## RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III (Special Types of Quadrilaterals) Ex 17.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1

**Other Exercises**

**Question 1.****Given below is a parallelogram ABCD. Complete each statement along with the definition or property used:****(i) AD = ………****(ii) ∠DCB = ……….****(iii) OC’ = …….****(iv) ∠DAB + ∠CDA = …….****Solution:**

(i) AD = BC

(ii) ∠DCB = ∠ADC

(iii) OC = OA

(iv) ∠DAB + ∠CDA = 180°

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

In a parallelogram, opposite angles are equal and sum of adjacent angle is 180°.

(i) In parallelogram ABCD,

∠B = 100°

∠A = ∠C = 180° (Co-interior angles)

⇒ x + 100° = 180°

⇒ x = 180° – 100°

⇒ x = 80°

But ∠A = ∠C and ∠B = ∠D (Opposite angles)

A = x

⇒ z = x

⇒ z = 80°

and ∠D = ∠B

⇒ y = 100°

x = 80°, y = 100° and z = 80°

(ii) In parallelogram PQRS, side PQ is produced to T.

∠S = 50°

∠PQR = ∠S (Opposite angles)

w = 50°

But ∠P + ∠PQR = 180° (Sum of adjacent angles)

⇒ x + 50° = 180°

⇒ x = 180° – 50° = 130°

⇒ But ∠P = ∠R (Opposite angles)

x = y

⇒ y = 130°

But w + z = 180° (A linear pair)

⇒ 50° + z = 180°

⇒ z = 180° – 50° = 130°

⇒ x = 130°, y = 130°, ∠ = 130

(iii) In parallelogram LMNP, PM is its diagonal ∠NPM = 30°, ∠PMN = 90°

PN || LM (Opposite sides of a parallelogram) and PM is its transversal

∠NPM = ∠PML

⇒ 30° = x

x = 30°

In ∆PMN,

∠P = 30°, ∠M = 90°

But ∠P + ∠M + ∠N = 180° (Sum of angles of a triangle)

⇒ 30° + 90° + z = 180°

⇒ 120° + z = 180°

⇒ z = 180° – 120° = 60°

But ∠L = ∠N (Opposite angles of a parallelogram)

y = z

⇒ y = 60°

Hence x = 30°, y = 60° and ∠ = 60°

(iv) In rhombus ABCD, diagonals AC and BD bisect each other at right angles.

x = 90°

In ∆OCD,

∠O + ∠C + ∠D = 180°

⇒ 90° + 30° + y = 180°

⇒ 120° + y = 180°

⇒ y = 180° – 120° = 60°

y = 60°

CD || AB, BD is the transversal

y = z (Alternate angles)

z = 60°

Hence x = 90°, y = 60° and z = 60°

(v) In parallelogram PQRS, side QR is produced to T

∠Q = 80°

∠P + ∠Q = 180° (Sum of adjacent angles)

⇒ x + 80° = 180°

⇒ x = 180° – 80° = 100°

∠Q = ∠S (Opposite angles of a parallelogram)

⇒ 80° = y

⇒ y = 80°

PQ || SR and QRT is transversal

∠TRS = ∠RQP (Corresponding angles)

⇒ ∠ = 80°

Hence x = 100°, y = 80° and z = 80°

(vi) In parallelogram TUVW, UW is its diagonal

∠TUW = 40° and ∠V = 112°

∠T = ∠V (Opposite angles)

y = 112°

In ∆TUW,

∠T + ∠V + ∠W = 180° (Sum of angles of a parallelogram)

⇒ y + 40° + x = 180°

⇒ 112° + 40° + x = 180°

⇒ 152° + x = 180°

⇒ x = 180° – 152° = 28°

UV || TW and UW is its transversal

∠WUV = ∠TWU (Alternate angles)

⇒ z = x

⇒ z = 28°

Hence x = 28°, y = 112°, z = 28°

**Question 3.****Can the following figures be parallelograms. Justify your answer.****Solution:**

(i) In quadrilateral PLEH

∠H = 100°, ∠L = 80°

But there are opposite angles

∠H ≠ ∠L

PLEH is not a parallelogram.

(ii) In quadrilateral GNIR,

RI = 8 cm, GN = 8 cm, RG = 5 cm and IN = 5 cm

PI = GH and RG = IN

But there are opposite sides of the quadrilateral.

