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These NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Exercise 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:
I. Draw a line segment DE = 4 cm.
II. At E, draw \(\overrightarrow{\mathrm{EX}}\) such that ∠DEX = 60°.
III. From \(\overrightarrow{\mathrm{EX}}\), cut-off EA = 5 cm.
IV. At A, draw ray \(\overrightarrow{\mathrm{AY}}\) such that ∠EAY = 90°.
V. From \(\overrightarrow{\mathrm{AY}}\), cut-off AR = 4.5 cm.
VI. Join R and D. Thus, DEAR is the required quadrilateral.

(ii)

Steps of construction:
I. Draw a line segment \(\overrightarrow{\mathrm{TR}}\) = 3.5 cm.
II. At R, draw a ray \(\overrightarrow{\mathrm{RX}}\), such that ∠TRX = 75°
III. From \(\overrightarrow{\mathrm{RX}}\), cut-off RU = 3 cm.
IV. At U, draw \(\overrightarrow{\mathrm{UY}}\) such that ∠RUY = 120°.
V. From UY, cut-off \(\overrightarrow{\mathrm{UE}}\) = 4 cm.
VI. Join E and T. Thus, TRUE is the required quadrilateral.

These NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Exercise 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
∠A = 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:
I. Draw a line segment MO = 6 cm.
II. At M, draw \(\overrightarrow{\mathrm{MX}}\), such that ∠OMX = 60°
III. At O, draw \(\overrightarrow{\mathrm{OY}}\), such that ∠MOY = 105°
IV. From \(\overrightarrow{\mathrm{OY}}\), cut off OR = 4.5 cm.
V. At R, draw RZ, such that ∠ORZ = 105°. Let \(\overrightarrow{\mathrm{RZ}}\) intersects \(\overrightarrow{\mathrm{MX}}\) at E. Thus, MORE is the required quadrilateral.

(ii)

Steps of construction:
I. Draw a line segment PL = 4 cm
II. At L, draw \(\overrightarrow{\mathrm{LX}}\) such that ∠PLX = 75°
III. From \(\overrightarrow{\mathrm{LX}}\), cut off LA = 6.5 cm
IV. At A, draw \(\overrightarrow{\mathrm{AY}}\) such that ∠LAY = 110°
V. At P, draw \(\overrightarrow{\mathrm{PZ}}\) such that ∠LPZ = 90°. Let \(\overrightarrow{\mathrm{PZ}}\) and \(\overrightarrow{\mathrm{AY}}\) intersect at N. Thus, PLAN is required quadrilateral.

(iii)

Steps of Construction:
I. Draw a line segment \(\overrightarrow{\mathrm{HE}}\) = 5 cm
II. At E, draw \(\overrightarrow{\mathrm{EX}}\) such that ∠HEA = 85°
III. From \(\overrightarrow{\mathrm{EX}}\), cut-off EA = 6 cm.
IV. With centre at A and radius = 5 cm, draw an arc.
V. With centre at H and radius = 6 cm, draw an arc such that it intersects the previous arc at R.
VI. Join RA and RH. Thus, HEAR is the required parallelogram.

(iv)

Steps of Construction:
I. Draw a line segment OK = 7 cm.
II. At O, draw \(\overrightarrow{\mathrm{OP}}\) such that ∠KOP = 90°.
III. From \(\overrightarrow{\mathrm{OP}}\), cut-off \(\overrightarrow{\mathrm{OY}}\) = 5 cm.
IV. At K, draw \(\overrightarrow{\mathrm{KQ}}\) such that ∠OKQ = 90°
V. From \(\overrightarrow{\mathrm{KQ}}\), cut-off \(\overrightarrow{\mathrm{KA}}\) = 5 cm.
VI. Join AY. Thus, OKAY is the required rectangle.

These NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Exercise 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:
I. Draw a line segment LI = 4 cm.
II. With centre at I and radius = 4 cm, draw an arc.
III. With centre at L and radius = 2.5 cm, draw another arc to intersect the previous arc at T.
IV. Join TI and TL.
V. With centre at L and radius = 4.5 cm, draw an arc on the side opposite to that of L.
VI. With centre at I and radius = 3 cm, draw another arc to intersect the previous arc at F.
VII. Join FL and FI.
Thus, LIFT is the required quadrilateral.

(ii)

Steps of construction:
I. Draw a line segment LD = 5 cm.
II. With centre at L and radius = 6 cm, draw an arc.
III. With centre at D and radius = 6 cm, draw another arc to intersect the previous arc at G.
IV. Join GL and GD.
V. With centre at D and radius = 10 cm, draw an arc.
VI. With centre at L at radius = 7.5 cm, draw another arc to intersect the previous arc at O.
VII. Join OL and OG.
Thus, GOLD is the required quadrilateral.

