Category: Class 7

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 12 Quadrilaterals Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 12th Lesson Quadrilaterals Exercise 1

Question 1.
In quadrilateral PQRS
(i) Name the sides, angles, vertices and diagonals.
(ii) Also name all the pairs of adjacent sides, adjacent angles, opposite sides and opposite angles. .

Solution:
ï) In quadrilateral PQRS,
sides = PQ, QR, RS and SP
angles = ∠P, ∠Q, ∠R, and ∠S
vertices = P, Q, R and S
diagonals = PR and QS
ii) Pairs of adjacent sides = (PQ, QR), (QR. RS), (RS, SP), (SP, PQ)
Pairs of adjacent angles (∠P, ∠Q). (∠Q, ∠R), (∠R, ∠S), (∠S, ∠P)
Pairs of opposite sides = (PQ, RS) and (QR, PS)
Pairs of opposite angles = (∠P,∠R) and (∠Q, ∠S)

Question 2.
The three angles of a quadrilateral are 60°, 80° and 1200. Find the fourth angle?
Solution:
Given three angles = 60°,80°,120°
Sum of the given three angles = 60°+ 80° + 120° = 260°
Sum of the four Interior angles of a quadrilateral = 360°
∴ The fourth angle = 360° — 260° = 100°

Question 3.
The angles of a quadrilateral are in the ratio 2 : 3 : 4: 6. Find the measure of each of the four angles.
Solution:

Given that ratio of the four angles = 2 : 3 : 4 : 6
Sum of the terms in the ratio = 2 + 3 + 4 + 6 = 15
Sum of the four angles = 360°
∴ 1st angle = \(\frac { 2 }{ 15 }\) x 360° = 48°
2nd angle = \(\frac { 3 }{ 15 }\) x 360° = 72°
3rd angle = \(\frac { 4 }{ 15 }\) x 360° = 96°
4th angle = \(\frac { 6 }{ 15 }\) x 360° = 144°

Question 4.
The four angles of a quadrilateral are equal. Draw this quadrilateral in your notebook.
Find each of them.
Solution:

Sum of the four angles in a quadrilateral = 360°
Each of the angle = \(\frac{360^{\circ}}{4}\) = 90°
∠A = ∠B = ∠C = ∠D = 90°

Question 5.
In a quadrilateral, the angles arex°, (x + l0)°, (i+ 20)°, (x + 30)°. Find the angles.
Solution:
Given angles are x°, (x + 10)°, (x + 20)°, (x + 30)°
Sum of the four angles = x + x + 10° + x + 20° + x + 30° = 4x + 60°
But the sum of the four angles = 360°
4 x 60° = 60° = 360°
4x = 360° – 60° = 300°
x = \(\frac{300^{\circ}}{4}\) = 75°
∴ The angles are x = 750
x + 10°- 75°+ 10° = 85°
x + 20° = 75° +20° = 95°
x + 30° = 75° + 30° = 105°

Question 6.
The angles of a quadrilateral cannot be in the ratio 1: 2 : 3 : 6. Why? Give reasons.
(Hint: Try to draw a rough diagram of this quadrilateral)
Solution:
Given that the angles of a quadrilateral cant be in the ratio 1: 2 : 3 : 6
If the ratio is 1 : 2 : 3 : 6 then
the sum of terms of ratio = I + 2 + 3 + 6 = 12
Sum of the angles = 360°
∴ 1st angle = \(\frac{1}{12}\) x 360° = 30°
2nd angle = \(\frac{2}{12}\)x 360° = 60°
3rd angle = \(\frac{3}{12}\) x 360° = 90°
4th angle = \(\frac{3}{12}\) x 360° = I80
i.e., 4th angle is 180°, a straight angle.
A quadrIlateral cant he formed with these angles.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 11 Exponents Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 11th Lesson Exponents Exercise 3