GNIR is a parallelogram.

(iii) In quadrilateral BEST,

BS and ET are its diagonals

But these diagonal do not bisect each other.

BEST is not a parallelogram

**Question 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:**

In the figure, HOPE is a parallelogram in which HG is produced and HP is the diagonal

∠EHP = 40° and ∠POQ = 70°

But ∠POQ + ∠POH = 180° (Linear pair)

⇒ 70° + w = 180°

⇒ w = 180° – 70° = 110°

But ∠E = ∠POE (Opposite angles of a parallelogram)

x = 110°

HE || OP and HP is its transversal.

∠EHP = ∠HPO (Alternate angles)

⇒40° = y

⇒ y = 40°

In ∆PHO,

Ext. ∠POQ = ∠PHO + ∠HPO

⇒ 70° = z + y

⇒ 70° = z + 40°

⇒ z = 70° – 40° = 30°

Hence x = 110°, y = 40°, z = 30°

**Question 5.****In the following figures GUNS and RUNS are parallelograms. Find x and y.****Solution:**

(i) In parallelogram GUNS,

Opposite sides are parallel and equal

3x = 18

⇒ x = 6

and 3y – 1 = 26

⇒ 3y = 26 + 1 = 27

⇒ y = 9

x = 6, y = 9

(ii) Diagonals of a parallelogram bisect each other.

y – 7 = 20

⇒ y = 20 + 7 = 27

and x – 27 = 16

⇒ x – 27 = 16

⇒ x = 16 + 27 = 43

x = 43, y = 27

**Question 6.****In the following figure RISK and CLUE are parallelograms. Find the measure of x.****Solution:**

In the figure, RISK and CLUE are parallelograms

∠K = 120° and ∠L = 70°

In parallelogram RISK

RK || IS

∠RKS = ∠ISU (Corresponding angles)

⇒ ∠ISU = 120°

In parallelogram CLUE,

∠E = ∠L (Opposite angles of a parallelogram)

∠E = 70° (∠L = 70°)

Now in ∆EOS,

Ext. ∠ISU = x + ∠E

⇒ 120° = x + 70°

⇒ x = 120° – 70° = 50°

x = 50°

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

In parallelogram ABCD,

∠A = (3x – 2)° and ∠C = (50 – x)°

∠A = ∠C (Opposite angles of a parallelogram)

⇒ 3x – 2° = 50° – x

⇒ 3x + x = 50° + 2°

⇒ 4x = 52°

x = 13°

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

Let the parallelogram is ABCD and ∠A of a parallelogram ABCD be x

Hence other angles will be

∠C = ∠A = 108° (Opposite angles)

and ∠D = ∠B = 72° (Opposite angles)

Hence ∠A = 108°, ∠B = 72°, ∠C = 108° and ∠D = 72°

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

Let in parallelogram ABCD,

∠A = 70°

But ∠A + ∠B= 180° (Sum of adjacent angles)

⇒ 70° + ∠B = 180°

⇒ ∠B = 180° – 70°

⇒ ∠B = 110°

But ∠C = ∠A and ∠D = ∠B (Opposite angles)

∠C = 70° and ∠D = 110°

Hence ∠B = 110°, ∠C = 70° and ∠D = 110°

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

Let in parallelogram ABCD,

∠A : ∠B = 1 : 2

Let ∠A = x, then ∠B = 2x

But ∠A + ∠B = 180° (Sum of adjacent angles)

⇒ x + 2x = 180°

⇒ 3x = 180°

∠A = x = 60°

and ∠B = 2x = 2 x 60°= 120°

But ∠C = ∠A and ∠D = ∠B (Opposite angles)

∠C = 60° and ∠D = 120°

Hence ∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120°

**Question 11.****In a parallelogram ABCD, ∠D = 135°, determine the measure of ∠A and ∠B.****Solution:**

In parallelogram ABCD,

∠D = 135°

But ∠A + ∠D= 180° (Sum of adjacent angles)

∠A + 135° = 180°

∠A = 180° – 135° = 45°

But ∠B = ∠D (Opposite angles)

∠B = 135°

Hence ∠A = 45° and ∠B = 135°

**Question 12.****ABCD is a parallelogram in which ∠A = 70°, compute ∠B, ∠C and ∠D.****Solution:**

In parallelogram ABCD,

∠A = 70°

But ∠A + ∠B = 180° (Sum of adjacent angles)

