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These NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 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:
(a) False
(b) True
(c) True
(d) False
(e) False
(f) True
(g) True
(h) True

Question 2.
Identify all the quadrilaterals that have
(a) four sides of equal length
(b) four right angles.
Solution:
(a) square, rhombus
(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 square is a 4-sided figure, so it is a quadrilateral.
(ii) The opposite sides of a square are parallel, so it is a parallelogram.
(iii) All the sides of a square are equal, so it is a rhombus.
(iv) Each angle of a square is 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) The diagonals of the following quadrilaterals bisect each other.
Square, rectangle, rhombus, and parallelogram
(ii) The diagonals are perpendicular bisectors of the following quadrilaterals.
Square and rhombus
(iii) The diagonals are equal in the case of square and rectangle.

Question 5.
Explain why a rectangle is a convex quadrilateral.
Solution:
A rectangle is a convex quadrilateral as both of its diagonals lies 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 B to D such that BO = OD.
Join AD and CD
Proof: AO = OC (As O is a midpoint of AC)
BO = DO (By construction)
∴ ABCD is a parallelogram.
(As diagonals bisect each other)
Also, ∠B = 90°
∴ ABCD is a rectangle
∴ Diagonal AC = BD
\(\frac{1}{2}\) AC = \(\frac{1}{2}\) BD
AO = OB
Also, AO = OC
∴ AO = OB = OC
∴ O is equidistant from A, B and C.

These NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 Questions and Answers are prepared by our highly skilled subject experts.

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

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

(i) AD = __________
(ii) ∠DCB = __________
(iii) OC = __________
(iv) m∠DAB + m∠CDA = __________
Solution:
(i) AD = BC (opposite sides are equal)
(ii) ∠DCB = ∠DAB (opposite angles are equal)
(iii) OC = OA (Diagonals bisect each other)
(iv) m∠DAB + m∠CDA = 180° (Adjacent angles are supplementary)

Question 2.
Consider the following parallelograms. Find the values of the unknown x, y, z.

Solution:
(i) ∠y = 100° (opposite angles of a parallelogram are equal)

∠x + 100° = 180° (adjacent angles in a parallelogram)
∠x = 180° – 100° = 80°
∠z = ∠x = 80° (opposite angles of parallelogram are equal)
∴ ∠x = 80°, ∠y = 100°, ∠z = 80°

(ii) Opposite angles are equal.

∠1 = 50° (opposite angles are equal)
∠1 + ∠z = 180° (Linear pair)
50 + ∠z = 180
∠z = 180° – 50°= 130°
x + ∠1 + y + 50° = 360°
⇒ x + 50° + y + 50° = 360°
⇒ x + y + 100° = 360°
⇒ x + y = 360° – 100°
⇒ x + y = 260°
∴ x = y (opposite angles of a parallelogram)
x = \(\frac{260^{\circ}}{2}\) = 130°
Thus x = 130°, y = 130° and z = 130°.

(iii) x = 90° (vertically opposite angles are equal)

x + y + 30° = 180°
(sum of the angles of a triangle = 180°)
90° + y + 30° = 180°
y + 120° = 180°
∠y = 180° – 120° = 60°
In the parallelogram ABCD
AD || BC and BD is a transversal.
∴ y = z (alternate angles are equal)
z = y = 60°
Thus x = 90°, y = 60° and z = 60°

(iv) In the parallelogram ABCD
y = 80° (opposite angles are equal)
AD || BC and CD is a transversal equal)

∴ z = 80° (corresponding angles are equal)
x + 80° = 180° (sum of the adjacent angle is 180°)
x = 180° – 80° = 100°
Thus, x = 100°, y = 80° and z = 80°

(v) y = 112° (in a parallelogram, opposite angles are equal)

In ΔACD,
x + y + 40° = 180° (sum of the angles of a triangle is 180°)
⇒ x + 112° + 40° = 180°
⇒ x + 152°= 180°
⇒ x = 180° – 152° = 28°
∠C + ∠B = 180° (sum of the adjacent angles of a parallelogram is 180°)
⇒ 40° + z + 112° = 180°
⇒ z + 152° = 180°
⇒ z = 180° – 152° = 28°
Thus, x = 28°, y = 112° and z = 28°.

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) In a quadrilateral ABCD
∠D + ∠B = 180° can be, but need not be
∴ The quadrilateral may be a parallelogram but not always

(ii) In a quadrilateral ABCD
AB = DC = 8 cm
AD = 4 cm
BC = 4.4 cm
∴ Opposite sides AD and BC are not equal.
∴ It cannot be a parallelogram.