Question 1.
Express the number appearing in the following statements in standard form.
(i) The distance between the Earth and the Moon is approximately 384,000,000m.
(ii) The universe is estimated to be about 12,000,000,000 years old.
(iii) The distance of the sun from the center of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000.,000 m.
(iv) The earth has approximately 1,353,000,000 cubic km of sea water.
Solution:
(i) 3.84 × 10⁸ m
(ii) 1.2 × 10¹⁰ years
(iii) 3 × 10²⁰ m
(iv) 1.353 × 10⁹cubic km

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 11 Exponents Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 11th Lesson Exponents Exercise 2

Question 1.
Simplify the following using laws of exponents.
(i) 2¹⁰ × 2⁴
(ii) (3²) × (3²)⁴
(iii) \(\frac{5^{7}}{5^{2}}\)
(iv) 9² × 9¹⁸ × 9¹⁰
(v) \(\left(\frac{3}{5}\right)^{4} \times\left(\frac{3}{5}\right)^{3} \times\left(\frac{3}{5}\right)^{8}\)
(vi) (-3)³ × (-3)¹⁰ × (-3)⁷
(vii) 3(²)²
(viii) 2⁴ × 3⁴
(ix) 2^4a × 2^5a
(x) (10²)³
(xi) \(\left[\left(\frac{-5}{6}\right)^{2}\right]^{5}\)
(xii) 2^3a+7 × 2^7a+3
(xiii) \(\left(\frac{2}{3}\right)^{5}\)
(xiv) (-3)³ × (-5)³
(xv) \(\frac{(-4)^{6}}{(-4)^{3}}\)
(xvi) \(\left(\frac{2}{3}\right)^{5}\)
(xvii) \(\frac{(-6)^{5}}{(-6)^{9}}\)
(xviii) (-7)⁷ × (-7)⁸
(xix) (-6⁴)⁴
(xx) a^x × a^y × a ^x
Solution:
(i) 2¹⁰ × 2⁴ = 2¹⁰⁺⁴ = 2¹⁴
[∵ a^m × aⁿ = a^m+n]

(ii) (3²) × (3²)⁴ = (3²)¹⁺⁴ = (3²)⁵
= 3^2×5
= 3¹⁰
[∵ a^m × aⁿ = a^m+n]
[∵ (a^m)ⁿ = (a)^mn]

(iii) \(\frac{5^{7}}{5^{2}}\) = 5^{7 – 2} = 5⁵
= 5 × 5 × 5 × 5 × 5 = 5⁵
[∵ \(\frac{a^{m}}{a^{n}}\) = a^m-n, m > n]

(iv) 9² × 9¹⁸ × 9¹⁰ = 9^2+18+10 = 9³⁰
[∵ a^m × aⁿ = a^m+n]

(v) \(\left(\frac{3}{5}\right)^{4} \times\left(\frac{3}{5}\right)^{3} \times\left(\frac{3}{5}\right)^{8}\) = \(\left(\frac{3}{5}\right)^{4+3+8}\) = \(\left(\frac{3}{5}\right)^{15}\)
[∵ a^m × aⁿ = a^m+n]

(vi) (-3)³ × (-3)¹⁰ × (-3)⁷ = (-3)^{3 + 10 + 7} = (-3)²⁰
[∵ a^m × aⁿ = a^m+n]

(vii) 3(²)² = 3^2×2 = 3⁴
[∵ (a^m)ⁿ = a^mn])

(viii) 2⁴ × 3⁴ = (2 × 3 )⁴ = 6⁴
[∵ a^m × b^m = (ab)^m]

(ix) 2^4a × 2^5a = 2^4a+5a = 2^9a
[∵ a^m × aⁿ = a^m+n]

(x) (10²)³ = 10^2×3 = 10⁶
[∵ (a^m)ⁿ = a^m×n ]

(xi) \(\left[\left(\frac{-5}{6}\right)^{2}\right]^{5}\)
\(\left[\left(\frac{-5}{6}\right)^{2}\right]^{5}=\left[\frac{-5}{6}\right]^{2 \times 5}\) = \(\left(\frac{-5}{6}\right)^{10}=\left(\frac{5}{6}\right)^{10}\) [∵ 10 even number]
[∵ (a^m)ⁿ = an]

