## RD Sharma Solutions Class 8 Chapter 17 Exercise 17.1

### Exercise 17.1

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

(i) AD =

(ii) \(\angle\)DCB =

(iii) OC =

(iv) \(\angle\)DAB + \(\angle\)CDA =

SOLUTION:

Let us Change the coordinates as follow to make figure correct:

(i) As opposite sides of a parallelogram are equal, So AD = BC

(ii) As opposite angles of a parallelogram are equal, So \(\angle\)DCB = \(\angle\)BAD

(iii) As diagonals of a parallelogram bisect each other, So OC = OA

(iv) The sum of two adjacent angles of a parallelogram is \(180^{\circ}\) ).

So \(\angle\)DAB +\(\angle\)CDA = \(180^{\circ}\)

Question 2. The following figures are parallelograms. Find the degree values of the unknowns x, y and z.

SOLUTION:

Keep all basic properties of parallelogram in mind before you start attempting this question.

(i) using property: Opposite angles of a parallelogram are same.

So, x = z and y = 100°

and y + z = 180° (As we know, sum of adjacent angles of quadrilaterals = 180 degrees)

Now,

z + 100° = 180°

x = 180° – 100°

=> x = 80°

From above, value for all the angles is: x = 80°, y = 100° and z = 80°

(ii) using property: Opposite angles of a parallelogram are same.

So, x = y and \(\angle\)ROP = 100°

\(\angle\)PSR + \(\angle\)SRQ = 180°

y + 50° = 180° and

x = 180° – 50°

This implies, x = 130°

So, x=130°, y=130°

here y and z are alternate angles, so, z = 130°.

(iii) using property: Sum of all angles in a triangle = 180°

So, 30° + 90° + z = 180°

=> z = 60°

Again, in a parallelogram, Opposite angles are equal

So, y = z = 60°and x=30° (alternate angles)

(iv) x = 90° (vertically opposite angles)

Sum of all angles in a triangle is 180°.

So, y + 90° + 30° = 180°

=> y=180°- (90°+30°)

=> y = 60°

y= z = 60° (alternate angles)

(v)Opposite angles are equal in a parallelogram.

So, y = 80°

y + x = 180°

=> x = 180° – 100° = 80°

z = y = 80° (alternate angles)

(vi) y = 112° (we know that, opposite angles are equal )

In triangleUTW :

x + y + 40° = 180° (using angle sum property of a triangle)

x = 180° – (112° – 40°) = 28°

Again,

Bottom left vertex = 180° – 112° = 68°

So, z = x = 28° (alternate angles)

Q 3. Can the following figures be parallelograms? Justify your answers.

SOLUTION:

(i) No.

Reason: Opposite angles of a parallelogram are not equal.

(ii) Yes.

Reason: The opposite sides are equal.

(iii) No.

Reason: The diagonals do not bisect each other.

Q 4. In the adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the geometrical truths you use to find them.

SOLUTION:

Using linear pair property, we have

\(\angle\)HOP + 70° = 180°

\(\angle\)HOP = 180° – 70° =110°

As opposite angles of a parallelogram are equal, So

x = \(\angle\)HOP = 110°

Again, sum of adjacent angles of a parallelogram is 180°

\(\angle\)EHP + \(\angle\)HEP = 180°

Now, let’s find the values of y and z.

110° + 40° + z = 180°

z = 180° – 150° = 30°

y = 40° (as this is an alternate angles)

Q 5. In the following figures GUNS and RUNS are parallelograms. Find x and y.

SOLUTION:

(i) Since opposite sides of a parallelogram are equal

So, 3y – 1 = 26

=> 3y = 27

y = 9.

Similarly, 3x = 18

x = 6.

(ii) Diagonals of a parallelogram bisect each other

So, y – 7 = 20

y = 27

x – y = 16

x – 27 = 16

x = 43

Q 6. In the following figure RISK and CLUE are parallelograms. Find the measure of x.

SOLUTION:

In the parallelogram CLUE:

\(\angle\)CEU = \(\angle\)CLU = 70° (opposite angles of a parallelogram are equal)

Similarly, in parallelogram RISK:

\(\angle\)ISK + \(\angle\)RKS = 180° (sum of adjacent angles of a parallelogram is 180°)

\(\angle\)ISK = 180° – 120° = 60°

Now, Find the value of x.

