# RD Sharma Solutions Class 8 Understanding Shapes Special Types Quadrilaterals Exercise 17.1

## 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.

(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)

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.

1. Arfa Ibrar

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