˆ A = 130Â°

In Î”DFC

âˆ DFC +Â âˆ FCD +âˆ CDF = 180Â° [Sum of angles of a triangle is 180Â°]

90Â° +Â âˆ FCD +Â 50Â°Â = 180Â°

âˆ FCD = 180Â°-140Â°

âˆ FCD = 40Â°

âˆ A =Â âˆ C = 130Â° [Opposite angles of a parallelogram are equal]

âˆ C =Â âˆ FCE +âˆ BCE +Â âˆ FCD

âˆ FCD +Â 40Â°Â +Â 40Â°Â = 130Â°

âˆ FCD = 130Â° – 80Â°

âˆ FCD = 50Â°

âˆ´ âˆ EBC = 50^{o}, âˆ ADC = 50^{o} and âˆ FCD = 50^{o}

**21. The angle between the altitudes of a parallelogram, through the same vertex of an obtuse angle of the parallelogram is 60Â°. Find the angles of the parallelogram.**

**Solution:**

Let us consider a parallelogram, ABCD. Where, DPâŠ¥AB and DQÂ âŠ¥BC.

GivenÂ âˆ PDQ = 60Â°

In quadrilateral DPBQ

âˆ PDQ +Â âˆ DPB +Â âˆ B +Â âˆ BQD = 360Â° [Sum of all the angles of a Quadrilateral is 360Â°]

60Â° + 90Â° +Â âˆ B + 90Â° = 360Â°

âˆ B = 360Â° â€“ 240Â°

âˆ B = 120Â°

âˆ B =Â âˆ D = 120Â° [Opposite angles of parallelogram are equal]

âˆ B +Â âˆ C = 180Â° [Sum of adjacent interior angles in a parallelogram is 180Â°]

120Â° +Â âˆ C = 180Â°

âˆ C = 180Â° â€“ 120Â° = 60Â°

âˆ A =Â âˆ C = 60Â° (Opposite angles of parallelogram are equal)

âˆ´ Angles of a parallelogram are 60^{o}, 120^{o}, 60^{o}, 120^{o}

**22. In Figure,Â ABCDÂ andÂ AEFGÂ are parallelograms. IfÂ âˆ CÂ =55Â°, what is the measure ofÂ âˆ F?
**

**Solution:**

In parallelogram ABCD

âˆ CÂ = âˆ A =Â 55Â° [In a parallelogram opposite angles are equal in a parallelogram]

In parallelogram AEFG

âˆ AÂ = âˆ F =Â 55Â° [In a parallelogram opposite angles are equal in a parallelogram]

âˆ´ Measure of âˆ F =Â 55Â°

**23. In Figure,Â BDEFÂ andÂ DCEFÂ are each a parallelogram. Is it true thatÂ BD = DC? Why or why not?**

**Solution:**

In parallelogram BDEF

BD = EF [In a parallelogram opposite sides are equal]

In parallelogram DCEF

DC = EF [In a parallelogram opposite sides are equal]

Since, BD = EF = DC

So, BD = DC

**24. In Figure, suppose it is known thatÂ DEÂ =Â DF. Then, is Î”ABCÂ isosceles? Why or why not?**

**Solution:**

In parallelogram BDEF

BD = EF and BF = DE [opposite sides are equal in a parallelogram]

In parallelogram DCEF

DC = EF and DF = CE [opposite sides are equal in a parallelogram]

In parallelogram AFDE

AF = DE and DF = AE [opposite sides are equal in a parallelogram]

So, DE = AF = BF

Similarly: DF = CE = AE

Given, DE = DF

Since, DF = DF

AF + BF = CE + AE

AB = AC

âˆ´ Î”ABC is an isosceles triangle.

**25. Diagonals of parallelogramÂ ABCDÂ intersect atÂ OÂ as shown in Figure. XYÂ containsÂ O, andÂ X, YÂ are points on opposite sides of the parallelogram. Give reasons for each of the following:**

**(i) OB = OD**

**(ii) âˆ OBY = âˆ ODX**

**(iii) âˆ BOY = âˆ DOX**

**(iv) Î”BOY = Î”DOX**

**Now, state if XY is bisected at O.**

**Solution:**

(i)Â OBÂ =Â OD

OB = OD. Since diagonals bisect each other in a parallelogram.

(ii)Â âˆ OBYÂ =âˆ ODX

âˆ OBYÂ =âˆ ODX. Since alternate interior angles are equal in a parallelogram.

(iii)Â âˆ BOY=Â âˆ DOX

âˆ BOY=Â âˆ DOX. Since vertical opposite angles are equal in a parallelogram.

(iv)Â Î”BOYÂ â‰…Â Î”DOX

Î”BOY and Î”DOX. Since OB = OD, where diagonals bisect each other in a parallelogram.

âˆ OBYÂ =âˆ ODX [Alternate interior angles are equal]

âˆ BOY=Â âˆ DOXÂ [Vertically opposite angles are equal]

Î”BOYÂ â‰…Î”DOX [by ASA congruence rule]

OX = OY [Corresponding parts of congruent triangles]

âˆ´Â XYÂ is bisected atÂ O.

