CBSE Class 8

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4.

• Practical Geometry Class 8 Ex 4.1
• Practical Geometry Class 8 Ex 4.2
• Practical Geometry Class 8 Ex 4.3
• Practical Geometry Class 8 Ex 4.5

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4

Question 1.
Construct the following quadrilaterals:
(i) Quadrilateral DEAR
DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°

(ii) Quadrilateral TRUE
TR = 3.5 cm
RU = 3 cm
UE = 4 cm
∠R = 75°
∠U= 120°
Solution.
(i) Steps of Construction

1. Draw DE = 4 cm.
2. At E, draw ray EX such that ∠DEX = 60°.
3. From ray EX, cut EA = 5 cm.
   
4. At A, draw ray AY such that ∠EAY = 90°.
5. Cut AR = 4.5 cm from ray AY.
6. Join RD.

Then, DEAR is the required quadrilateral.

(ii) Steps of Construction

1. Draw TR = 3.5 cm.
2. At R, draw ray RX such that ∠TRX = 75°.
   
3. Cut RU = 3 cm from ray RX.
4. At U, draw ray UY such that ∠RUY = 120°.
5. Cut UE = 4 cm from ray UY.
6. Join ET.

Then, TRUE is the required quadrilateral.

We hope the NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3.

• Practical Geometry Class 8 Ex 4.1
• Practical Geometry Class 8 Ex 4.2
• Practical Geometry Class 8 Ex 4.4
• Practical Geometry Class 8 Ex 4.5

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3

Question 1.
Construct the following quadrilaterals :
(i) Quadrilateral MORE
MO = 6 cm
OR = 4.5 cm
∠M = 60°
∠O = 105°
∠R = 105°

(ii) Quadrilateral PLAN
PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 110°
∠N – 85°.

(iii) Parallelogram HEAR
HE = 5 cm
EA = 6 cm
∠R = 85°

(iv) Rectangle OKAY
OK = 7 cm
KA = 5 cm.
Solution.
(i) Steps of Construction

1. Draw MO = 6 cm.
2. At 0, draw ray OX such that Z MOX = 105°.
   
3. Cut OR = 4.5 cm from ray OX.
4. At M, draw ray MY such that ∠OMY = 60°.
5. At R, draw ray RZ such that ∠ORZ = 105°.
6. Let the rays MY and RZ meet at E.

Then, MORE is the required quadrilateral.

(ii) Steps of Construction

1. Draw PL = 4 cm.
2. At L, draw ray LX such that ∠PLX = 75°.
   
   By Angle-sum property of a quadrilateral,
   ∠P + ∠A + ∠N + ∠L = 360°
   ⇒ 90° + 110° + 85° + Z L = 360°
   ⇒ 285° + ∠ L = 360°
   ⇒ ∠L = 360° – 285°
   ⇒ ∠L = 75°.
3. Cut LA = 6.5 cm from ray LX.
4. At A, draw ray AY such that ∠LAY = 110°.
5. At P, draw ray PZ such that ∠LPZ = 90°.
   Let the rays AY and PZ meet at N.

Then, PLAN is the required quadrilateral.

(iii) Steps of Construction

1. Draw HE = 5 cm.
2. At E, draw ray EX such that ∠HEX = 85°
   ∴ Opposite angles of a parallelogram are equal.
   ∵ ∠E = ∠R = 85°
3. Cut EA = 6 cm from the ray EX.
   
4. With A as centre and radius AR = 5 cm, draw an arc.
5. With H as centre and radius HR = 6 cm; draw another arc to intersect the arc drawn in step 4 at R.
   ∴ opposite sides of a parallelogram are equal in length
   ∵ AR = EH = 5 cm
   and HR = EA = 6 cm
6. Join AR and HR.

Then, HEAR is the required parallelogram.

(iv) Steps of Construction
[We know that each angle of a rectangle
is 90°.
∴ ∠O=∠K=∠A=∠Y= 90°.
Also, opposite sides of a rectangle are equal in length.
∴ OY = KA = 5 cm and AY = KO = 7 cm]

1. Draw OK = 7 cm.
2. At K, draw ray KX such that ∠OKX = 90°.
3. Cut KA – 5 cm from ray KX.
   
4. Taking A as centre and radius AY = 7 cm, draw an arc.
5. Taking O as centre and radius OY = 5 cm, draw another arc to intersect the arc drawn in step 4 at Y.
6. Join AY and OY.