(xii) 2^3a+7 × 2^7a+3
2^3a+7+7a+3 = = 2^10a+10 = 2^10(a+1)
[∵ a^m × aⁿ = (a)^m+n]

(xiii) \(\left(\frac{2}{3}\right)^{5}\) = \(\frac{2^{5}}{3^{5}}\)
[∵ \(\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}}\) ]

(xiv) (-3)³ × (-5)³ = [(-3) × (-5)]³ = (15)³
[∵ a^m × b^m = (ab)^m]

(xv) \(\frac{(-4)^{6}}{(-4)^{3}}\) = (-4)³
[∵ \(\frac{a^{m}}{a^{n}}\) = a^m-n, m > n]

(xvi) \(\left(\frac{2}{3}\right)^{5}\) = \(\frac{1}{9^{15-7}}\) = \(\frac{1}{9^{8}}\) [∵ \(\frac{a^{m}}{a^{n}}=\frac{1}{a^{n-m}}\), n > m]

(xvii) \(\frac{(-6)^{5}}{(-6)^{9}}\) = \(\frac{1}{(-6)^{9-5}}\) [∵ \(\frac{a^{m}}{a^{n}}=\frac{1}{a^{n-m}}\), n > m]
\(\frac{1}{(-6)^{4}}=\frac{1}{6^{4}}\) [∵4 is even number ]

(xviii) (-7)⁷ × (-7)⁸ = (-7)⁷⁺⁸ [∵ a^m × aⁿ = a ^m+n]
= (-7)¹⁵ = -(7)¹⁵ [∵ 15 is odd number ]

(xix) (-6⁴)⁴ [∵ (a^m)ⁿ = a^mn]
= (-6)^4×4 = (-6)¹⁶ = 6¹⁶
[∵ 16 is even number ]

(xx) a^x × a^y × a ^x = a^x+y+z
[a^m × aⁿ × a^p = a^m+n+p]

Question 2.
By what number should 3 be multiplied so that the product is 729’?
Solution:
Given number = 3^-4
Given product = 729 [∵ 3⁶ = 729]
Let the number to be multiplied be x then
⇒ (3^-4) . (x) = 3⁶ ⇒ \(\frac{x}{3^{4}}\) =3⁶
[∵ a^m × nⁿ = a^m+n]
⇒ x = 3⁶ × 3⁴ = 3¹⁰

Question 3.
1f 5⁶ × 5^2x = 5¹⁰ then find x.
Solution:
Given that 5⁶ × 5^2x = 5¹⁰
⇒ 5^6+2x = 5¹⁰ [∵ a^m × aⁿ = a^m+n]
Since bases are equal, we equate the exponents
6 + 2x = 10
2x = 10 – 6 = 4
x = \(\frac{4}{2}\) = 2

Question 4.
Evaluate 2⁰ + 3⁰
Solution:
2⁰ + 3⁰ = 1 + 1 = 2 [∵ a⁰ = 1]

Question 5.
Simplify \(\left(\frac{x^{a}}{x^{b}}\right)^{a} \times\left(\frac{x^{b}}{x^{a}}\right)^{a} \times\left(\frac{x^{a}}{x^{a}}\right)^{b}\)
Solution:
Given \(\left(\frac{x^{\mathrm{a}}}{\mathrm{x}^{\mathrm{b}}}\right)^{\mathrm{a}} \times\left(\frac{\mathrm{x}^{\mathrm{b}}}{\mathrm{x}^{\mathrm{a}}}\right)^{\mathrm{a}} \times\left(\frac{\mathrm{x}^{\mathrm{a}}}{\mathrm{x}^{\mathrm{a}}}\right)^{\mathrm{b}}=\left(\frac{\mathrm{x}^{\mathrm{a}}}{\mathrm{x}^{\mathrm{b}}} \times \frac{\mathrm{x}^{\mathrm{b}}}{\mathrm{x}^{\mathrm{a}}}\right)^{\mathrm{a}} \times(1)^{\mathrm{b}}\)
[∵ a^m × b^m = (ab)^m]
= 1^a x 1^b = 1^a+b = 1