From diagram:

x + \(\angle\)ISK + \(\angle\)CEU = 180°

x = 180° – 70° + 60° = 50°.

The value of x is 50 degrees.

Q 7. Two opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of each angle of the parallelogram.

SOLUTION:

As we know, opposite angles of a parallelogram are congruent.

So, 3x – 2° = 50 – x°

3x° – 2° = 50° – x°

4x° = 52°

x° = 13°

Put the measure of x in 3x° – 2°

3x° – 2° = 39° – 2° = 37°

And second angle is as:

50° – x° = 37°

(Opposite angles are congruent)

We are known with two angles.

Let us find the remaining two angles, say y and z.

Angles y and z are congruent because they are also opposite angles.

So, y = z

Sum of adjacent angles of a parallelogram = 180°

So, 37° + y = 180°

y = 180° – 37° = 143°

So, all angles are: 37°, 37°, 143° and 143°. Answer!

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

SOLUTION:

Sum of Adjacent two angles of a parallelogram = 180 degrees

Let say, one angle is x, then second angle be 2 /3 x.

x + \(\frac{2x}{3}\) = 180°

\(\frac{5x}{3}\) = 180°

x = 72° and

\(\frac{2x}{3}\) = \(\frac{2(72°)}{3}\) = 108°

Therefore, two of the angles are 108° and the other two are 72°.

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

SOLUTION:

Given : One angle of a parallelogram is 70°

Since opposite angles have the same value.

Step 1: One angle is 70°, then the one directly opposite will also be 70°

Step 2: Find remaining angles

Let us say, one angle be x°, so

x° + 70° = 180°

{using property: sum of adjacent angles of a parallelogram is 180°}

x° = 180° – 70°

x° = 110°

Hence, remaining angles are 70°, 110° and 110°.

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

SOLUTION:

Consider A and B are two angles. And the ratio is given as 1 : 2.

Measures of \(\angle\)A and \(\angle\)B are x° and 2x°.

As we know that the sum of adjacent angles of a parallelogram is 180°, we have

\(\angle\)A + \(\angle\)B = 180°

=>x° + 2x° = 180°

=> x° = 60°

Therefore, measure of \(\angle\)A = 60°, \(\angle\)B = 120°, \(\angle\)C = 60° and \(\angle\)D = 120°.

Q 11. In a parallelogram ABCD, \(\angle\)D=135°, determine the measure of \(\angle\)A and \(\angle\)B.

SOLUTION:

As we know, opposite angles of a parallelogram have the same value.

So, \(\angle\)D = \(\angle\)B = 135°

Again, \(\angle\)A + \(\angle\)B +\(\angle\)C + \(\angle\)D = 360° and \(\angle\)A + \(\angle\)D = 180°

\(\angle\)A = 180° – 135° = 45°

Q 12. ABCD is a parallelogram in which \(\angle\)A = 70°. Compute \(\angle\)B, \(\angle\)C and \(\angle\)D.

SOLUTION:

Opposite angles are equal.

So, \(\angle\)C = 70° = \(\angle\)A

\(\angle\)B = \(\angle\)D

Using property that, the sum of the adjacent angles of a parallelogram = 180 degrees

\(\angle\)A + \(\angle\)B = 180°

70° + \(\angle\)B = 180°

\(\angle\)B = 110°

\(\angle\)C = 70°

\(\angle\)D = 110°

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

SOLUTION:

Consider angles be A, B, C and D.

Sum of two opposite angles = 130°

So, \(\angle\)A + \(\angle\)C = 130°

\(\angle\)A + \(\angle\)A = 130° (using property : opposite angles are equal)

\(\angle\)A = 65°and \(\angle\)C = 65°

The sum of adjacent angles = 180°, we have

\(\angle\)A + \(\angle\)B =180°

65° + \(\angle\)B = 180°

\(\angle\)B = 180° – 65° = 115°

So, \(\angle\)A = 65°, \(\angle\)B = 115°, \(\angle\)C = 65° and \(\angle\)D = 115°.Answer!

Q 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’s say required angle is x.

All the angles are equal, then they sum up to 360 degrees.

x + x + x + x = 360°

4x = 360°

Or x = 90°

So, each angle is 90° and quadrilateral is a parallelogram.