Then OKAY is the required rectangle.

We hope the NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2.

• Practical Geometry Class 8 Ex 4.1
• Practical Geometry Class 8 Ex 4.3
• Practical Geometry Class 8 Ex 4.4
• Practical Geometry Class 8 Ex 4.5

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2

Question 1.
Construct the following quadrilaterals:
(i) Quadrilateral LIFT
LI = 4 cm
IF = 3 cm
TL = 2.5 cm
LF = 4.5 cm
IT = 4 cm

(ii) Quadrilateral GOLD
OL = 7.5 cm
GL = 6 cm
GD = 6 cm
LD = 5 cm
OD = 10 cm

(iii) Rhombus BEND
BN – 5.6 cm
DE = 6.5 cm
Solution.
(i) Steps of Construction

1. Draw LI = 4 cm.
2. With L as center and radius LT = 2.5 cm, draw an arc.
   
3. With I as center and radius, IT = 4 cm, draw another arc to intersect the arc drawn in step 2 at T.
4. With I as center and radius IF = 3 cm, draw an arc.
5. With L as center and radius LF = 4.5 cm, draw another arc to intersect the arc drawn in step 4 at F.
6. Join IF, FT, TL, LF and IT.

Then, LIFT is the required quadrilateral.

(ii) Steps of Construction

1. Draw LD = 5 cm.
2. With L as center and radius LG = 6 cm, draw an arc.
   
3. With D as center and radius DG = 6 cm, draw another arc to intersect the arc drawn in step 2 at G.
4. With L as center and radius LO = 7.5 cm, draw an arc.
5. With D as center and radius DO = 10 cm, draw another arc to intersect the arc drawn in step 4 at O.
6. Join DG, GO, OL, LG and DO.

Then GOLD is the required quadrilateral.

(iii) Steps of Construction

1. Draw DE = 6.5 cm.
2. Draw perpendicular bisector PQ of DE so as to intersect DE at M. Then M is the mid-point of DE.
3. With M as centre and radius
   \(=\frac { 1 }{ 2 } \times \left( 5.6 \right) =2.8 cm\)
   opposite sides of DE to intersect MP at N and MQ at B.
   
4. Join DN, NE, EB, and BD.

Then, BEND is the required rhombus.

We hope the NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4.

• Understanding Quadrilaterals Class 8 Ex 3.1
• Understanding Quadrilaterals Class 8 Ex 3.2
• Understanding Quadrilaterals Class 8 Ex 3.3

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4

Question 1.
State whether True or False :
(a) All rectangles are squares
(b) All rhombuses are parallelograms
(c) All squares are rhombuses and also rectangles
(d) All squares are not parallelograms
(e) All kites are rhombuses
(f) All rhombuses are kites
(g) All parallelograms are trapeziums
(h) All squares are trapeziums.
Solution.
(b), (c), (f), (g), (h) are true;
others are false.

Question 2.
Identify all the quadrilaterals that have.
(a) four sides of equal length
(b) four right angles
Solution.
(a) Rhombus; square
(b) Square; rectangle

Question 3.
Explain how a square is
(i) a quadrilateral
(ii)a parallelogram
(iii) a rhombus
(iv) a rectangle.
Solution.
(i) a quadrilateral
A square is 4 sided, so it is a quadrilateral.

(ii) a parallelogram
A square has its opposite sides parallel; so it is a parallelogram.

(iii) a rhombus
A square is a parallelogram with all the 4 sides equal, so it is a rhombus.

(iv) a rectangle
A square is a parallelogram with each angle a right angle; so it is a rectangle.

Question 4.
Name the quadrilaterals whose diagonals :
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal.
Solution.
(i) bisect each other
The names of the quadrilaterals whose diagonals bisect each other are parallelogram; rhombus; square; rectangle.

(ii) are perpendicular bisectors of each other
The names of the quadrilaterals whose diagonals are perpendicular bisectors of each other are rhombus; square.