⇒ 70° + ∠B = 180°

⇒ ∠B = 180° – 70° = 110°

But ∠C = ∠A and ∠D = ∠B (Opposite angles)

∠C = 70° and ∠D = 110°

Hence ∠B = 110°, ∠C = 70° and ∠D = 110°

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

Let in parallelogram

∠A + ∠C = 130°

But ∠A = ∠C (Opposite angles)

⇒ ∠C = \\(\\frac { 130 }{ 2 }) = 65°

∠B + ∠D = 180° (Sum of adjacent angles)

⇒ 65° + ∠B = 180°

⇒ ∠B = 180° – 65° = 115°

⇒ ∠B = 115°

But ∠D = ∠B (Opposite angles)

∠D = 115°

Hence ∠A = 65°, ∠B = 115°, ∠C = 65° and ∠D = 115°

**Question 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:**

All the angles of a quadrilateral are equal and sum of the four angles = 360°

Each angle will be = \\(\\frac { 360 }{ 4 }) = 90°

Let in quadrilateral ABCD,

∠A = ∠B = ∠C = ∠D = 90°

It is a parallelogram as opposite angles are equal

i.e., ∠A = ∠C and ∠B = ∠D.

Each angle is of 90°

This parallelogram is a rectangle.

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

Length of two adjacent sides = 4 cm and 3 cm

i.e., l = 4 cm and b = 3 cm

Perimeter = 2 (l + b) = 2 (4 + 3) cm = 2 x 7 = 14 cm

**Question 16.****The perimeter of a parallelogram is 150 cm. One of its sides is greater ii.au the other by 25 cm. Find the length of the sides of the parallelogram.****Solution:**

Perimeter of a parallelogram = 150 cm

Let l he the longer side and b be the shorter side

l = b + 25 cm.

⇒ 2 (l + b) = 150

⇒ l + b = 75

⇒ b + 25 + b = 75

⇒ 2b = 75 – 25 = 50

⇒ b = 25

l = b + 25 = 25 + 25 = 50

Sides are 50 cm, 25 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:**

In a parallelogram shorter side (b) = 4.8 cm.

longer side (l) = 4.8 + \\(\\frac { 1 }{ 2 }) x 4.8

= 4.8 + 2.4 = 7.2 cm

Perimeter = 2 (l + b) = 2 (7.2 + 4.8) cm = 2 x 12.0 = 24 cm

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

Let in parallelogram ABCD,

∠A = (3x + 4)° and ∠B = (3x + 10)°

But ∠A + ∠B = 180° (Sum of adj adjacent angles)

⇒ 3x – 4 + 3x + 10 = 180°

⇒ 6x + 6° = 180°

⇒ 6x = 180° – 6° = 174°

⇒ x = 29°

∠A = 3x – 4 = 3 x 29 – 4 = 87° – 4° = 83°

∠B = 3x + 10 = 3 x 29° + 10° = 87° + 10° = 97°

But ∠C = ∠A and ∠D = ∠B (Opposite angles)

∠C = 83° and ∠D = 97°

Hence ∠A = 83°, ∠B = 97°, ∠C = 83° and ∠D = 97°

**Question 19.****In a parallelogram ABCD, the diagonals bisect each other at O. If ∠ABC = 30°, ∠BDC = 10° and ∠CAB = 70°.****Find : ∠DAB, ∠ADC, ∠BCD, ∠AOD, ∠DOC, ∠BOC, ∠AOB, ∠ACD, ∠CAB, ∠ADB, ∠ACB, ∠DBC and ∠DBA.****Solution:**

In parallelogram ABCD, diagonal AC and ED bisect each other at O.