It is a rectangle, which is a special type of parallelogram.

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

SOLUTION:

As opposite sides of a parallelogram are equal.

4 cm and 3 cm be the two given sides.

Remaining sides will also be 4 cm and 3 cm.

Again,

Perimeter = Sum of all the sides of a parallelogram

= 4 + 3 + 4 + 3 = 14

Perimeter is 14 cm.

Q 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:

Consider, 2 of the sides of the parallelogram be x and y.

As per statement, x = y + 25

As, Opposite sides of a parallelogram are same.

Perimeter = x + y + x + y = 150

y + 25 + y + y + 25 + y = 150

4y = 100

Or y = 25

Therefore, x = y + 25 = 25 +25 = 50

Answer: 50 cm and 25 cm are the lengths of the sides of the parallelogram.

Q 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:

From the statement, we are given

Shorter side = 4.8 cm

Longer side = \(\frac{4.8}{2}\) + 4.8 = 7.2 cm

Perimeter of parallelogram = 4.8 + 4.8 + 7.2 + 7.2 = 24

Therefore, perimeter is 24 cm

Q 18. Two adjacent angles of a parallelogram are (3x – 4)° and (3x + 10)°. Find the angles of the parallelogram.

SOLUTION:

Given that , two adjacent angles of a parallelogram are 3x – 4° and 3x + 10°

{adjacent angles of a parallelogram are supplementary i.e. sum of angles is 180 degrees}

3x + 10° + 3x – 4° = 180°

6x° = 174°

x = 29°

1st angle measure = 3x+10° = 3(29°) + 10° = 97°

2nd angle measure = 3x – 4° = 83°

Angles of the parallelogram are 97°, 83°, 97° and 83°. Answer!

Q 19. In a parallelogram ABCD, the diagonals bisect each other at O. If \(\angle\)ABC = 30°, \(\angle\)BDC =10° and \(\angle\)CAB =70°. Find:

\(\angle\)DAB, \(\angle\)ADC, \(\angle\)BCD, \(\angle\)AOD, \(\angle\)DOC, \(\angle\)BOC, \(\angle\)AOB, \(\angle\)ACD, \(\angle\)CAB, \(\angle\)ADB, \(\angle\)ACB, \(\angle\)DBC and \(\angle\)DBA.

SOLUTION:

Given: \(\angle\)ABC=30°

\(\angle\)ADC = 30° and \(\angle\)BDA = \(\angle\)ADC – \(\angle\)BDC = 30°-10° = 20°

Since opposite angles and alternate angles of the parallelogram are same.

\(\angle\)BAC = \(\angle\)ACD = 70° (alternate angle)

From triangle ABC:

Sum of all the angles = 180 degrees

\(\angle\)CAB + \(\angle\)ABC + \(\angle\)BCA = 180°

70° + 30° + \(\angle\)BCA = 180°

or \(\angle\)BCA = 80°

Again,

\(\angle\)DAB = \(\angle\)DAC + \(\angle\)CAB = 70° + 80° = 150°

Since opposite angles of the parallelogram

So, \(\angle\)BCD = 150°

\(\angle\)DCA = \(\angle\)CAB = 70°

From triangle DOC:

Sum of all the angles = 180°

10° + 70° + \(\angle\)DOC = 180°

or \(\angle\)DOC=100°

From figure, use linear angle property, we get

\(\angle\)DOC + \(\angle\)BOC = 180°

\(\angle\)BOC = 180° – 100° = 80°

\(\angle\)AOB = \(\angle\)DOC = 100°( vertically opposite angles)

\(\angle\)AOD = \(\angle\)BOC = 80° (vertically opposite angles)

From figure, \(\angle\)CAB = 70°

Given \(\angle\)ADB = 20°

\(\angle\)DBC = \(\angle\)ADB = 20° (alternate angle)

\(\angle\)BDC = \(\angle\)DBA = 10° (alternate angles)

Q 20. Find the angles marked with a question mark shown in Fig. 17.27.

