# Class 8 RD Sharma Solutions – Chapter 17 Understanding Shapes Special Types Of Quadrilaterals – Exercise 17.1 | Set 2

### Chapter 17 Understanding Shapes Special Types Of Quadrilaterals – Exercise 17.1 | Set 1

**Question 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 that,

Perimeter of the parallelogram = 150 cm

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

and other side as = (x + 25) cm

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

Therefore, Perimeter = Sum of all sides

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

4x + 50 = 150

4x = 150 â€“ 50

x = 100/4 = 25

Hence, Sides of the parallelogram are (x) = 25 cm and (x+25) = 50 cm.

**Question 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 that,

Shorter side of the parallelogram = 4.8 cm

and longer side of the parallelogram = 4.8 + 4.8/2 = 4.8 + 2.4 = 7.2cm

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

Therefore, Perimeter = Sum of all sides

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

Hence, Perimeter of the parallelogram is 24 cm.

**Question 18. Two adjacent angles of a parallelogram are (3x-4)o and (3x+10)Â°. Find the angles of the parallelogram.**

**Solution: **

As we know that adjacent angles of a parallelogram are equal.

Therefore, (3x â€“ 4)Â° + (3x + 10)Â° = 180Â°

3xÂ° + 3xo â€“ 4 + 10 = 180Â°

6x = 180Â° â€“ 6Â°

x = 174Â°/6 = 29Â°

The adjacent angles are,

(3x â€“ 4)Â° = 3Ã—29 â€“ 4 = 83Â°

(3x + 10)Â° = 3Ã—29 + 10 = 97Â°

As we know that Sum of adjacent angles = 180Â°

Hence, each angle is 83Â°, 97Â°, 83Â°, 97Â°.

**Question 19. In a parallelogram ABCD, the diagonals bisect each other at O. If âˆ ABC =30Â°, âˆ BDC= 10Â° and âˆ CAB =70Â°. Find:**

**âˆ DAB, âˆ ADC, âˆ BCD, âˆ AOD, âˆ DOC, âˆ BOC, âˆ AOB, âˆ ACD, âˆ CAB, âˆ ADB, âˆ ACB, âˆ DBC, and âˆ DBA.**

**Solution:**

Given that,

âˆ ABC = 30Â°,

âˆ ABC = âˆ ADC = 30Â° (We know that measure of opposite angles are equal in a parallelogram)

âˆ BDC = 10Â°

âˆ CAB =70Â°

âˆ BDA = âˆ ADB = âˆ ADC â€“ âˆ BDC = 30Â° â€“ 10Â° = 20Â° (From figure we concluded)

âˆ DAB = 180Â° â€“ 30Â° = 150Â°

âˆ ADB = âˆ DBC = 20Â° (alternate angles)

âˆ BCD = âˆ DAB = 150Â° (we know, opposite angles are equal in a parallelogram)

âˆ DBA = âˆ BDC = 10Â° (we know, Alternate interior angles are equal)

In Î”ABC âˆ CAB + âˆ ABC + âˆ BCA = 180Â° (since, sum of all angles of a triangle is 180Â°)

70Â° + 30Â° + âˆ BCA = 180Â°

âˆ BCA = 180Â° â€“ 100Â° = 80Â°

âˆ DAB = âˆ DAC + âˆ CAB = 70Â° + 80Â° = 150Â°

âˆ BCD = 150Â° (opposite angle of the parallelogram)

âˆ DCA = âˆ CAB = 70Â°

In Î”DOC âˆ BDC + âˆ ACD + âˆ DOC = 180Â° (since, sum of all angles of a triangle is 180Â°)

10Â° + 70Â° + âˆ DOC = 180Â°

âˆ DOC = 180Â°- 80Â°

âˆ DOC = 100Â°

Therefore, âˆ DOC = âˆ AOB = 100Â° (Vertically opposite angles are equal)

âˆ DOC + âˆ AOD = 180Â° [Linear pair]

100Â° + âˆ AOD = 180Â°

âˆ AOD = 180Â°- 100Â°

âˆ AOD = 80Â°

Therefore, âˆ AOD = âˆ BOC = 80Â° (Vertically opposite angles are equal)

âˆ CAB = 70o

âˆ ABC + âˆ BCD = 180Â° (In a parallelogram sum of adjacent angles is 180Â°)

30Â° + âˆ ACB + âˆ ACD = 180Â°

30Â° + âˆ ACB + 70Â° = 180Â°

âˆ ACB = 180Â° â€“ 100Â°

âˆ ACB = 80Â°

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

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

**Solution:**

In Î”BEC âˆ BEC + âˆ ECB +âˆ CBE = 180Â° (Sum of angles of a triangle is 180Â°)

90Â° + 40Â° + âˆ CBE = 180Â°

âˆ CBE = 180Â°-130Â°

âˆ CBE = 50Â°

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

âˆ B = âˆ D = 50Â° (opposite angles of a parallelogram are equal)

âˆ A + âˆ B = 180Â° (Sum of adjacent angles of a triangle is 180Â°)

âˆ A + 50Â° = 180Â°

âˆ A = 180Â°-50Â°

therefore, âˆ A = 130Â°

In Î”DFC âˆ DFC + âˆ FCD +âˆ CDF = 180Â° (Sum of angles of a triangle is 180Â°)

90Â° + âˆ FCD + 50Â° = 180Â°

âˆ FCD = 180Â°-140Â°

âˆ FCD = 40Â°

âˆ A = âˆ C = 130Â° (Opposite angles of a parallelogram are equal)

âˆ C = âˆ FCE +âˆ BCE + âˆ FCD

âˆ FCD + 40Â° + 40Â° = 130Â°

âˆ FCD = 130Â° â€“ 80Â°

âˆ FCD = 50Â°

Hence âˆ EBC = 50Â°, âˆ ADC = 50Â° and âˆ FCD = 50Â°.