Question 6.
State true or false and justify your answer.
(i) 100 × 10¹¹ = 10¹³
(ii) 3² × 4³ = 12⁵
(iii)) 5° = (100000)°
(iv) 4³ = 8²
(v) 2³ > 3²
(vi) (-2)⁴ > (-3)⁴
(vii) (-2)⁵ > (-3)⁵
Solution:
(i) 100 × 10¹¹ = 10¹³ – True
as 100 × 10¹¹ = 10² × 10¹¹ = 10²⁺¹¹ = 10¹³

(ii) 3² × 4³ = 12⁵ – False
as 3² × 4³ ≠ 12⁵

(iii)) 5° = (100000)° – True as 5° = 1 and 100000° = 1

(iv) 4³ = 8² – True, as 4³ = 4 × 4 × 4 = 64 and 8² = 8 × 8 = 64

(v) 2³ > 3² – False as 2³ = 8 and 3² = 9 and 8 < 9

(vi) (-2)⁴ > (-3)⁴ – False, as (-2)⁴ = (-2) × (-2) × (-2) × (-2) = 16
(-3)⁴ = (-3) × (-3) × (-3) × (-3) = 81

(vii) (-2)⁵ > (-3)⁵ – True, as (-2)5 = (-2) × (-2) × (-2) × (-2) × (-2) = -32
(-3)⁵ = (-3) × (-3) × (-3) × (-3) × (-3) = -243

Students can go through AP Board 7th Class Maths Notes Chapter 15 Symmetry to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 15 Symmetry

→ Line of symmetry: The line which divides a figure into two identical parts is called the line of symmetry or axis of symmetry.
Ex: In the adjacent figure the dotted lines are the line of symmetry.

→ An object can have one or more than one lines of symmetry or axes of symmetry.
Ex: In the above figure there are two lines of symmetry.

→ If we rotate a figure, about a fixed point by a certain angle and the figure looks exactly the same as before, we say that the figure has rotational symmetry.
Ex: An equilateral triangle; a square etc.

→ The angle of turning during rotation is called the angle of rotation (or) the minimum angle rotation of a figure to get exactly the same figure as original is called the angle of rotation.
Ex: i) Angle of rotation of an equilateral triangle = 120°.
ii) Angle of rotation of a square = 90°.

→ All figures having rotational symmetry of order 1, can be rotated completely through 360° to come back to their original position. So we say that an object has rotational symmetry only when the order of symmetry is more than 1.
Eg: The order of rotational symmetry for an equilateral triangle is 3.
ii) For a square is 4.

→ Some shapes only have line symmetry and some have only rotational symmetry and some have both. Squares, equilateral triangles and circles have both line symmetry and rotational symmetry.

Students can go through AP Board 7th Class Maths Notes Chapter 14 Understanding 3D and 2D Shapes to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 14 Understanding 3D and 2D Shapes

→ A net is a sort of skeleton – outline in 2-D, which, when folded, results in a 3-D shape. Each shape can also have more than one net according to the way we cut it.
Eg:

→ 3-D shapes can be visualised by drawing their nets on 2-D surfaces.

→ Oblique sketches are drawn on a grid paper to visualise 3-D shapes.

→ Isometric sketches can be drawn on a dot isometric paper to visualise 3-D shapes.

Students can go through AP Board 7th Class Maths Notes Chapter 13 Area and Perimeter to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 13 Area and Perimeter

→ The area of a parallelogram is equal to the product of its base (b) and corresponding height (h) A = bh.

Any side of the parallelogram can be taken as its base.

→ The area of a triangle is equal to half the product of its base and height.

A = \(\frac{1}{2}\) bh
A triangle = Half a parallelogram

→ The area of a Rhombus is equal to half the product of Its diagonals.

A = \(\frac{1}{2}\) d₁d₂

→ The circumference of a circle = 2πr = πd where π = \(\frac{22}{7}\) or 3.14, d = \(\frac{r}{2}\)

Students can go through AP Board 7th Class Maths Notes Chapter 12 Quadrilaterals to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 12 Quadrilaterals

→ Quadrilateral: A closed figure bounded by four line segments is called a quadrilateral.
In the figure, ABCD is a quadrilateral.