(iii) are equal
The names of the quadrilaterals whose diagonals are equal are square; rectangle.

Question 5.
Explain why a rectangle is a convex quadrilateral.
Solution.
A rectangle is a convex quadrilateral because both of its diagonals lie wholly in its interior.

Question 6.
ABC is a right-angled triangle and O is the mid-point of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).

Solution.
Construction: Produce BO to D such that BO = OD. Join AD and CD.
Proof. AO = OC ∵ O is the mid-point of AC
BO = OD By construction
∴ Diagonals of quadrilateral ABCD bisect each other.
∴ Quadrilateral ABCD is a parallelogram.
Now, ∠ABC = 90° given
∴ ABCD is a rectangle.
Since the diagonals of a rectangle bisect each other, therefore,
O is the mid-point of AC and BD both. But AC = BD
∵ Diagonals of a rectangle are equal
∴ OA = OC =\(\frac { 1 }{ 2 } \)AC =\(\frac { 1 }{ 2 } \)BD = OB
⇒ OA = OB = OC.

We hope the NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3.

• Understanding Quadrilaterals Class 8 Ex 3.1
• Understanding Quadrilaterals Class 8 Ex 3.2
• Understanding Quadrilaterals Class 8 Ex 3.4

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3

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

(i) AD = ………
(ii) ∠DCB = …………….
(iii) OC = ……………….
(iv) m∠DAB + m∠CDA = …………..
Solution.
(i) AD = BC
Opposite sides of a parallelogram are equal

(ii) ∠DCB = ∠DAB
Opposite angles of a parallelogram are equal

(iii) OC = OA
∵ Diagonals of a parallelogram bisect each other

(iv) m∠DAB + m∠CDA = 180°
∵ Adjacent angles of a parallelogram are supplementary.

Question 2.
Consider the following parallelo¬grams. Find the values of the unknowns x, y, z.

Solution.
(i) y = 100°
Opposite angles of a parallelogram are equal
x + 100° = 180°
Adjacent angles in a parallelogram are supplementary
⇒ x = 180° – 100°
⇒ x = 80°
⇒ z – x = 80°
Opposite angles of a parallelogram are of equal measure

(ii) x + 50° = 180°
Adjacent angles in a parallelogram are supplementary
⇒ x = 180° – 50° = 130°
⇒ y = x = 130°
The opposite angles of a parallelogram are of equal measure
180° – z = 50°
Opposite angles of a parallelogram are of equal measure
⇒ z = 180° – 50° = 130°

(iii) x = 90°
Vertically opposite angles are equal
x + y + 30° = 180°
By angle sum property of a triangle
⇒ 90° + y + 30° = 180°
⇒ 120° + y = 180°
⇒ y = 180° – 120° = 60° z + 30° + 90° – 180°
By angle sum property of a triangle
z = 60°

(iv) y = 80°
Opposite angles of a parallelogram are of equal measure
x + 80° = 180°
Adjacent angles in a parallelogram are supplementary
⇒ x = 180° – 80°
⇒ x = 100°
⇒ 180°-2+ 80°= 180°
Linear pair property and adjacent angles in a parallelogram are supplementary.
z = 80°

(v) y = 112°
Opposite angles of a parallelogram are equal
x + y + 40° = 180°
By angle sum property of a triangle
⇒ x + 112° + 40° = 180°
⇒ x + 152° = 180°
⇒ x = 180°- 152°
⇒ x = 28°
z = x = 28°.
Alternate interior angles

Question 3.
Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180° ?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm
(iii) ∠A = 70° and ∠C = 65°?
Solution.
(i) Can be, but need not be
(ii) No: in a parallelogram, opposite sides are equal; but here, AD ≠ BC.
(iii) No: in a parallelogram, opposite angles are of equal measure; but here ∠A ≠ ∠C.

Question 4.
Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution.
A kite, for example

Question 5.
The measures of two adjacent angles of a parallelogram are in the ratio 3:2. Find the measure of each of the angles of the parallelogram.
Solution.
Let the two adjacent angles be 3x° and 2x°.
Then,
3x° + 2x° = 180°
∴ Sum of the two adjacent angles of a parallelogram is 180°
⇒ 5x° = 180°
⇒ \({ x }^{ \circ }=\frac { { 180 }^{ \circ } }{ 5 } \)
⇒ x° = 36°
⇒ 3x° = 3 x 36° = 108°
and
2x° = 2 x 36° = 72°.
Since, the opposite angles of a parallelogram are of equal measure, therefore the measures of the angles of the parallelogram are 72°, 108°, 72°, and 108°.