(iii) In a quadrilateral ABCD
∠A = 70° and ∠C = 65°
∵ Opposite angles ∠A ≠ ∠C
∴ It cannot be a parallelogram.

Question 4.
Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution:
In the adjoining figure, ABCD is not a parallelogram such that opposite angles ∠B and ∠D are equal. It is a kite.

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 ABCD be a parallelogram in which adjacent angles ∠A and ∠B are 3x and 2x respectively.
Since adjacent angles are supplementary.
∴ ∠A + ∠B = 180°
⇒ 3x + 2x = 180°
⇒ 5x = 180°
⇒ x = \(\frac{180^{\circ}}{5}\) = 36°
∠A = 3 × 36° = 108°
and ∠B = 2 × 36° = 72°
Since opposite angles are equal.
∴ ∠D = ∠B = 72° and ∠C = ∠A = 108°
∴ ∠A = 108°, ∠B = 72°, ∠C = 108° and ∠D = 72°

Question 6.
Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram such that adjacent angles ∠A = ∠B
Since ∠A + ∠B = 180°
∠A = ∠B = \(\frac{180^{\circ}}{2}\) = 90°
Since opposite angles of a parallelogram are equal
∴ ∠A = ∠C = 90° and ∠B = ∠D = 90°
Thus, ∠A = 90°, ∠B = 90°, ∠C = 90° and ∠D = 90°

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:
y + z = 70°
(An exterior angles of a triangle is equal to the sum of the opposite interior angles)
∵ In ∆HOP
∠HOP = 180° – 70° = 110° (sum of the adjacent angle is 180°)
Now, x = ∠HOP = 110° (opposite angles of a parallelogram are equal)
EH || OP and PH is a transversal
∴ y = 40° (alternate angles are equal)
From (1),
y + z = 70°
⇒ 40° + z = 70°
⇒ z = 70° – 40° = 30°
Thus x = 110°, y = 40° and z = 30°

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

Solution:
(i) GUNS is a parallelogram.
GS = NU (opposite sides are equal)
3x = 18
⇒ x = \(\frac{18}{3}\) = 6
In the parallelogram GUNS
GU = SN (opposite sides are equal)
⇒ 3y – 1 = 26
⇒ 3y = 26 + 1
⇒ 3y = 27
⇒ y = \(\frac{27}{3}\) = 9
Thus, x = 6 cm and y = 9 cm.

(ii) RUNS is a parallelogram and the di-agonal RN and US bisect each other.
∴ y + 7 = 20
⇒ y = 20 – 7 = 13
x + y = 16
⇒ x + 13 = 16
⇒ x = 16 – 13 = 3
Thus, x = 3 cm and y = 13 cm

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

Solution:
RISK is a parallelogram
∠R + ∠K = 180° (adjacent angles of a parallelogram are supplementary)
∴ ∠R + 120° = 180°
∠R = 180° – 120° = 60°
In the parallelogram RISK
∠R = ∠S (opposite angles are equal)
∠S = 60°
CLUE is also a parallelogram.
∠E = ∠L = 70° (opposite angle of a parallelogram)
∠E = 70°
Now, in triangle ESO
∠E + ∠S + x = 180°
⇒ 70° + 60° + x = 180°
⇒ 130° + x = 180°
⇒ x = 180° – 130°
⇒ x = 50°

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

Solution:
∠KLM + ∠NML = 80° + 100° = 180°
∴ KL || NM (The sum of consecutive interior angles is 180°)
∴ The given figure KLMN is a trapezium.

Question 11.
Find m∠C in the figure given below if AB || DC.

Solution:
AB || DC and BC is a transversal,
∵ m∠B + m∠C = 180°
sum of interior angles is 180°
m∠C = 180° – m∠B
∴ m∠C = 180° – 120° = 60°

Question 12.
Find the measure of ∠P and ∠S if SP || QR in figure (if you find m∠R, is there more than one method to find m∠P)

Solution:
PQRS is a trapezium such that SP || RQ and PQ is a transversal.
∴ m∠P + m∠Q = 180° (Interior angles are supplementary)
m∠P + 130° = 180°
m∠P = 180° – 130° = 50°
Also, m∠S + m∠R = 180°
⇒ m∠S + 90° = 180°
⇒ m∠S = 180° – 90°
⇒ m∠S = 90°
m∠P + m∠Q + m∠R + m∠S = 360° (sum of the angles of a quadrilateral is 360°)
⇒ m∠P + 130° + 90° + 90° = 360°
⇒ m∠P + 130° + 90° + 90° = 360°
⇒ m∠P + 130° + 90° + 90° = 360°
⇒ m∠P + 310° = 360°
⇒ m∠P = 360° – 310°
⇒ m∠P = 50°
Hence, m∠P = 50° and m∠S = 90°

These NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.3

Question 1.
If m – 2, find the value of:
(i) m – 2
(ii) 3m – 5
(iii) 9 – 5m
(iv) 3m² – 2m – 7
(v) \(\frac { 5m }{ 2 }\) – 4
Answer:
(i) m- 2 = 2 – 2 (m = = 2)
= 0

(ii) 3m – 5 = 3 (2) – 5 (m = – 2)
= 6 – 5 = 1

(iii) 9 – 5m = 9 – 5(2) (m = = 2)
= 9- 10
= -1

(iv) 3m² – 2m – 7
= 3 (2)² – 2(2) – 7
(m = 2)
= 3 (4) – 4 – 7
= 12 – 4 – 7
= 12 – 11
= 1

(v) \(\frac { 5m }{ 2 }\) – 4 = \(\frac{5(2)}{2}\) – 4 (m = 2)
= 5 – 4
= 1

Question 2.
If p = -2, find the value of:
(i) 4p + 7
(ii) – 3p² + 4p + 7
(iii) – 2p² – 3p² + 4p + 7
Answer:
(i) 4p + 7 = 4(-2) + 7
= – 8 + 7 = -1

(ii) -3p² + 4p + 7
= – 3 (-2)² + 4(-2) + 7
(when p = -2)
= – 3 (4) + (- 8) + 7
= -12 – 8 + 7
= -20 +7
= -13

(iii) -2p³ – 3p² + 4p + 7 =
-2 (-2)³ – 3 (-2)² + 4 (-2) + 7
(when p = -2)
= – 2(- 8) — 3 (4) + (— 8) + 7 = + 16 – 12 – 8 + 7 = 23 – 20 = 3

Question 3.
Find the value of the following expressions, when x = -1:
(i) 2x – 7
(ii) -x + 2
(iii) x² + 2x + 1
(iv) 2x² – x -2
Answer:
(i) 2x – 7 = 2(-1) – 7 (whenx = -1)
= -2 – 7 = -9

(ii) -x + 2 = -(-1) + 2 (When x = -1)
= 1 + 2 = 3

(iii) x² + 2x + 1 = (-1)² + 2 (-1) + 1
(When x = – 1)
= 1 – 2 + 1
= 2 – 2 = 0

(iv) 2x² – x – 2 =2 (-1)² – (-1) – 2
(When x = -1)
= 2 + 1 – 2
= 3 – 2 = 1

Question 4.
If a = 2, b = – 2, find the value of:
(i) a² + b²
(ii) a² + ab + b²
(iii) a² – b²
Answer:
(i) a² + b² = (2)² + (-2)²
(When a = 2, b = -2)
= 4 + 4 = 8

(ii) a² + ab + b²
= (2)² + 2(-2) + (-2)²
(When a = 2, b = – 2)
= 4 – 4 + 4
= 8 – 4 = 4

(iii) a2 – b2 = 22 – (-2)2
(When a = 2, b = -2) = 4-(4) = 4- 4 = 0

Question 5.
When a = 0, b = – 1, find the value of the given expressions:
(i) 2a + 2b
(ii) 2a² + b² + 1
(iii) 2a²b + 2ab² + ab
(iv) a² + ab + 2
Answer:
(i) 2a + 2b = 2 (0) + 2(—1)
(When a = 0, b = -1)
= 0 – 2 = -2

(ii) 2a² + b² +1
= 2 (0)² + (-1)² + 1
(When a = 0, b = -1)
= 0 + (1) + (1)
= 1 + 1 = 2

(iii) 2a²b + 2ab² + ab
= 2(0)² (-1) + 2(0) (-1)² + (0) (-1)
(When a = 0, b = -1)
= 2(0) (-1) + 2(0) (1) + 0
= 0 + 0 + 0
= 0

(iv) a² + ab + 2 = (0)² + 0 (-1) + 2
(When a = 0, b = -1)
= 0 – 0 + 2 = 2

Question 6.
Simplify the expressions and find the value if x is equal to 2.
(i) x + 7 + 4 (x – 5)
(ii) 3 (x + 2) + 5x – 7
(iii) 6x + 5 (x – 2)
(iv) 4(2x – l) + 3x+ 11
Answer:
(i) x + 7 + 4 (x -5)
= x + 7 + 4x – 20
= x + 4x + 7 – 20
= 5x – 13 (When x = 2)
= 5 (2) – 13
= 10- 13
= -3

(ii) 3 (x + 2) + 5x – 7 = 3x + 6 + 5x – 7
= 3x + 5x + 6 – 7
= 8x – 1 (When x = 2)
= 16 – 1 = 15