(iii)

Steps of construction:
I. Draw a line segment DE = 6.5 cm.
II. Draw perpendicular bisector of DE such that M is the mid-point of DE.
III. With centre M and radius = 2.8 cm, draw arcs to intersect the perpendicular bisector at B and N.
IV. Join ND, NE, BE, and BD.
Thus, BEND is the required rhombus.

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.2

Question 1.
Give first step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y – 4 = -7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4
Answer:
(a) x – 1 = 0
Adding 1 on both sides, we get.
x – 1 + 1 = 0 + 1
x = 1
The solution is x = 1.

(b) x + 1 = 0
Subtracting 1 from both sides, we get.
x + 1 – 1 = 0 – 1
x = -1.
The solution is x = – 1

(c) x – 1 = 5
Adding 1 to both sides, we get.
x – 1 + 1 = 5 + 1
x = 6
The solution is x = 6.

(d) x + 6 = 2
Subtracting 6 from both sides, we get
x + 6 – 6 = 2 – 6
x = – 4
The solution is x = – 4.

(e) y – 4 = – 7
Adding 4 on both sides, we get
y – 4 + 4 = -7 + 4
y = – 3
The solution is y = – 3.

(f) y – 4 = 4
Adding + 4 on both sides, we get
y – 4 + 4 = 4 + 4
y = 8
The solution is y = 8.

(g) y + 4 = 4
Subtracting 4 from both sides, we get
y + 4 – 4 = 4 – 4
y = 0
The solution is y = 0.

(h) y + 4 = – 4
Adding – 4 on both sides, we get
y + 4 – 4 = -4 -4
y = – 8
The solution is y = – 8.

Question 2.
Give first step you will use to separate the variable and then solve the equation:
(a) 31 = 42
(b) \(\frac{\mathrm{b}}{2}\) = 6
(c) \(\frac{\mathrm{p}}{7}\) = 4
(d) 4x = 25
(e) 8y = 36
(f) \(\frac{z}{3}=\frac{5}{4}\)
(g) \(\frac{a}{5}=\frac{7}{15}\)
(h) 20t = -10
Answer:
(a) 31 = 42
Dividing both sides by 3, we get
\(\frac{31}{3}=\frac{42}{3}\)
1 = 14
The solution is 1 = 14.

(b) \(\frac{\mathrm{b}}{2}\) = 6
Multiplying both sides by 2, we get. b
\(\frac{\mathrm{b}}{2}\) × 2 = 6 × 2
b = 12
The solution is b = 12.

(c) \(\frac{\mathrm{p}}{7}\) = 4
Multiplying both sides by 7, we get
\(\frac{\mathrm{p}}{7}\) × 7 = 4 × 7
p = 28
The solution is p = 28.

(d) 4x = 25
Dividing both sides by 4, we get
\(\frac{4 x}{4}=\frac{25}{4}\)
x = \(\frac{25}{4}\)
The solution is x = \(\frac{25}{4}\) or x = 6 \(\frac{1}{4}\) .

(e) 8y = 36
Dividing both sides by 8, we get
\(\frac{8 y}{8}=\frac{36}{8}\)
y = \(\frac{9}{2}\)
The solution is y = \(\frac{9}{2}\) or y = 4 \(\frac{1}{2}\)

(f) (f) \(\frac{z}{3}=\frac{5}{4}\)
Multiplying both sides by 3, we get
\(\frac{z}{3}\) × 3 = \(\frac{5}{4}\) × 3
z = \(\frac{25}{4}\)
The solution is z = \(\frac{15}{4}\) or y = 3 \(\frac{3}{4}\)

(g) \(\frac{a}{5}=\frac{7}{15}\)
Multiplying both sides by 5, we get
\(\frac{a}{5}\) × 5 = \(\frac{7}{15}\) × 5
a = \(\frac{7}{3}\)
The solution is a = \(\frac{7}{3}\) or a = 2 \(\frac{1}{3}\).

(h) 20t = – 10
Dividing both sides by 20, we get
\(\frac{20 \mathrm{t}}{20}=\frac{-10}{20}\)
t = \(\frac{-1}{2}\)
The solution is t = \(\frac{-1}{2}\).

Question 3.
Give the steps you will use to separate the variable and then solve the equation:
(a) 3n – 2 = 46
(b) 5m +7 = 17
(c) \(\frac{20 \mathrm{p}}{3}\) = 40
(d) \(\frac{3 \mathrm{p}}{10}\) = 6
Answer:
(a) 3n – 2 = 46
Adding 2 on both sides, we get
3n – 2 + 2 =46 + 2
3n = 48
Dividing both sides by 3, we get
\(\frac{3 n}{3}=\frac{48}{3}\)
n = 16
The solution is n = 16.