**Question 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 parallelogram ABCD, where DPâŠ¥ AB and DQ âŠ¥ BC.

Given that âˆ PDQ = 60Â°

In quadrilateral DPBQ âˆ PDQ + âˆ DPB + âˆ B + âˆ BQD = 360Â° (Sum of all the angles of a Quadrilateral is 360Â°)

60Â° + 90Â° + âˆ B + 90Â° = 360Â°

âˆ B = 360Â° â€“ 240Â°

âˆ B = 120Â°

âˆ B = âˆ D = 120Â° (Opposite angles of parallelogram are equal)

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

120Â° + âˆ C = 180Â°

âˆ C = 180Â° â€“ 120Â° = 60Â°

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

Hence, Angles of a parallelogram are 60Â°, 120Â°, 60Â°, 120Â°

**Question 22. In Figure, ABCD and AEFG are parallelograms. If âˆ C =55Â°, what is the measure of âˆ F?**

**Solution: **

From figure, we conclude that,

In parallelogram ABCD âˆ C = âˆ A = 55Â° (In a parallelogram opposite angles are equal in a parallelogram)

In parallelogram AEFG âˆ A = âˆ F = 55Â° (In a parallelogram opposite angles are equal in a parallelogram)

Hence âˆ F = 55Â°

**Question 23. In Figure, BDEF and DCEF are each a parallelogram. Is it true that BD = DC? Why or why not?**

**Solution: **

From figure, we conclude that,

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

Therefore,

BD = DC

**Question 24. In Figure, suppose it is known that DE = DF. Then, is **Î”**ABC isosceles? Why or why not?**

**Solution: **

From figure we conclude that,

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)

therefore, DE = AF = BF

similarly, DF = CE = AE

Given that, DE = DF

Since, DF = DF

AF + BF = CE + AE

AB = AC

Hence, Î”ABC is an isosceles triangle.

**Question 25. Diagonals of parallelogram ABCD intersect at O as shown in Figure. XY contains O, and X, Y 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) OB = ODOB = OD. Since diagonals bisect each other in a parallelogram.

(ii) âˆ OBY =âˆ ODXâˆ OBY =âˆ ODX. Since alternate interior angles are equal in a parallelogram.

(iii) âˆ BOY= âˆ DOXâˆ BOY= âˆ DOX. Since vertical opposite angles are equal in a parallelogram.

(iv) Î”BOY â‰… Î”DOXÎ”BOY and Î”DOX. Since OB = OD, where diagonals bisect each other in a parallelogram.

âˆ OBY =âˆ ODX (Alternate interior angles are equal)

âˆ BOY= âˆ DOX (Vertically opposite angles are equal)

Î”BOY â‰…Î”DOX (by ASA congruence rule)

OX = OY (Corresponding parts of congruent triangles)

Hence XY is bisected at O.

**Question 26. In Fig., ABCD is a parallelogram, CE bisects âˆ C and AF bisects 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) âˆ A = âˆ CTrue, Since âˆ A =âˆ C = 55Â° [opposite angles are equal in a parallelogram]

(ii) âˆ FAB = Â½ âˆ ATrue, Since AF is the angle bisector of âˆ A.

(iii) âˆ DCE= Â½ âˆ CTrue, Since CE is the angle bisector of angle âˆ C.

(iv) âˆ CEB= âˆ FABTrue, Since âˆ DCE = âˆ FAB (opposite angles are equal in a parallelogram).

âˆ CEB = âˆ DCE (alternate angles)

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

(v) CE || AFTrue, since one pair of opposite angles are equal, therefore quad. AEFC is a parallelogram.

**Question 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:**

Given that,

AL and CM are perpendiculars on diagonal BD.

In Î”AOL and Î”COM,

âˆ AOL = âˆ COM (vertically opposite angle) ———–(i)

âˆ ALO = âˆ CMO = 90Â° (each right angle) ——–(ii)

By using angle sum property,

âˆ AOL + âˆ ALO + âˆ LAO = 180Â° ———(iii)

âˆ COM + âˆ CMO + âˆ OCM = 180Â° ———- (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)

therefore, Î”AOL is congruent to Î”COM

Hence AL = CM (Corresponding parts of congruent triangles)

**Question 28. Points E and F lie on diagonals AC of a parallelogram ABCD such that AE = CF. what type of quadrilateral is BFDE?**

**Solution:**

From figure, we conclude that,

In parallelogram ABCD,

AO = OC ———- (i) (Diagonals of a parallelogram bisect each other)

AE = CF ———-(ii) Given

On subtracting (ii) from (i)

AO â€“ AE = OC â€“ CF

EO = OF ——-(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 SAS congruence Î”DOE â‰… Î”BOF

therefore, 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 SAS congruence Î”DOE â‰… Î”BOF

Hence, DF = BE (Corresponding parts of congruent triangles).

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

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

**Solution:**

From figure we concluded that,

In a parallelogram ABCD,

Given, AB = 10 cm, AD = 6 cm

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)

Hence CF = 4cm.