# RD Sharma Solutions for Class 8 Chapter 17 Understanding Shapes- III (Special Types of Quadrilaterals)

In Chapter 17 of RD Sharma Class 8 Maths, we shall learn concepts based on some special types of quadrilaterals and their properties. The experienced faculty team at BYJUâ€™S has explained the concepts in a step by step format to help students analyze their problem-solving ability when preparing for their board exams. Students are advised to practice on a regular basis which helps them crack even difficult questions at ease in their examinations. RD Sharma Class 8 is one of the best study materials one opts for their exam preparation, which yields good results. Students can download the PDF from the links given below and start practising.

Chapter 17- Understanding Shapes-III (Special Types of Quadrilaterals) contains three exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

• Various types of quadrilaterals – trapezium, isosceles trapezium, parallelogram, rhombus, rectangle, square.
• Properties of a parallelogram.
• Properties of a rhombus.
• Properties of a rectangle.
• Properties of a square.
• Properties of a trapezium.

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

EXERCISE 17.1 PAGE NO: 17.9

1. Given below is a parallelogramÂ ABCD.Â Complete each statement along with the definition or property used.

(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 = 100o (opposite angles are equal in a parallelogram)

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

âˆ x + 100o = 180o

âˆ x = 180o – 100o

= 80o

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

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

âˆ y + 50o = 180o

âˆ y = 180o – 50o

= 130o

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

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

âˆ RQP + âˆ z = 180o (linear pair)

50o + âˆ z = 180o

âˆ z = 180o â€“ 50o

= 130o

âˆ´ âˆ x = 130o âˆ y = 130o âˆ z = 130o

(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 = 30o âˆ y = 60o âˆ z = 60o

(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 = 90o âˆ y = 60o âˆ z = 60o

(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 = 100o âˆ y = 80o âˆ z = 80o

(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 = 28o âˆ y = 112o âˆ z = 28o

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 z =Â 60Â°

In parallelogram CLUE

âˆ CEU =Â âˆ z =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 (3xÂ – 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°

3xo – 2Â° = 50Â° – xo

3xÂ° + xo = 50Â° + 2Â°

4xo = 52Â°

xo = 52o/4

= 13o

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 37o, 143o, 37o, 143o

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 = 2xo/3

We know that sum of adjacent angles = 180Â°

So,

xo + 2xo/3 = 180o

(3xo + 2xo)/3 = 180o

5xo/3 = 180o

xo = 180oÃ—3/5

= 108o

Other angle is = 180o â€“ 108o = 72o

âˆ´ Angles of a parallelogram are 72o, 72o, 108o, 108o

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°

We know that sum of adjacent angles = 180Â°

So,

xo + 70o = 180o

xo = 180o â€“ 70o

= 110o

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°

We know that sum of adjacent angles = 180Â°

So,

xo + 2xo = 180o

3xo = 180o

xo = 180o/3

= 60o

So other angle is 2x = 2Ã—60 = 120o

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°

xo + 135o = 180o

xo = 180o â€“ 135o

= 45o

âˆ A = xo = 45o

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

So, âˆ A = âˆ C = 45o

And âˆ D = âˆ B = 135o

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 = 70o

Other adjacent angle âˆ B be = xo

We know that sum of adjacent angles = 180Â°

xo + 70o = 180o

xo = 180o â€“ 70o

= 110o

âˆ B = xo = 110o

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

So, âˆ A = âˆ C = 70o

And âˆ D = âˆ B = 110o

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

Solution:

Given, one of the adjacent angle âˆ A = 130o

Other adjacent angle âˆ B be = xo

We know that sum of adjacent angles = 180Â°

xo + 130o = 180o

xo = 180o â€“ 130o

= 50o

âˆ B = xo = 110o

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

So, âˆ A = âˆ C = 130o

And âˆ D = âˆ B = 50o

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 xo

We know that sum of angles = 360o

xo + xo + xo + xo = 360o

4 xo = 360o

xo = 360o/4

= 90o

âˆ´ Measure of each angle is 90o

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 (3x-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° = 180o

3xÂ° + 3xo â€“ 4 + 10 = 180Â°

6x = 180Â° – 6o

x = 174o/6

= 29o

(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 83o, 97o, 83o, 97o

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 = 30o,

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 = 20o (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°]

70o + 30o + âˆ BCA = 180Â°

âˆ BCA = 180o â€“ 100o

= 80o

âˆ DAB = âˆ DAC + âˆ CAB = 70o + 80o = 150o

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

âˆ DCA = âˆ CAB = 70o

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 = 70o

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

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 = 50o, âˆ ADC = 50o and âˆ FCD = 50o

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°

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 = 70Â° (Opposite angles of parallelogram are equal)

âˆ´ Angles of a parallelogram are 60o, 120o, 60o, 120o

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 = 90o (each right angle) â€¦â€¦. (ii)

By using angle sum property

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

âˆ COM + âˆ CMO + âˆ OCM = 180o 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.