SOLUTION:

From a triangle FDC:

\(\angle\)FDC + \(\angle\)DCF + \(\angle\)DCF = 180°

50° + 90° + \(\angle\)DCF = 180°

\(\angle\)DCF = 40°

From triangle CEB:

\(\angle\)ECB + \(\angle\)CBE + \(\angle\)BEC = 180°

40° + 90° + \(\angle\)EBC = 180°

\(\angle\)EBC = 50°

And, \(\angle\)EBC = \(\angle\)ADC = 50°( opposite angles of a parallelogram)

Again,

\(\angle\)BCE + \(\angle\)ECF + \(\angle\)FCD + \(\angle\)FDC = 180°

{Sum of alternate angles of a parallelogram = 180° }

50° + 40° + \(\angle\)ECF + 40°=180°

\(\angle\)ECF = 180° – 50° + 40° – 40° = 50°

\(\angle\)ECF = 50 degrees. Answer!

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

SOLUTION:

Construction: Draw a parallelogram ABCD.

Drop a perpendicular from B to the side AD, at the point E.

Drop a perpendicular from B to the side CD, at the point F.

The figure looks like:

In the quadrilateral BEDF: \(\angle\)EBF = 60°, \(\angle\)BED = 90°, \(\angle\)BFD=90°

\(\angle\)EDF = 360° – (60° + 90° + 90°) = 120°

Since, opposite angles are congruent and adjacent angles are supplementary, we have

From figure ABCD:

\(\angle\)A = \(\angle\)C = 180° – 120° = 60°

\(\angle\)B = \(\angle\)D =120°

Q 22. In Fig. 17.28, ABCD and AEFG are parallelograms. If \(\angle\)C = 55°, what is the measure of \(\angle\)F?

SOLUTION:

Since ABCD and AEFG, both the parallelograms are similar.

So, \(\angle\)C = \(\angle\) A = 55°

{Using property, opposite angles of a parallelogram are equal}

So, \(\angle\)A = \(\angle\)F = 55°

{Using property, opposite angles of a parallelogram are equal}

Q 23. In Fig. 17.29, BDEF and DCEF are each a parallelogram. Is it true that BD = DC? Why or why not?

SOLUTION:

Given: BDEF and DCEF are parallelograms.

Choose parallelogram BDEF

So, BD = EF ………..(1) (opposite sides are equal)

From parallelogram DCEF

CD = EF ……………………(2) (opposite sides are equal)

From equation (1) and equation (2)

BD = CD

Q 24. In Fig. 17.29, suppose it is known that DE = DF. Then, is triangle ABC isosceles? Why or why not? (figure mentioned in the above question)

SOLUTION:

From triangle FDE:

DE = DF

Angles opposite to equal sides: \(\angle\)FED = \(\angle\)DFE ………. (1)

Opposite angles of a parallelogram are equal) : In figure BDEF: \(\angle\)FBD = \(\angle\)FED …….. (2)

In figure DCEF:

Opposite angles of a parallelogram are equal: \(\angle\)DCE = \(\angle\)DFE ……….. (3)

From equation (1), equation (2) and equation (3):

\(\angle\)FBD = \(\angle\)DCE

Again,

From triangle ABC:

If \(\angle\)DCE = \(\angle\)FBD, then AC = AB

{Sides opposite to the equal angles}

Hence, triangle ABC is an isosceles triangle.

Q 25. Diagonals of parallelogram ABCD intersect at O as shown in Fig. 17.30. XY contain, O, and X, Y are points on opposite sides of the parallelogram. Give reasons for each of the following:

(i) OB =OD

(ii) \(\angle\)OBY = \(\angle\)ODX

(iii) \(\angle\)BOY = \(\angle\)DOX

(iv) \(\Delta BOY\cong \Delta DOX\)

Now, state if XY is bisected at O.

SOLUTION:

(i) Diagonals bisect each other.

(ii) These are Alternate angles.

(iii) These are vertically opposite angles.

(iv) Form \(\Delta\)BOY and \(\Delta\)DOX

OD = OB (diagonals bisect each other)

\(\angle\)BOY = \(\angle\)DOX (vertically opposite angles)

\(\angle\)OBY = \(\angle\)ODX (alternate angles)

Triangle Congruence ASA:

YO = XO (c.p.c.t)

So, XY is bisected at O.

Q 26. In fig. 17.31, ABCD is a parallelogram, CE bisects \(\angle\)C and AF bisects \(\angle\)A. In each of the following, if the statement is true, give a reason for the same:

(i) \(\angle\)A = \(\angle\)C

(ii) \(\angle\)FAB = 1/2\(\angle\)A

(iii) \(\angle\)DCE = 1/2\(\angle\)C

(iv) \(\angle\)CEB = \(\angle\)FAB

(v) CE||AF

SOLUTION:

(i) True.