∠ABC = 30°, ∠CAB = 70° and ∠BDC = 10°

∠ADC = ∠ABC = 30° (Opposite angles)

and ∠ADB = ∠ADC – ∠BDC = 30° – 10° = 20°

AB || DC and AC is the transversal

∠ACD = ∠CAB = 70°(Altemate angles)

AB || DC and BD is transversal

∠CDB = ∠ABD = 10°(Altemate angles)

In ∆ABC

∠CAB + ∠ABC + ∠BCA = 180° (Sum of anlges of a triangle)

⇒ 70° + 30° + ∠BCA = 180°

⇒ 100° + ∠BCA = 180°

∠BCA = 180° – 100° = 80°

∠BCD = ∠BCA + ∠ACD = 80° + 70° = 150°

∠BCD = ∠DAB (Opposite angles)

⇒ ∠DAB = 150° and ∠CAD = 150° – 70° = 80°

In ∆OCD,

∠ODC + ∠OCD + ∠COD = 180° (Angles of a triangle)

⇒ ∠ACD + ∠ACD + ∠COD = 180°

⇒ 70° + 10° + ∠COD = 180°

⇒ 80° + ∠COD = 180°

⇒ ∠COD = 180° – 80° = 100°

∠COD = 100°

But ∠AOD + ∠COD = 180° (Linear pair)

⇒ ∠AOD + 100° = 180°

⇒ ∠AOD = 180° – 100° = 80°

⇒ ∠AOD = 80°

But ∠AOB = ∠COD

and ∠BOC = ∠AOD (Vertically opposite angles)

∠AOB = 100° and ∠BOC = 80°

Hence ∠DAB = 150°, ∠ADC = 30°, ∠BCD = 150°

∠AOD = 80° ∠DOC = 100°

∠BOC = 80° ∠AOB = 100°

∠ACD = 70°, ∠CAB = 70°, ∠AD = 20°, ∠ACB = 80°, ∠DBC = 20°, ∠DBA = 10

**Question 20.****Find the angles marked with a question mark shown in the figure.****Solution:**

In parallelogram ABCD,

CE ⊥ AB and CF ⊥ AD

∠BCE = 40°

In ∆BCE,

∠BCE + ∠CEB + ∠EBC = 180° (Sum of angles of a triangle)

⇒ 40° + 90° + ∠EBC = 180°

⇒ 130° + ∠EBC = 180°

⇒ ∠EBC = 180° – 130° = 50°

or ∠B = 50°

But ∠D = ∠B (Opposite angles)

∠D = 50° or ∠ADC = 50°

Similarly in ∆DCF,

∠DCF + ∠CFD + ∠FDC = 180°

⇒ ∠DCF + 90° + 50° = 180°

⇒ ∠DCF + 140° = 180°

⇒ ∠DCF = 180° – 140° = 40°

But ∠C + ∠B= 180° (Sum of adjacent angles)

⇒ ∠BCE + ∠ECF + ∠DCF + ∠B = 180°

⇒ 40° + ∠ECF + 40° + 50° = 180°

⇒ ∠ECF + 130° = 180°

⇒ 40° + ∠ECF + 40° + 50° = 180°

⇒ ∠ECF + 130° = 180°

⇒ ∠ECF = 180° – 130° = 50°

∠ECF = 50°

**Question 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:**

In parallelogram ABCD,

∠A is an obtused angle

AE ⊥ BC and AF ⊥ DC

∠EAF = 60°

In quadrilateral AECF,

∠EAF + ∠F + ∠C + ∠E = 360° (Sum of angles of a quadrilateral)

⇒ 60° + 90° + ∠C + 90° = 360°

⇒ 240° + ∠C = 360°

⇒ ∠C = 360° – 240° = 120°

∠C = 120°

But ∠A = ∠C (Opposite angles)

∠A = 120°

But ∠A + ∠B = 180° (Sum of adjacent angles)

⇒ 120° + ∠B = 180°

⇒ ∠B = 180° – 120° = 60°

∠B = 60°

But ∠D = ∠B (Opposite angles)

∠D = 60°

Hence ∠A = 120°, ∠B = 60°, ∠C = 120° and ∠D = 60°

**Question 22.****In the figure, ABCD and AEFG are parallelograms. If ∠C = 55°, what is the measure of ∠F ?****Solution:**

In the parallelogram ABCD,

∠A = ∠C …..(i) (Opposite angles of a parallelogram)

Similarly in parallelogram AEFG,

∠A = ∠F …(ii)

From (i) and (ii),

∠C = ∠F = 55° (∠C = 55°)

Hence ∠F = 55°

**Question 23.****In the figure, BDEF and DCEF are each a parallelogram. It is true that BD = DC ? Why or why not ?****Solution:**

In parallelogram BDEF,

BD = EF ……(i) (Opposite sides of a parallelogram)

Similarly, in parallelogram DCEF

DC = EF

From (i) and (ii),

BD = DC

Hence it is true that BD = DC.