(iii) 6x + 5 (x – 2) = 6x + 5x – 10
= 1 lx – 10 (When x = 2)
= 11 (2) – 10 = 22-
= 12

(iv) 4(2x – 1)+ 3x + 11
= 8x + 3x – 4 + 11
= 8x + 3x + 11 – 4
= 11x + 7 (When x = 2)
h = 11(2) + 7
= 22 + 7 = 29

Question 7.
Simplify these expressions and find their values if x = 3, a = -1, b = -2.
(i) 3x – 5 – x + 9
(ii) 2 – 8x + 4x + 4
(iii) 3a + 5 – 8a + 1
(iv) 10 – 3b – 4 – 5b
(v) 2a – 2b – 4 – 5 + a
Answer:
(i) 3x-5-x + 9
= 3x-x-5 + 9
= 2x + 4 (When x = 3)
= 2(3)+ 4
= 6 + 4
= 10

(ii) 2-8x + 4x + 4
= -8x + 4x + 4 + 2
= (- 8 + 4) x + 6
= – 4x + 6
(When x = 3)
= – 4(3) + 6
= -12+ 6 =-6

(iii) 3a + 5 – 8a + 1
= 3a – 8a + 5 + 1
= (3 – 8) a + 6
= -5a + 6
(When a = – 1)
= -5 (-1) + 6
= 5 + 6= 11

(iv) 10 – 3b – 4 – 5b = -3b -5b +10-4
= (-3 -5) b + 6 = – 8 b + 6
(When b = – 2) = – 8 (-2) + 6
= 16 + 6 = 22

(v) 2a – 2b – 4 – 5 + a =
2a + a – 2b – 4 -5 = 3a – 2b – 9
(When a = – 1 and b = -2) = 3 (-1) -2 (-2) -9 = -3 + 4 -9 = -12+ 4 =-8

Question 8.
(i) If z = 10, find the value of z³ – 3 (z – 10).
(ii) If p = – 10, find the value of p² – 2p – 100
Answer:
(i) z³ – 3 [z – 10] = 10² – 3 [10- 10]
(When z = 10)
= 1000 – 3 (0)= 1000

(ii) p² – 2p – 100
= (-10)² – 2 (-10) – 100
(When p = -10)
= 100 + 20 – 100
= 120 – 100 = 20

Question 9.
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0? Ans: 2x²+ x – a = 5 (Given: x = 0)
2(0)² + 0 – a = 5
0 + 0 – a = 5
-a =5
a = – 5
The value of a = – 5

Question 10.
Simplify the expression and find its value when a = 5 and b = – 3.
2(a² + ab) + 3 – ab
Answer:
2(a² + ab) + 3 – ab
= 2 [5² + 5(-3)] + 3 -(5) (-3)
= 2 [25- 15] +3 + 15
= 2(10) + 18
= 20 + 18 = 38

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.1

Question 1.
Complete the last column of the table.
Answer:

Question 2.
Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (n = 1)
(b) 7n + 5 = 19 (n = – 2)
(c) 7n + 5 = 19 (n = 2)
(d) 4p – 3 = 13 (p = 1)
(e) 4p – 3 = 13 (p = – 4)
(f) 4p – 3 = 13 (p = 0)
Answer:
(a) n + 5 = 19
1 + 5 =19 (Putting n = 1)
6 ≠ 19
∴ n = 1 is not a solution.

(b) 7n + 5 =19
7(- 2) + 5 =19 (Put n = – 2)
-14 + 5 = 19
-9 ≠ 19
n = -2 is not a solution.

(c) 7n + 5 =19
7(2) + 5=19 (Put n = 2)
14 + 5 = 19
19 = 19
∴ n = 2 is a solution.

(d) 4p – 3 = 13
4(1) – 3 =13 (Put p = 1)
4 – 3 = 13
1 ≠ 13
∴ p = 1 is not a solution.

(e) 4p – 3 = 13
4(- 4) – 3 = 13 (Put p = – 4)
-16 -3 = 13
-19 ≠ 13
∴ p = – 4 is not a solution.

(f) 4p – 3 = 13
4(0) – 3 = 13 (Put p = 0)
0 – 3 = 13
-3 ≠ 13
p = 0 is not a solution.