→ A quadrilateral divides a plane into three parts.
i) Interior of the quadrilateral
ii) Exterior of the quadrilateral
iii) Boundary of the quadrilateral

→ In the figure the points P, Q are in the interior of the quadrilateral. i3r In the figure the points R, S are in the exterior of the quadrilateral.

→ In the figure the points A, B, C, D are on the boundary of the quadrilateral.

→ A quadrilateral is said to be a convex quadrilateral if all line segments joining points in the interior of it also lie in its interior completely.
□ BELT is a convex quadrilateral.

→ A quadrilateral is said to be a concave quadrilateral if all line segments joining points in the interior of it do not necessarily lie in its interior completely.

In □ RING, the line segment \(\overline{\mathrm{AB}}\) does not lie completely in its interior, as such the quadrilateral RING is a concave quadrilateral.

→ Sum of the interior angles of a quadrilateral is 360°.
∠A + ∠B + ∠C + ∠D = 360°

→ A quadrilateral in which one pair of opposite sides are parallel is called a trapezium.
In □ ABCD ; AB // CD

→ A kite has four sides. There are exactly two distinct pairs of equal length.
In quadrilateral KITE,

KI = KE and IT = ET

→ A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram. In quadrilateral ABCD,
AB // CD and AD // BC. Hence □ ABCD is a parallelogram.

→ In a parallelogram,

• Opposite sides are parallel and equal [AB = CD and AD = BC]
• Diagonals bisect each other (AO = OC and BO = OD)
• Opposite angles are equal (∠A = ∠C and ∠B = ∠D)
• Adjacent angles are supplementary (∠A + ∠B = ∠B + ∠C = ∠C + ∠D = ∠D + ∠A = 180°)

→ A parallelogram in which adjaœnt sides are equal is called a Rhombus.
In quadrilateral ABCD,
AB = BC = CD = DA and hence □ ABCD is a Rhombus.

In a rhombus diagonals bisect each other at right angles,
(i.e.) AC ⊥ BD and AO = OC, BO = OD

→ A rectangle is a parallelogram with equal angles (OR)
A parallelogram in which one angle is a right angle is called a rectangle.

In fig. ∠A = ∠B = ∠C = ∠D = 90° and □ ABCD is a rectangle.
In a rectangle the diagonals are equal.
In a rectangle the diagonals bisect each other.

(AC = BD and AO = OC; BO = OD)

→ A square is a rectangle with equal adjacent sides.

In the figure AB = BC = CD = DA
∠A = ∠B = ∠C = ∠D = 90°
In a square the diagonals are equal and bisect at right angles. Also they are equal.
[(AO = OC ; BO = OD), (AC ⊥ BD) and (AC = BD)]

Flow chart of family of quadrilaterals

Students can go through AP Board 7th Class Maths Notes Chapter 11 Exponents to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 11 Exponents

→ Very large numbers are easier to read, write and understand when expressed in exponential form.
Eg : 10000 = 10⁴
8 × 8 × 8 × 8 × 8 × ….. × 8 (16 times) = 8¹⁶.

→ When a number is multiplied by itself for many number of times (repeated multiplication) then we write it in exponential form.
Eg : 2 × 2 × 2 × 2 = 2⁴ Here 2 is base 4 is exponent.
3 × 3 × 3 × 3 × 3 = 3⁵ Here 3 is base 5 is exponent.

→ a . a . a . a ….. a (m times) = a^m.

→ Here ‘a’ is called the base and ‘m’ is called the exponent.