Reason: Opposite angles of a parallelogram are equal.

(ii) True.

Reason: AF is the bisector of angle A.

(iii) True.

Reason: CE is the bisector of angle C.

(iv) True.

Reason:

Alternate angles: \(\angle\)CEB = \(\angle\)DCE ………….(1)

Opposite angles are equal: \(\angle\)DCE = \(\angle\)FAB ………….(2)

From equation (1) and equation (2):

\(\angle\)FAB = \(\angle\)CEB

(v) True.

Reason:

Corresponding angles are equal

Q 27. Diagonals of a parallelogram ABCD intersect at O. AL and CM are drawn perpendiculars to BD such that L and M lie on BD. Is AL = CM? Why or why not?

SOLUTION:

In \(\Delta\)CMO and \(\Delta\)AOL :

Vertically opposite angles : \(\angle\)AOL = \(\angle\)COM …….(1)

Right angles : \(\angle\) ALO = \(\angle\)CM0 = 90° ……….. (2)

Using angle sum property, we have

\(\angle\)AOL + \(\angle\)LAO + \(\angle\)ALO = 180° …………. (3)

\(\angle\)COM + \(\angle\)OCM + \(\angle\)CMO = 180°………. (4)

From equations (3) and (4):

\(\angle\)AOL + \(\angle\)LAO + \(\angle\)ALO = \(\angle\)COM + \(\angle\)OCM + \(\angle\)CMO

from equation (1) and (2)

\(\angle\)LAO = \(\angle\)OCM

In \(\Delta\)CMO and \(\Delta\)AOL:

right angles : \(\angle\)ALO = \(\angle\)CMO

Diagonals bisect each other : AO=OC

\(\angle\)LAO = \(\angle\)OCM

Now by SAS congruence:

\(\Delta\)AOL is congruent to \(\Delta\)CMO.

AL = CM [cpct]

Q 28. Points E and F lie on diagonal AC of a parallelogram ABCD such that AE = CF. What type of quadrilateral is BFDE ?

SOLUTION:

From ABCD parallelogram:

Diagonals bisect each other

AO = OC…………. (i)

And AE = CF……….(ii)

(i) – (ii), we get

AO – AE = OC – CF

EO = OF…………(iii)

In \(\Delta\)BOF and \(\Delta\)DOE:

EO = OF

diagonals of a parallelogram bisect each other, so DO = OB

\(\angle\)BOF = \(\angle\)DOE (vertically opposite angles)

By SAS congruence, we get

\(\Delta\)DOE \(\cong\) \(\Delta\)BOF

So, DE = BF (c.p.c.t)

Again in \(\Delta\)DOF and \(\Delta\)BOE:

EO = OF

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

\(\angle\)DOF = \(\angle\)BOE (vertically opposite)

By SAS congruence, we have

\(\Delta\)DOE \(\cong\) \(\Delta\)BOF

So, DF = BE (c.p.c.t)

Which shows that, the pair of opposite sides are equal.

Therefore, BFDE is a parallelogram.

Q 29. In a parallelogram ABCD, AB =10 cm, AD = 6 cm. The bisector of \(\angle\)A meets DC in E, AE and BC produced meet at F. Find the length CF.

SOLUTION:

Given: AE is the bisector of \(\angle\)DAE = \(\angle\)BAE = x

\(\angle\)BAE = \(\angle\)AED = x (because alternate angles are equal)

Triangle ADE is an isosceles triangle. (As opposite angles in triangle ADE are equal)

From the above figure:

AD = DE = 6 cm (sides opposite to equal angles)

AB = CD = 10 cm

CD = DE + EC

EC = CD –DE

EC = 10 – 6 = 4 cm

Again,

Vertically opposite angles: \(\angle\)DEA = \(\angle\)CEF = x

Alternate angles: \(\angle\)EAD = \(\angle\)EEFC = x

Since opposite angle in triangle EFC are equal, Triangle EFC is an isosceles triangle.

So, CF = CE = 4 cm

Therefore, the length of CF is 4cm.

Good!!!,😁😀😉