Question 6.
Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution.
Let the two adjacent angles of a parallelogram be x° each.
Then,
x° + x° = 180°
∴ Sum of the two adjacent angles of a parallelogram is 180°.
⇒ 2x° = 180°
⇒ \({ x }^{ \circ }=\frac { { 180 }^{ \circ } }{ 2 } \)
⇒ x° = 90°.
Since the opposite angles of a parallelogram are of equal measure, therefore the measure of each of the angles of the parallelogram is 90°, i.e., each angle of the parallelogram is a right angle.

Question 7.
The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.
Solution.
x = 180° – 70° = 110°
Linear pair property and the opposite angles of a parallelogram are of equal measure.
∵ HOPE is a || gm
∴ HE || OP
and HP is a transversal
∴ y = 40°
alternate interior angles
40° + z + x = 180°
The adjacent angles in a parallelogram are supplementary

⇒ 40° + z + 110° = 180°
⇒ z + 150° = 180°
⇒ z = 180° – 150°
⇒ z = 30°.

Question 8.
The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
(i)

(ii)

Solution.
(i)
For Figure GUNS
Since the opposite sides of a parallelogram are of equal length, therefore,
⇒ 3x = 18
⇒ \(x=\frac { 18 }{ 3 } =6\)
and, 3y – 1 = 26
⇒ 3y = 26 + 1
⇒ 3y = 27
\(y=\frac { 27 }{ 3 } =9\)
Hence, x = 6; y = 9.

(ii)
For Figure RUNS
Since the diagonals of a parallelogram bisect each other, therefore,
⇒ x + y = 16 …(1)
and, y + 7 = 20 …(2)
From (2),
⇒ y – 20 – 7 = 13
Putting y = 13 in (1), we get
⇒ x + 13 = 16 ⇒ x = 16 – 13 = 3.
Hence, x = 3; y = 13.

Question 9.
In the below figure both RISK and CLUE are parallelograms. Find the value of x.

Solution.

Question 10.
Explain how this figure is a trapezium. Which of its two sides is parallel?
Solution.

∵ ∠KLM + ∠NML = 80° + 100° = 180°
∴ KL || NM
∵ The sum of consecutive interior angles is 180°
∴ Figure KLMN is a trapezium.
Its two sides \(\overline { KL } \) and \(\overline { NM } \) are parallel.

Question 11.
Find m∠C in the figure, if \(\overline { AB } \) || \(\overline { DC } \).

Solution.
∵ \(\overline { AB } \) || \(\overline { DC } \)
∴ m∠C + m∠B = 180°
∵ The sum of consecutive interior angles is 180°
m∠C+ 120° = 180°
⇒ m∠C = 180° – 120° = 60°.

Question 12.
Find the measure of ∠P and ∠S, if \(\overline { SP } \) || \(\overline { RQ } \) in the figure. (If you find mZ R, is there more than one method to find m∠P ?)
Solution.

∵ \(\overline { SP } \) || \(\overline { RQ } \)
∴ m∠P+m∠Q = 180°
∵ The sum of consecutive interior angles is 180°
⇒ m∠P + 130° = 180°
⇒ m∠P = 180° – 130°
⇒ m∠P = 50°
Again, m∠R + m∠S = 180°
∵ The sum of consecutive interior angles is 180°
⇒ 90° + m Z S = 180°
⇒ m∠S = 180° – 90° = 90°
Yes; there is one more method of finding m∠P if m∠R is given and that is by using the angle sum property of a quadrilateral.
We have,
m∠P + m∠Q + m∠R + m∠S = 360°
⇒ m∠P + 130° + 90° + 90° = 360°
⇒ m∠P + 310° = 360°
⇒ m∠P = 360° – 310° = 50°.

We hope the NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3, drop a comment below and we will get back to you at the earliest.