Question 3.
Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m – 14 = 4
Answer:
(i) 5p + 2 = 17
Put p = 0
L.H.S = 5 (0) + 2
= 0 + 2
= 2 ≠ RHS

Put p = 1
L.H.S = 5(1)+ 2
= 5 + 2
= 7 ≠ RHS

Put p = -1
L.H.S = 5 (-1) + 2
= -5 + 2
= -3 ≠ RHS

Put p = 2
L.H.S = 5 (2) + 2
= 10 + 2
= 12 ≠ RHS

Put p = -2
L.H.S = 5 (-2) + 2
= -10 + 2
= -8 ≠ 17 RHS

Put p = 3
L.H.S = 5 (3) + 2
= 15 + 2
= 17 = RHS
p = 3 is the solution of 5P + 2 = 17

(ii) 3m – 14 =4
Put m = 0
L.H.S = 3m – 14
= 3 (0) – 14
= – 14
– 14 ≠ 4
L.H.S ≠ R.H.S

Put m = 1
L.H.S = 3m -14
= 3(1) – 14
= 3 – 14
= – 11 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 2
L.H.S = 3m – 14
= 3 (2) – 14
= 6 – 14
= – 8 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 3
L.H.S = 3(3) – 14
= 9 – 14
= – 5 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 4
L.H.S = 3(4) – 14
= 12 – 14
= – 2 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 5
L.H.S = 3(5) – 14
= 15 – 14
= 1 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 6
L.H.S = 3(6) – 14
= 18 – 14
= 4 = R.H.S
L.H.S = R.H.S
m = 6 is the solution to 3m – 14 = 4

Question 4.
Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of t is 15.
(vi) Seven times m plus 7 gets you 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.
Answer:
(i) x + 4 = 9
(ii) y – 2 = 8
(iii) 10 a = 70
(iv) \(\frac { b }{ 5 }\) = 6
(v) \(\frac { 3t }{ 4 }\) = 15
(vi) 7m + 7 = 77
(vii) \(\frac { x }{ 4 }\) x – 4 = 4
(viii) 6 y – 6 = 60
(ix) \(\frac { z }{ 3 }\) + 3 = 30

Question 5.
Write the following equations in statement forms:
(i) p + 4 = 15
(ii) m – 7 = 3
(iii) 2m = 7
(iv) \(\frac { m }{ 5 }\) = 3
(v) \(\frac { 3m }{ 5 }\) = 6
(vi) 3p + 4 = 25
(vii) 4p – 2 = 18
(viii) \(\frac { p }{ 2 }\) + 2 = 8
Answer:
(i) The sum of p and 4 is 15.
(ii) 7 subtracted from m is 3.
(hi) Twice a number m is 7.
(iv) One-fifth of a number m is 3.
(v) Three-fifth of a number m is 6.
(vi) Three times a number p when added to 4 gives 25.
(vii) 2 subtracted from four times a number p is 18.
(viii) 2 added to half of a number p is 8.

Question 6.
Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than live times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)

(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be 1.)

(iv) In an isosceles triangle, the verte x angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).
Answer:
(i) Let Parmit has m marbles.
Then, five time the marbles Parmit has = 5 m.
Irfan has 7 marbles more than five times the marbles parmit has
So, Irfan has (5m + 7) marbles
But it is given that Irfan has 37 marbles.
5m + 7 = 37

(ii) Let Laxmi’s age = y years
3 times Laxmi’s age = 3y years.
Age of Laxmi’s father = 3 times
Laxmi’s age + 4 years
= 3y + 4 years
But Laxmi’s father is 49 years old.
3y + 4 = 49

(iii) Let the lowest score (marks) = 1
Twice the lowest marks = 2 l
Since highest marks = (twice the lowest marks) + 7 = 2l + 7
But the highest marks = 87
2l + 7 = 87

(iv) Let the base angle be b degrees
The base angle of an isosceles triangle are equal.
The other base angle = b degrees
Since the vertex angle = Twice either base angle = 2b degrees.
Also, the sum of three angles of triangle = 180°
b + b + 2b = 180°
(OR)
4b = 180°

These NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.2

Question 1.
Simplify combining like terms:
(i) 21b – 32 + 7b – 20b
(ii) – z² + 13z² – 5z + 7z³ – 15z
(iii) p – (p – q) – q – (q – p)
(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
(vi) (3y² + 5y – 4) – (8y – y² – 4)
Answer:
(i) 21b – 32 + 76 – 20b
Combining the like terms, we have
(21b + 7b – 20b) + (- 32) = (21 + 7 – 20) b + (- 32)
= 8b – 32

(ii) -z² + 13z² – 5z + 7z³ – 15z
Combining like terms, we get
= 7z³ + (-z² + 13z²) + (-5z – 15z)
= 7z³ + (-1 + 13)z² + (- 5 – 15)z
= 7z³ + 12z² – 20z