(b) 5m + 7 = 17
Adding – 7 on both sides, we get
5m + 7 – 7 = 17 – 7
5m = 10
Dividing both sides by 5, we get
\(\frac{5 m}{5}=\frac{10}{5}\)
m = 2
The solution is m = 2

(c) \(\frac{20 \mathrm{p}}{3}\) = 40
Multiplying both sides by 3, we get.
\(\frac{20 \mathrm{p}}{3}\) × 3 = 40 × 3
20 p = 120
Dividing both sides by 20 we get.
\(\frac{20 p}{20}=\frac{120}{20}\)
p = 6
The solution is p = 6.

(d) \(\frac{3 \mathrm{p}}{10}\) = 6
Multiplying both sides by 10, we get 3p
\(\frac{3 \mathrm{p}}{10}\) × 10 = 6 × 10
3p = 60
Dividing both sides by 3, we get
\(\frac{3 p}{3}=\frac{60}{3}\)
p = 20
The solution is p = 20.

Question 4.
Solve the following equations:
(a) 10p = 100
(b) 10p + 10 = 100
(c) \(\frac{\mathrm{p}}{4}\)
(d) \(\frac{-p}{3}\) = 5
(e) \(\frac{3 \mathrm{p}}{4}\) = 6
(f) 3s = – 9
(g) 3s + 12 = 0
(h) 3s = 0
(i) 2q = 6
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
Answer:
(a) 10p = 100
Dividing both sides by 10, we get
\(\frac{10 \mathrm{p}}{10}=\frac{100}{10}\)
p = 10 10p = 90
The solution is p = 10.

(b) 10p + 10 = 100
10p + 10 – 10= 100 – 10
Dividing both sides by 10, we get
\(\frac{10 p}{10}=\frac{90}{10}\)
p = 9
The solution is p = 9.

(c) \(\frac{\mathrm{p}}{4}\) = 5
Multiplying both sides by 4, we get
\(\frac{\mathrm{p}}{4}\) × 4 = 5 × 4
p = 20
The solution is p = 20.

(d) \(\frac{-p}{3}\) = 5
Multiplying both sides by 3, we get
-p = 15
\(\frac{-p}{3}\) × 3 = 5 × 3
Multiplying both sides by (- 1)
– p × (- 1)= 15 × (- 1)
p = – 15
The solution is p = -15.

(e) \(\frac{3 \mathrm{p}}{4}\) = 6
Multiplying both sides by 4, we get
\(\frac{3 \mathrm{p}}{4}\) × 4 = 6 × 4
3p = 24
Dividing both sides by 3, we get
\(\frac{3 p}{3}=\frac{24}{3}\)
p = 8
The solution is p = 8.

(f) 3s = – 9
Dividing both sides by 3, we get
\(\frac{3 s}{3}=\frac{-9}{3}\)
s = – 3
The solution is s = – 3.

(g) 3s + 12 = 0
Subtracting 12 from both sides, we get
3s + 12- 12 = 0 – 12
3s = – 12
Dividing both sides by 3, we get
\(\frac{3 s}{3}=\frac{-12}{3}\)
s = – 4
The solution is s = – 4.

(h) 3s =0
Dividing both sides by 3, we get
\(\frac{3 s}{3}=\frac{0}{3}\)
s = 0
The solution is s = 0.

(i) 2q = 6
Dividing both sides by 2, we get
\(\frac{2 \mathrm{q}}{2}=\frac{6}{2}\)
q = 3
The solution is q = 3.

(j) 2q – 6 = 0
Adding 6 on both sides, we get
2q – 6 + 6= 0 + 6
2q = 6
Dividing both sides by 2, we get
\(\frac{2 \mathrm{q}}{2}=\frac{6}{2}\)
q = 3
The solution is q = 3.

(k) 2q + 6 =0
Subtracting 6 from both sides, we get
2q + 6 – 6 = 0 – 6
2q = – 6
Dividing both sides by 2, we get
\(\frac{2 q}{2}=\frac{-6}{2}\)
q = -3
The solution is q = – 3.

(l) 2q + 6 =12
Subtracting 6 from both sides, we get
2q + 6 – 6 = 12 – 6
2q = 6
Dividing both sides by 2, we get
\(\frac{2 \mathrm{q}}{2}=\frac{6}{2}\)
q = 3
The solution is q = 3.

These NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Exercise 4.1

Question 1.
Construct the following quadrilaterals.
(i) Quadrilateral ABCD
AB = 4.5 cm
BC = 5.5 cm
CD = 4 cm
AD = 6 cm
AC = 7 cm

(ii) Quadrilateral JUMP
JU = 3.5 cm
UM = 4 cm
MP = 5 cm
PJ = 4.5 cm
PU = 6.5 cm

(iii) Parallelogram MORE
OR = 6 cm
RE = 4.5 cm
EO = 7.5 cm

(iv) Rhombus BEST
BE = 4.5 cm
ET = 6 cm
Solution:
(i)

Steps of Construction:
I. Draw a line segment AB = 4.5 cm.
II. With centre at A and radius = 7 cm, draw an arc.
III. With centre at B and radius = 5.5 cm, draw another arc to intersect the previous arc at C.
IV. With centre at A and radius = 6 cm, draw an arc on the side opposite to that of B.
V. With centre at C and radius = 4 cm, draw another arc to intersect the previous arc at D.
VI. Join AC, BC, DA, and DC. Thus, ABCD is the required quadrilateral.

(ii)

Steps of Construction:
I. Draw a line segment JU = 3.5 cm.
II. With centre at U and radius UP = 6.5 cm, draw an arc.
III. With centre at J and radius = 4.5 cm, draw another arc to intersect the previous arc at P.
IV. With centre at U and radius = 4 cm, draw an arc on the side opposite to that of J.
V. With centre at P and radius = 4 cm, draw another arc intersecting the previous arc at M.
VI. Join PJ, PU, PM, and UM. Thus, JUMP is the required quadrilateral.

(iii)

Note: In a parallelogram opposite sides are equal.
Steps of Construction:
I. Draw a line segment MO = 4.5 cm
II. With centre O and radius = 7.5 cm, draw an arc.
III. With centre M and radius = 6 cm, draw another arc to intersect the previous arc at E.
IV. With centre at O and radius = 6 cm, draw an arc on the side opposite to E
V. With centre at E and radius = 4.5 cm, draw another arc to intersect the previous arc at R.
VI. Join OR, RE, EO, and EM. Thus, MORE is the required parallelogram.

(iv)

Steps of constructions:
I. Draw a line segment BE = 4.5 cm.
II. With centre at E and radius = 6 cm, draw an arc.
III. With centre at B and radius = 4.5 cm, draw another arc to intersect the previous arc at T.
Note: All 4 sides of a rhombus are equal.
IV. Join TB and TE.
V. With centre at T and radius = 4.5 cm, draw an arc on the side opposite to that of B.
VI. With centre at E and radius = 4.5 cm, draw another arc to intersect the previous arc at S.
VII. Join ST and SE. Thus, BEST is the required rhombus.

These NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals InText Questions

Try These (Page No. 43)

Take a regular hexagon Fig 3.10.

Question 1.
What is the sum of the measures of its exterior angles x, y, z, p, q, r?
Solution:
∠x + ∠y + ∠x + ∠z + ∠p + ∠q + ∠r = 360°
[∵ Sum of exterior angles of a polygon = 360°]

Question 2.
Is x = y = z = p = q = r? Why?
Solution:
Since, all the sides of the polygon are equal.
∴ It is a regular hexagon.
So, its interior angles are equal.
∴ x = (180° – a)
y = (180° – a)
z = (180° – a)
p = (180° – a)
q = (180° – a)
r = (180° – a)
∴ x = y = z = p = q = r

Question 3.
What is the measure of each?
(i) exterior angle
(ii) interior angle
Solution:
(i) x + y + z = p + q = r = 360° [sum of exterior angles = 360°]
and all these angles are equal
∴ Measure of each exterior angles = \(\frac{360^{\circ}}{6}\) = 60°

(ii) ∵ Exterior angle = 60°
∴ 180° – 60° = Interior angle
or 120° = Interior angle
or Measure of interior angle = 120°

Question 4.
Repeat this activity for the cases of
(i) a regular octagon
(ii) a regular 20-gon
Solution:
(i) In a regular octagon, number of sides (n) = 8
∴ Each exterior angle = \(\frac{360^{\circ}}{8}\) = 45°
∴ Each interior angle = 180° – 45° = 135°

(ii) For a regular 20-gon, the number of sides (n) = 20
∴ Each exterior angle = \(\frac{360^{\circ}}{20}\) = 18°
Thus, each interior angle = 180° – 18° = 162°

Try These (Page No. 47)

Question 5.
Take two identical set squares with angles 30° – 60° – 90° and place them adjacently to form a parallelogram as shown in Fig 3.20. Does this help you to verify the above property?

Solution:
The given figure helps us to verify that opposite sides of a parallelogram are of equal length.

Try These (Page No. 48)

Question 6.
Take two identical 30° – 60° – 90° set-squares and form a parallelogram as before. Does the figure obtain the help you to confirm the above property?
Solution:
The above figure also helps us to confirm that: opposite angles of a parallelogram are equal.