EXERCISE 17.2 PAGES NO: 17.16

1. Which of the following stat
ements are true for a rhombus?
(i) It has two pairs of parallel sides.
(ii) It has two pairs of equal sides.
(iii) It has only two pairs of equal sides.
(iv) Two of its angles are at right angles.
(v) Its diagonals bisect each other at right angles.
(vi) Its diagonals are equal and perpendicular.
(vii) It has all its sides of equal lengths.
(viii) It is a parallelogram.
(x) It can be a square.
(xi) It is a square.

Solution:

(i) It has two pairs of parallel sides.

True, Rhombus is a parallelogram having two pairs of parallel sides.

(ii) It has two pairs of equal sides.

True, Rhombus has two pairs of equal sides .

(iii) It has only two pairs of equal sides.

False, Rhombus has all four sides equal.

(iv) Two of its angles are at right angles.

False, rhombus has no angle of right angle.

(v) Its diagonals bisect each other at right angles.

True, in rhombus diagonals bisect each other at right angles.

(vi) Its diagonals are equal and perpendicular.

False, in rhombus diagonals are of unequal length.

(vii) It has all its sides of equal lengths.

True, Rhombus has all four sides equal.

(viii) It is a parallelogram.

True, Rhombus is a parallelogram since opposite sides equal and parallel.

True, Rhombus is a quadrilateral since it has four sides.

(x) It can be a square.

True, Rhombus becomes square when any one angle is 90Â°.

(xi) It is a square.

False, Rhombus is never a square. Since in a square each angle is 90Â°.

2. Fill in the blanks, in each of the following, so as to make the statement true:
(i) A rhombus is a parallelogram in which _______.
(ii) A square is a rhombus in which _________.
(iii) A rhombus has all its sides of ______ length.
(iv) The diagonals of a rhombus _____ each other at ______ angles.
(v) If the diagonals of a parallelogram bisect each other at right angles, then it is a ______.

Solution:

(i) A rhombus is a parallelogram in which adjacent sides are equal.

(ii) A square is a rhombus in which one angle is right angle.

(iii) A rhombus has all its sides of equal length.

(iv) The diagonals of a rhombus bisect each other at right angles.

(v) If the diagonals of a parallelogram bisect each other at right angles, then it is a rhombus.

3. The diagonals of a parallelogram are not perpendicular. Is it a rhombus? Why or why not?

Solution:

No, Diagonals of a rhombus bisect each other at 90Â°. Where, diagonals must be perpendicular.

Solution:

No, because diagonals of a rhombus are perpendicular and bisect each other and its sides are equal. Below in the diagram we can see that diagonals are perpendicular to each other but they donâ€™t bisect each other.

5. ABCDÂ is a rhombus. IfÂ âˆ ACBÂ = 40Â°, findÂ âˆ ADB.

Solution:

ABCD is a rhombus. Diagonals are perpendicular.

So, âˆ BOC = 90o

From Î”BOC, the sum of angles is 180o

âˆ CBO + âˆ BOC + âˆ OBC = 180o

âˆ CBO = 180o – âˆ BOC – âˆ OBC

= 180o â€“ 40o â€“ 90o

= 50o

âˆ´ âˆ ADB = âˆ CBO = 50o (alternate angles are equal)

6. If the diagonals of a rhombus are 12 cm and 16 cm, find the length of each side.

Solution:

We know in rhombus diagonals bisect each other at right angle.

In Î”AOB

AO = 12/2 = 6cm, BO = 16/2 = 8cm

Using Pythagoras theorem in Î”AOB

AB2Â = AO2Â + BO2

AB2Â = 62Â + 82

AB2Â = 36 + 64

AB2Â = 100

AB =âˆš100 = 10cm

âˆ´ Each side of a rhombus is 10cm.