**26. In Fig. 17.31,Â ABCDÂ is a parallelogram,Â CEÂ bisectsÂ âˆ CÂ and AFÂ bisectsÂ âˆ A. In each of the following, if the statement is true, give a reason for the same:**

**(i) âˆ A = âˆ C**

**(ii) âˆ FAB = Â½ âˆ A**

**(iii) âˆ DCE = Â½ âˆ C**

**(iv) âˆ CEB = âˆ FAB**

**(v) CE âˆ¥ AF**

**Solution:**

(i)Â âˆ AÂ =Â âˆ C

True, Since âˆ AÂ =âˆ C =Â 55Â° [opposite angles are equal in a parallelogram]

(ii)Â âˆ FAB = Â½ âˆ A

True, Since AF is the angle bisector of âˆ A.

(iii)Â âˆ DCE= Â½ âˆ C

True, Since CE is the angle bisector of angle âˆ C.

(iv)Â âˆ CEB=Â âˆ FAB

True,

Since âˆ DCEÂ =Â âˆ FAB (opposite angles are equal in a parallelogram).

âˆ CEBÂ =Â âˆ DCE (alternate angles)

Â½ âˆ C = Â½ âˆ AÂ [AF and CE are angle bisectors]

(v)Â CEÂ ||Â AF

True, since one pair of opposite angles are equal, therefore quad. AEFC is a parallelogram.

**27. Diagonals of a parallelogramÂ ABCDÂ intersect atÂ O. ALÂ andÂ CMÂ are drawn perpendiculars toÂ BDÂ such thatÂ LÂ andÂ MÂ lie onÂ BD. IsÂ ALÂ =Â CM? Why or why not?**

**Solution:**

Given, AL and CM are perpendiculars on diagonal BD.

In Î”AOL and Î”COM

âˆ AOL = âˆ COM (vertically opposite angle) â€¦.. (i)

âˆ ALO = âˆ CMO = 90^{o} (each right angle) â€¦â€¦. (ii)

By using angle sum property

âˆ AOL + âˆ ALO + âˆ LAO = 180^{o} â€¦â€¦â€¦ (iii)

âˆ COM + âˆ CMO + âˆ OCM = 180^{o} â€¦â€¦. (iv)

From (iii) and (iv)

âˆ AOL + âˆ ALO + âˆ LAO = âˆ COM + âˆ CMO + âˆ OCM

âˆ LAO = âˆ OCM (from (i) and (ii))

In Î”AOL and Î”COM

âˆ ALO = âˆ CMO (each right angle)

AO = OC (diagonals of a parallelogram bisect each other)

âˆ LAO = âˆ OCM (proved)

So, Î”AOL is congruent to Î”COM

âˆ´ AL = CM (Corresponding parts of congruent triangles)

**28. PointsÂ EÂ andÂ FÂ lie on diagonalsÂ ACÂ of a parallelogramÂ ABCDÂ such thatÂ AEÂ =Â CF. what type of quadrilateral isÂ BFDE?**

**Solution:**

In parallelogram ABCD:

AO = OCâ€¦â€¦â€¦â€¦â€¦.. (i) (Diagonals of a parallelogram bisect each other)

AE = CFâ€¦â€¦.. (ii) Given

On subtracting (ii) from (i)

AO – AE = OC – CF

EO = OF â€¦.. (iii)

In Î”DOE and Î”BOF

EO = OF (proved)

DO = OB (Diagonals of a parallelogram bisect each other)

âˆ DOE = âˆ BOF (vertically opposite angles are equal in a parallelogram)

By the rule of SAS congruence Î”DOE â‰… Î”BOF

So, DE = BF (Corresponding parts of congruent triangles)

In Î”BOE and Î”DOF

EO = OF (proved)

DO = OB (diagonals of a parallelogram bisect each other)

âˆ DOF = âˆ BOE (vertically opposite angles are equal in a parallelogram)

By the rule of SAS congruence Î”DOE â‰… Î”BOF

âˆ´ DF = BE (Corresponding parts of congruent triangles)

âˆ´ BFDE is a parallelogram, since one pair of opposite sides are equal and parallel.

**29. In a parallelogramÂ ABCD, AB =Â 10cm,Â ADÂ = 6 cm. The bisector ofÂ âˆ AÂ meetsÂ DCÂ inÂ E, AEÂ andÂ BCÂ produced meet atÂ F.Â Find the lengthÂ CF.**

**Solution:**

In a parallelogram ABCD

Given, AB = 10 cm, AD = 6 cm

â‡’Â CD = AB = 10 cm and AD = BC = 6 cm [In a parallelogram opposite sides are equal]

AE is the bisector of âˆ DAE = âˆ BAE = x

âˆ BAE = âˆ AED = x (alternate angles are equal)

Î”ADE is an isosceles triangle. Since opposite angles in Î”ADE are equal.

AD = DE = 6cm (opposite sides are equal)

CD = DE + EC

EC = CD â€“ DE

= 10 â€“ 6

= 4cm

âˆ DEA = âˆ CEF = x (vertically opposite angle are equal)

âˆ EAD = âˆ EFC = x (alternate angles are equal)

Î”EFC is an isosceles triangle. Since opposite angles in Î”EFC are equal.

CF = CE = 4cm (opposite side are equal to angles)

âˆ´ CF = 4cm.