(iii) p-(p-q)-q-(q-p) = p- p + q – q – q + p
Combining like terms, we get
= p – p + p + q – q – q
= (1 – 1 + 1) p + (1 – 1 – 1) q
= p + (-1)q
= p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b -a
Combining like terms, we get
= 3a – a – a -2b + b + b – ab – ab + 3ab
= (3 – 1 – 1) a + (-2 + 1 + 1) b + (-1 -1 + 3)ab
= (3 – 2) a + (-2 + 2) b + (-2 + 3) ab
= (1) a + (0) b + (1) ab
= a + ab

(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
Combining the like terms, we get
(5x²y + 3yx²) + 8xy² + (-5x² + x² ) + (-3y² – y² – 3y² )
(5 + 3) x²y + 8xy² + (-5 + 1) x² + (-3 – 1 – 3)y²
8x²y + 8xy² + (- 4) x² + (-7) y²
8x²y + 8xy² – 4x² – 7y²

(vi) (3y² + 5y – 4)-(8y – y² – 4)
= 3y² + 5y – 4 – 8y + y² + 4
Combining the like terms, we get
= 3y² + y² + 5y – 8y – 4 + 4
= (3 + 1)y² + (5-8)y + (-4 + 4)
= 4y² – 3y

Question 2.
Add:
(i) 3mn, – 5mn, 8mn, – 4mn
(ii) t – 8tz, 3tz – z, z – t
(iii) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
(iv) a + b-3, b-a + 3, a-b + 3
(v) 14x + lOy – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii) 4x²y, – 3xy², – 5xy², 5x²y
(viii) 3p²q² – 4pq + 5, – 10p²q², 15 + 9pq + 7p²q²
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x² – y² – 1, y² -1 – x², 1 – x² – y²
Answer:
(i) 3mn, – 5mn, 8mn, – 4mn
3mn + (-5mn) + 8mn + (- 4mn)
= 3mn – 5mn + 8mn – 4mn
= (3 – 5 + 8 – 4) mn = 2mn

(ii) t – 8tz, 3tz – z; z -t
t – 8tz + 3tz + (-z) + z + (-t)
= t – 8tz + 3tz – z + z – t
Combining like terms, we get
= t – t-z + z – 8tz + 3tz
= (1 – 1) t + (- 1 + 1) z + (- 8 + 3) tz
= (0) t + (0) z + (-5) tz
= 0 + 0 – 5tz
= -5tz

(iii) -7mn + 5; 12mn + 2; 9mn – 8; – 2mn -3
-7mn + 5 + 12mn + 2 + 9mn – 8 + (-2mn – 3)
– 7mn + 5 + 12mn + 2 + 9mn – 8 -2mn – 3
Combining like terms, we get
– 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3
= (-7 + 12 + 9-2) mn + (5 + 2 – 8 -3)
= (21-9) mn + (7 – 11)
= (12) mn + (- 4)
= 12mn – 4

(iv) a + b-3;b-a + 3;a-b + 3
a + b – 3 + b – a + 3 + a – b + 3
Combining like terms, we get
= (a – a + a) + (b + b – b) + (-3 + 3 + 3)
= (1 – 1 + 1) a + (1 + 1 – 1)b + (6 – 3)
= (2 – 1) a + (2 -1) b + 3
= a + b + 3

(v) 14x + 10y – 12xy – 13; 18 – 7x – lOy + 8xy, 4xy
14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
Combining like terms, we get
= (14x – 7x) + (10y – 10y) + (-12xy + 8xy + 4xy)
= (14 – 7)x + (10 – 10) y + (-12 + 8 + 4) xy+ (-13 + 18)
= 7x + (0)y + (0)xy + (+5)
= 7x + 5

(vi) 5m – 7n, 3n – 4m + 2; 2m – 3mn – 5
5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
Combining like terms, we get
= (5m – 4m + 2m) + (-7n + 3n) + (-3mn) + (2 – 5)
= (5 – 4 + 2) m + (-7 + 3) n – 3mn + (-3)
= (7 – 4) m + (- 4) n – 3mn -3
= 3m – 4n – 3mn – 3

(vii) 4x²y; – 3xy², – 5xy²; 5x²y
4x²y + (-3xy²) + (-5xy²) + 5x²y
= 4x²y – 3xy² – 5xy² + 5x²y
Combining like terms, we get
= 4x²y + 5x²y – 5xy² – 5xy²
= (4 + 5) x²y + (- 3 – 5) xy²
= 9x²y + (- 8) xy²
= 9x²y – 8xy²