7. Construct a rhombus whose diagonals are of length 10 cm and 6 cm.

Solution:

Steps to Construct a rhombus:

(i) Draw diagonal AC of length 10 cm.

(ii) With O as centre, draw perpendicular bisector of AC to point O.

(iii) From point â€˜Oâ€™ cut two arcs of length 3cm each to get points OB and OD.

(iv) Now join AD and AB, BC and CD to get rhombus ABCD.

8. Draw a rhombus, having each side of length 3.5 cm and one of the angles as 40Â°.

Solution:

Steps to construct a rhombus:

(i) Draw a line segment AB of length 3.5 cm.

(ii) From point A and B draw angles of 40 and 140 respectively.

(iii) From points A and B cut two arcs of length 3.5 cm each to get points D and C.

(iv) Now, join AD, BC, CD to get rhombus ABCD.

9. One side of a rhombus is of length 4 cm and the length of an altitude is 3.2 cm. Draw the rhombus.

Solution:

Steps to construct a rhombus:

(i) Draw a line segment of 4 cm

(ii) From point A draw a perpendicular line and cut a length of 3.2 cm to get point E.

(iii) From point E draw a line parallel to AB.

(iv) From points A and B cut two arcs of length 4 cm on the drawn parallel line to get points D and C.

(v) Join AD, BC and CD to get rhombus ABCD.

10. Draw a rhombusÂ ABCD, ifÂ ABÂ = 6 cm andÂ ACÂ = 5 cm.

Solution:

Steps of construction:

(i) Draw a line segment AB of length 6 cm.

(ii) From point â€˜Aâ€™ cut an arc of length 5 cm and from point B cut an arc of length 6 cm. Such that both the arcs intersect at â€˜Câ€™.

(iii) Join AC and BC.

(iv) From point A cut an arc of length 6 cm and from point C cut an arc of 6cm, so that both the arcs intersect at point D.

(v) Joint AD and DC to get rhombus ABCD.

11. ABCDÂ is a rhombus and its diagonals intersect atÂ O.
(i) is Î”BOC â‰… Î”DOC? State the congruence condition used?
(ii) Also state, ifÂ âˆ BCOÂ = âˆ DCO.

Solution:

(i) Yes,

In Î”BOCÂ and Î”DOC

BO = DO [In a rhombus diagonals bisect each other]

CO = CO Common

BC = CD [All sides of a rhombus are equal]

By using SSS Congruency, Î”BOCâ‰…Î”DOC

(ii)Â Yes,

âˆ BCOÂ = âˆ DCO, by corresponding parts of congruent triangles.

12. Show that each diagonal of a rhombus bisects the angle through which it passes.

Solution:

(i) In Î”BOCÂ and Î”DOC

BO = DO [In a rhombus diagonals bisect each other]

CO = CO Common

BC = CD [All sides of a rhombus are equal]

By using SSS Congruency, Î”BOCâ‰…Î”DOC

âˆ BCOÂ = âˆ DCO, by corresponding parts of congruent triangles

âˆ´ Each diagonal of a rhombus bisect the angle through which it passes.

13. ABCDÂ is a rhombus whose diagonals intersect atÂ O. IfÂ AB=10 cm, diagonalÂ BDÂ = 16 cm, find the length of diagonalÂ AC.

Solution:

We know in rhombus diagonals bisect each other at right angle.

In Î”AOB

BO = BD/2 = 16/2 = 8cm

Using Pythagoras theorem in Î”AOB

AB2Â = AO2Â + BO2

102Â = AO2Â + 82

100-64 = AO2

AO2Â = 36

AO =Â âˆš36Â = 6cm

âˆ´ Length of diagonal AC is 6 Ã— 2 = 12cm.

14. The diagonal of a quadrilateral are of lengths 6 cm and 8 cm. If the diagonals bisect each other at right angles, what is the length of each side of the quadrilateral?

Solution:

We know in rhombus diagonals bisect each other at right angle.

In Î”AOB

BO = BD/2 = 6/2 = 3cm

AO = AC/2 = 8/2 = 4cm

Using Pythagoras theorem in Î”AOB

AB2Â = AO2Â + BO2

AB2Â = 42Â + 32

AB2Â = 16 + 9

AB2Â = 25

AB =Â âˆš25Â = 5cm

âˆ´ Length of each side of a Quadrilateral ABCD is 5cm.