**Question 24.****In the figure, suppose it is known that DE = DF. Then is ∆ABC isosceles ? Why or why not ?****Solution:**

In parallelogram BDEF,

∠B = ∠E ……(i) (Opposite angles of a parallelogram)

Similarly, in parallelogram DCEF,

∠C = ∠F ……(ii)

But DE = DF (Given)

In ∆DEF

∠E = ∠F

From (i) and (ii),

∠B = ∠C

AC = AB (Sides opposite to equal angles)

∆ABC is an isosceles triangle.

**Question 25.****Diagonals of parallelogram ABCD intersect at O as shown in the 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) ∠OBY = ∠ODX****(iii) ∠BOY = ∠DOX****(iv) ∆BOY = ∆DOX****Now, state if XY is bisected at O.****Solution:**

In parallelogram ABCD,

diagonals AC and BP intersect each other at O

O is the mid-point of AC and BD.

Through O, XY is draw such that X lies on AD and Y, on BC.

(i) OB = OD (O is mid-point of BD)

(ii) AD || BC and BD is transversal

∠OBY = ∠ODX (Alternate angles)

(iii) ∠BOY = ∠DOX (Vertically opposite angles)

(iv) Now in ∆BOY and ∆DOX,

OB = OD

∠OBY = ∠ODX

∠BOY = ∠DOX

∆BOY = ∆DOX (ASA axiom)

OY = OX (c.p.c.t.)

Hence XY is bisected at O.

**Question 26.****In the figure ABCD is a parallelogram, CE bisects ∠C and AF bisects ∠A. In each of the following, if the statement is true, give a reason for the same.****(i) ∠A = ∠C****(ii) ∠FAB = \\(\\frac { 1 }{ 2 }) ∠A****(iii) ∠DCE = \\(\\frac { 1 }{ 2 }) ∠C****(iv) ∠CEB = ∠FAB****(v) CE || AF.****Solution:**

In parallelogram ABCD,

CE is the bisector of ∠C and and AF is the bisector of ∠A.

(i) ∠A = ∠C (Opposite angles of a parallelogram)

(ii) AF is the bisector of ∠A

∠FAB = \\(\\frac { 1 }{ 2 }) ∠A

(iii) CE is the bisector of ∠C

∠DCE = \\(\\frac { 1 }{ 2 }) ∠C

(iv) From (i), (ii) and (iii)

∠FAB = ∠DCE

(v) ∠FAB = ∠DCE

But these are opposite angles of quadrilateral AECF

AB or AE || DC or FC

AECF is a parallelogram

CE || AF

Hence proved

**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:**

In parallelogram ABCD, diagonals AC and BD intersect each other at O.

O is the mid-point of AC and BD.

AL ⊥ BD and CM ⊥ BD.

In ∆ALO and ∆CMO

∠L = ∠M (Each 90°)

∠AOL = ∠COM (Vertically opposite angles)

AO = CO (O is mid-point of AC)

∆ALO = ∆CMO (AAS axiom)

AL = CM (c.p.c.t.)

Hence proved

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

In parallelogram ABCD,

AC is its diagonal. E and F are points on AC such that AE = CF

Join EB, BF, FD and DE .

Join also diagonal BD which intersects AC at O

O is the mid-point of AC and BD

AO = OC

But AE = CF

⇒ AO – AE = CO – CF

⇒ EO = OF

But BO = OD (O is mid-point of BD)

Diagonals EF and BD of quadrilateral bisect each other at O.

BFDE is a parallelogram.

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

AB || DC

∠DEA = ∠EAB (Alternate angles)

= ∠DAE (EA is bisector of ∠A)

In ∆DAE,

∠DEA = ∠DAE

AD = DE = 6 cm

But DE = AB = 10 cm.

EC = DC – DE = 10 – 6 = 4 cm

AD || BC or BF and AF is transversal

∠DAE = ∠EFC (Alternate angle)

But ∠DAE = ∠DEA Prove

= ∠FEC (DEA = FEC vertically opposite angles)

In ∆ECF,

CE = CF = 4 cm (CE = 4 cm)

Hope given RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 are helpful to complete your math homework.

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