(viii) 3p²q² – 4pq + 5; – 10p²q²; 15 + 9pq + 7p²q²
3p²q² + (- 4pq) + 5 + (-10p²q²) + 15 + 9pq + 7p²q²
= 3p²q² – 4pq + 5 – 10p²q² + 15 + 9pq + 7p²q²
Combining like terms, we get
= 3p²q² – 10p²q² + 7p²q² – 4pq + 9pq + 5 + 15
= (3 – 10 + 7) p²q² + (- 4 + 9) pq + (5 + 15)
= (10 – 10) p²q² + (5) pq + 20
= (0) p²q² + 5pq + 20
= 5pq + 20

(ix) ab – 4a; 4b – ab; 4a – 4b
ab – 4a + 4b – ab + 4a – 4b = (ab – ab) + (- 4a + 4a) + (4b – 4b)
= (1 – 1) ab + (- 4 + 4) a + (4 – 4) b
= (0) ab + (0) a + (0) b
= 0 + 0 + 0
= 0

(x) x² – y² – 1; y² – 1 – x²; 1 – x² – y²
x²– y² – 1 + y² – 1 – x² + 1 – x² – y²
= (x² – x² – x²) + (- y² + y² – y²) + (-1 – 1 + 1)
= (1 – 1 – 1)x² + (-2 + 1)y² + (-1)
= (1 – 2) x² + (-2 + 1)y² + (-1)
= (- 1) x² + (-1) y² +(-1)
= – x² – y² – 1

Question 3.
Subtract:
(i) – 5y² from y²
(ii) 6xy from -12xy
(iii) (a – b) from (a + b)
(iv) a (b – 5) from b (5 – a)
(v) -m² + 5 mn from 4m² – 3mn + 8
(vi) – x² + 10x – 5 from 5x – 10
(vii) 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
Answer:
(i) Subtract – 5y² from y²
y² – (-5y² ) = y² + 5y² – 12xy – (6xy) = 6y²

(ii) Subtract 6xy from – 12xy = – 12xy – 6xy
= (-12 – 6)xy = -18xy

(iii) Subtract (a – b) from (a + b)
(a + b)-(a-b) = a + b- a + b =a-a+b+b
= (1 – 1)a + (1 + 1)b
= (0)a + 2b
= 0 + 2b
= 2b

(iv) Subtract a (b – 5) from b (5 – a) b (5 – a) – a (b – 5)
= 5b – ab – ab + 5a
= 5b + 5a – ab – ab
= 5a + 5b + (-1 -1) ab
= 5a + 5b – 2ab

(v) Subtract -m² + 5mn from 4m² – 3mn + 8
4m² – 3mn + 8 – (-m² + 5mn)
= 4m² – 3mn + 8 + m² – 5mn
= 4m² + m² – 3mn – 5mn + 8
= (4 + 1) m² + (-3 – 5) mn + 8 = 5m² + (- 8) mn + 8
= 5m² – 8mn + 8

(vi) Subtract -x² + 10x – 5 from 5x – 10
5x – 10 – (-x² + 10x – 5)
= 5x – 10 + x² – 10x + 5
= x² + 5x – 10x – 10 + 5
= x² + (5 – 10) x + (-10 + 5)
= x² + (-5) x + (-5)
= x² – 5x – 5

(vii) Subtract 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
3ab – 2a² – 2b² – (5a² – 7ab + 5b²)
= 3ab – 2a² – 2b² – 5a² + 7ab – 5b²
= -2a² – 5a² – 2b² – 5b² + 3ab + 7ab
= (-2 -5) a² + (-2 – 5) b² + (3 + 7) ab
= -7a² – 7b² + 10ab (or) 10ab – 7a² – 7b²

(viii) Subtract 4pq – 5q² – 3p² from 5p² + 3q² – pq
5p² + 3q² – pq – (4pq – 5q² – 3p²)
= 5p² + 3q² – pq – 4pq + 5p² + 3p²
= (5 + 3) p² + (3 + 5)q² – pq – 4pq
= 8p² + 8q² -5pq

Question 4.
(a) What should be added to x² + xy + y² to obtain 2x² + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16?
Answer:
(a) The required expression is
2x² + 3xy – (x² + xy + y²)
= 2x² + 3xy – x² – xy – y²
= (2x² – x²) – y² + 3xy – xy
= (2 – 1) x² – y² + (3 – 1) xy
= (1) x² – y² + (2) xy
= x² + 2xy – y²

(b) The required expression is
(2a + 8b + 10) – (- 3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= (2a + 3a) + (8b – 7b) + 10- 16
= (2 + 3) a + (8 – 7) b + (10 – 16)
= (5)a + (l)b + (-6)
= 5a + b – 6