EXERCISE 17.3 PAGES NO: 17.22

1. Which of the following statem
ents are true for a rectangle?
(i) It has two pairs of equal sides.
(ii) It has all its sides of equal length.
(iii) Its diagonals are equal.
(iv) Its diagonals bisect each other.
(v) Its diagonals are perpendicular.
(vi) Its diagonals are perpendicular and bisect each other.
(vii) Its diagonals are equal and bisect each other.
(viii) Its diagonals are equal and perpendicular, and bisect each other.
(ix) All rectangles are squares.
(x) All rhombuses are parallelograms.
(xi) All squares are rhombuses and also rectangles.
(xii) All squares are not parallelograms.

Solution:

(i) It has two pairs of equal sides.

True, in a rectangle two pairs of sides are equal.

(ii) It has all its sides of equal length.

False, in a rectangle only two pairs of sides are equal.

(iii) Its diagonals are equal.

True, in a rectangle diagonals are of equal length.

(iv) Its diagonals bisect each other.

True, in a rectangle diagonals bisect each other.

(v) Its diagonals are perpendicular.

False, Diagonals of a rectangle need not be perpendicular.

(vi) Its diagonals are perpendicular and bisect each other.

False, Diagonals of a rectangle need not be perpendicular. Diagonals only bisect each other.

(vii) Its diagonals are equal and bisect each other.

True, Diagonals are of equal length and bisect each other.

(viii) Its diagonals are equal and perpendicular, and bisect each other.

False, Diagonals are of equal length and bisect each other. Diagonals of a rectangle need not be perpendicular

(ix) All rectangles are squares.

False, in a square all sides are of equal length.

(x) All rhombuses are parallelograms.

True, all rhombuses are parallelograms, since opposite sides are equal and parallel.

(xi) All squares are rhombuses and also rectangles.

True, all squares are rhombuses, since all sides are equal in a square and rhombus. All squares are rectangles, since opposite sides are equal and parallel.

(xii) All squares are not parallelograms.

False, all squares are parallelograms, since opposite sides are parallel and equal.

2. Which of the following sta
tements are true for a square?
(i) It is a rectangle.
(ii) It has all its sides of equal length.
(iii) Its diagonals bisect each other at right angle.
(v) Its diagonals are equal to its sides.

Solution:

(i) It is a rectangle.

True. Since, opposite sides are equal and parallel where, each angle is right angle.

(ii) It has all its sides of equal length.

True. Since, ides of a square are of equal length.

(iii) Its diagonals bisect each other at right angle.

True. Since, diagonals of a square bisect each other at right angle.

(v) Its diagonals are equal to its sides.

False. Since, diagonals of a square are of equal length. Length of diagonals is not equal to the length of sides

3. Fill in the blanks in each of the following, so as to make the statement true :
(i) A rectangle is a parallelogram in which ________.
(ii) A square is a rhombus in which __________.
(iii) A square is a rectangle in which ___________.

Solution:

(i) A rectangle is a parallelogram in which one angle is a right angle.

(ii) A square is a rhombus in which one angle is a right angle.

(iii) A square is a rectangle in which adjacent sides are equal.

4. A window frame has one diagonal longer than the other. Is the window frame a rectangle? Why or why not?

Solution:

No, diagonals of a rectangle are equal length.

5. In a rectangleÂ ABCD, prove that Î”ACBÂ â‰…Î”CAD.

Solution:

Let us draw a rectangle,

In rectangle ABCD, AC is the diagonal.

AB = CD [Opposite sides of a rectangle are equal]

BC = DA

AC = CA [Common]

By using SSS congruency

6. The sides of a rectangle are in the ratio 2 : 3, and its perimeter is 20 cm. Draw the rectangle.

Solution:

In rectangle ABCD,

Given, perimeter of a rectangle = 20cm

Ratio = 2:3

So, let us consider the side as â€˜xâ€™

Length of rectangle (l) = 3x

Breadth of the rectangle (b) = 2x

We know that,

Perimeter of the rectangle = 2(length + breadth)

20 = 2(3x + 2x)

10x = 20

x = 20/10 = 2

Length of the rectangle = 3Ã—2 = 6cm

Breadth of the rectangle = 2Ã—2 = 4cm

Here, is the diagram of rectangle

7. The sides of a rectangle are in the ratio 4 : 5. Find its sides if the perimeter is 90 cm.

Solution:

In rectangle ABCD,

Given, perimeter of a rectangle = 90cm

Ratio = 4:5

So, let us consider the side as â€˜xâ€™

Length of rectangle (l) = 5x

Breadth of the rectangle (b) = 4x

We know that,

Perimeter of the rectangle = 2(length + breadth)

90 = 2(5x + 4x)

18x = 90

x = 90/18 = 5

Length of the rectangle = 5Ã—5 = 25cm

Breadth of the rectangle = 4Ã—5 = 20cm

Here, is the diagram of rectangle

8. Find the length of the diagonal of a rectangle whose sides are 12 cm and 5 cm.

Solution:

In rectangle ABCD,

Given, sides of a rectangle ABCD are 5cm and 12cm

In Î”ABC using Pythagoras theorem,

AC2Â = AB2Â + BC2

AC2Â = 122Â + 52

AC2Â = 144 + 25

AC2Â = 169

AC =Â âˆš169

AC = 13cm

âˆ´ Length of the diagonal AC is 13cm.

9. Draw a rectangle whose one side measures 8 cm and the length of each of whose diagonals is 10 cm.

Solution:

Given, one side of the rectangle is 8cm.

Length of the diagonal = 10cm

Now let us construct a rectangle,

Steps to construct a rectangle,

(i) Draw a line segment AB of length 8 cm

(ii) From point â€˜Aâ€™ cut an arc of length 10 cm and mark that point as C.

(iii) From point B draw an angle of 90Â°, and join the arc from point A which cuts at point C.

(iv) now join AC and BC

(v) From point A draw an angle of 90Â° and from point C cut an arc of length 8 cm to get point D.

(vi) Join CD and AD to form required rectangle.

Here, is the constructed diagram of rectangle

10. Draw a square whoâ€™s each side measures 4.8 cm.

Solution:

Given, side of a square is 4.8cm.

Now let us construct a square,

Steps to construct a square,

(i) Draw a line segment AB of length 4.8 cm.

(ii) From points A and B draw perpendiculars at 90o each.

(iii) Cut and arc of 4.8 cm from point A and B on the perpendiculars to get point D and C.

(iv) Join DC, AD and BC to form the required square.

Here, is the constructed diagram of square

11. Identify all the quadrilaterals that have:

(i) Four sides of equal length

(ii) Four right angles

Solution:

(i) Four sides of equal length

The quadrilaterals which has all the four sides of equal length are Square and Rhombus.

(ii) Four right angles

The quadrilaterals which has four right angles are Square and Rectangle.

12. Explain how a square is
(ii) A parallelogram?
(iii) A rhombus?
(iv) A rectangle?

Solution:

(i) A square is a quadrilateral since it has all sides of equal length.

(ii) A square is a parallelogram since itâ€™s opposite sides are equal and parallel.

(iii) A square is a rhombus since it has all sides of equal length and opposite sides are parallel.

(iv) A square is a rectangle since itâ€™s opposite sides are equal and each angle is a 90o.

13. Name the quadrilaterals whose diagonals:
(i) bisect each other
(ii) are perpendicular bisector of each other
(iii) are equal.

Solution:

(i) Quadrilaterals whose diagonals bisect each other are: Parallelogram,

Rectangle, Rhombus and Square.

(ii) Quadrilaterals whose diagonals are perpendicular bisector of each other are: Rhombus and Square.

(iii) Quadrilaterals whose diagonals are equal in Square and Rectangle.

14. ABCÂ is a right angled triangle andÂ OÂ is the mid-point of the side opposite to the right angle. Explain whyÂ OÂ is equidistant fromÂ A,Â B, andÂ C.

Solution:

ABC is a right angled triangle. O is the midpoint of hypotenuse AC, such that OA = OC

Now, draw CD||AB and join AD, such that AB = CD and AD = BC.

Now, quadrilateral ABCD is a rectangle, since each angle is a right angle and opposite sides are equal and parallel.

We know in a rectangle diagonals are of equal length and they bisect each other.

So, AC = BD

And also, AO = OC = BO = OD

Hence, O is equidistant from A, B and C.

15. A mason has made a concrete slab. He needs it to be rectangular. In what different ways can he make sure that it is rectangular?

Solution:

For a concrete slab to be rectangular the mason has to check,

(i) By measuring each angle and

(ii) By measuring the lengths of diagonals.

#### 1 Comment

1. Ritesh raj

thank u it helps me every time