Question 5.
What should be taken away from 3x² – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20?
Answer:
The required expression is
3x² – 4y² + 5xy + 20 –
(-x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 – (-x2 -y2 + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20)
= (3x² + x²) + (- 4y² + y²) +
(+ 5xy – 6xy) + (20 – 20) = (3 + 1) x2 + (- 4 + 1) y2 +
(+ 5 – 6) xy + (0)
= (4) x² + (-3) y² + (-1xy)
= 4x² – 3y² – xy

Question 6.
(a) From the sum of 3x – y + 11 and – y -11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and -x² + 2x + 5.
Answer:
(a) The required expression is (3x-y+ 11) + (- y-11) –
(3x – y -11)
= 3x – y + 11 – y – 11 – 3x + y + 11
= (3x – 3x) + (-y – y + y) + (11 – 11 + 11)
= (3 – 3)x + (-1 -1 + 1) y + (11 – 11 + 11)
= (3 – 3)x + (-2 + 1)y + (22 – 11)
– (0) x + (-1) y + (11)
= 0 – y + 11 = -y + 11

(b) The required expression is
(4 + 3x) + (5 – 4x + 2x²) –
[(3x² – 5x + (-x² + 2x + 5)]
= 4 + 3x + 5 – 4x + 2x² – [3x² – 5x – x2 + 2x + 5]
= 4 + 3x + 5 – 4x + 2x² – 3x² +5x + x2 – 2x – 5
= (2x² – 3x² + x²) + (3x -4x + 5x- 2x) + (4 + 5 – 5)
= (2 – 3 + 1) x² + (3 – 4 + 5 – 2) x + (9-5)
= (3 – 3) x² + (8 – 6) x + 4
= (0)x² + (2) x + 4
= 0 + 2x + 4
= 2x + 4

These NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.2

Question 1.
Find x in the following figures.

Solution:
(a) Sum of all the exterior angles of a polygon = 360°
⇒ 125° + 125° + x° = 360°
⇒ 250° + x = 360°
⇒ x = 360° – 250°
⇒ x = 110°

(b) x + 90° + 60° + 90° + 70° = 360°
⇒ x + 310° = 360°
⇒ x = 360° – 310°
⇒ x = 50°

Question 2.
Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
(ii) 15 sides
Solution:
(i) Number of sides (n) = 9
Number of exterior angles = 9
The given polygon is a regular polygon
All the exterior angles are equal
Measure of an exterior angle = \(\frac{360^{\circ}}{9}\) = 40°

(ii) Number of sides of regular polygon = 15
Number of equal exterior angles =15
The sum of all the exterior angles = 360°
The measure of each exterior angle = \(\frac{360^{\circ}}{15}\) = 24°

Question 3.
How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Solution:
For a regular polygon, measure of each angle is equal
Sum of all the exterior angles = 360°
Measure of an exterior angle = 24°
Number of sides = \(\frac{360^{\circ}}{24^{\circ}}\) = 15
Thus, there are 15 sides of the regular polygon.

Question 4.
How many sides does a regular polygon have if each of its interior angles is 165°?
Solution:
The given polygon is a regular polygon.
Each interior angle = 165°
Each exterior angle =180° – 165° = 15°
Number of sides = \(\frac{360^{\circ}}{15^{\circ}}\) = 24
Thus, there are 24 sides of the polygon.

Question 5.
(a) Is it possible to have a regular polygon with measure of each exterior angle 22°?
(b) Can it be an interior angle of a regular polygon? Why?
Solution:
(a) Each exterior angle = 22°
∴ Number of sides = \(\frac{360^{\circ}}{22^{\circ}}=\frac{180^{\circ}}{11^{\circ}}\)
If it is a regular polygon, then its number of sides must be a whole number.
Here \(\frac{180^{\circ}}{11^{\circ}}\) is not a whole number.
∴ 22° cannot be an exterior angle of a regular polygon.

(b) If 22° is an interior angle, then
(180° – 22°) = 158° is an exterior angle.
∴ Number of sides = \(\frac{360^{\circ}}{158^{\circ}}=\frac{180^{\circ}}{79^{\circ}}\)
\(\frac{180^{\circ}}{79^{\circ}}\) is not a whole number
∴ 22° cannot be an interior angle of a regular polygon.

Question 6.
(a) What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?
Solution:
(a) The minimum number of sides of a polygon = 3
The regular polygon of 3 sides is an equilateral triangle.
∴ Each interior angle of an equilateral triangle is 60°.

(b) The sum of an exterior angle and its corresponding interior angle is 180°.
Minimum interior angle of a regular polygon is 60°.
∴ The maximum exterior angle of a regular polygon = 180° – 60° = 120°