→ Laws of exponents

i) a^m × aⁿ = a^m+n

ii)  (a^m)ⁿ = a^mn

iii) (ab)^m = a^m . b^m

iv) a^m = aⁿ ⇒ m = n

v) a^-n = \(\frac{1}{a^{n}}\)

vi) \(\frac{\mathrm{a}^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}\) = a^m-n  if (m > n)
= \(\frac{1}{a^{n-m}}\)  if (m < n)
= 1 if (m = n)

vii) \(\left(\frac{a}{b}\right)^{m}\) = \(\frac{a^{m}}{b^{m}}\)

viii) a⁰ = 1 where a ≠ 0

Students can go through AP Board 7th Class Maths Notes Chapter 10 Algebraic Expressions to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 10 Algebraic Expressions

→ Variable: It takes different values.
Ex: x, y, z, a, b, c, m etc.
Constant: The value of constant is fixed.
Ex: 1, 2, \(\frac{-2}{3}\), \(\frac{4}{5}\) etc.

→ Algebraic Expression: An algebraic expression is a single term or a combination of terms connected by the symbols ‘+’ (plus) or(minus).
Ex: 2x + 3, \(\frac{2}{5}\)p, 3x – 1 etc.

→ Numerical Expression : If every term of an expression is a constant term, then the expression is called a Numerical expression.
Ex: 2 + 1, -5 × 3, (12 + 4) ÷ 3.
Note: In the expression 2x + 9, ‘2x’ is an algebraic term. ‘9’ is called numeric term.

→ Like terms are terms which contain the same variables with the same exponents.
Ex: 12x, 25x, -7x are like terms.
2xy², 3xy², 7xy² are like terms.

→ Coefficient: In a.xⁿ, ‘a’ is called the numerical coefficient and ‘x’ is called the literal coefficient.
Types of algebraic expressions.

→ Degree of a monomial: The sum of all exponents of the variables present in a monomial is called the degree of the term or degree of the monomial.
Ex: The degree of 9x²y² is 4 [∵ 2 + 2 = 4]
Note: Degree of constant term is zero.
The highest of the degrees of all the terms of an expression is called the degree of the expression.
Ex: The degree of the expression ax + bx² + cx³ + dx⁴ + ex⁵ is 5.

→ The difference between two like terms is a like term with a numerical coefficient equal to the difference between the numerical coefficients of the two like terms.
Note:

1. If no two terms of an expression are alike then it is said to be in the simplified form.
2. In an expression, if the terms are arranged in such a way that the degrees of the terms are in descending order then the expression is said to be in standard form.
3. Addition (or) subtraction of expressions should be done in two methods, they are
   i) Column or Vertical method.
   ii) Row or Horizontal method.

Students can go through AP Board 7th Class Maths Notes Chapter 9 Construction of Triangles to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 9 Construction of Triangles

→ A triangle can be drawn of any three of its elements are known.

→ To construct a triangle, we need

1. Three sides
2. Any two sides and the angle included between them.
3. Two angles and the side included between them.
4. Hypotenuse and one adjacent side of a right angled triangle.
5. Two sides and a non-included angle.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 11 Exponents Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 11th Lesson Exponents Exercise 1

Question 1.
Write the base and the exponent in each case. Also, write the term in the expanded form.
(i) 34
(ii) (7x)²
(iii) (5ab)³
(iv) (4y)⁵
Solution:

Question 2.
Write the exponential form of each expression.
(i) 7 × 7 × 7 × 7 × 7
(ii) 3 × 3 × 3 × 5 × 5 × 5 × 5
(iii) 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5
Solution:
(i) 7 × 7 × 7 × 7 × 7 = 7⁵
(ii) 3 × 3 × 3 × 5 × 5 × 5 × 5 = 3³ × 5⁴
(iii) 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5 = 2³ × 3⁴ × 5³

Question 3.
Express the following as the product of exponents through prime factorization.
(i) 288
(ii) 1250
(iii) 2250
(iv) 3600
(v) 2400
Solution:
(i) 288
1) 288 = 2 × 144
= 2 × 2 × 72
= 2 × 2 × 2 × 36
= 2 × 2 × 2 × 2 × 18
= 2 × 2 × 2 × 2 × 2 × 9
= (2 × 2 × 2 × 2× 2) × (3 × 3)
= 2⁵ × 3²

(ii) 1250 = 2 × 625
= 2 × 5 × 125
= 2 × 5 × 5 × 25
= 2 × 5 × 5 × 5 × 5
= 2 × 5⁴

(iii) 2250 = 2 × 1125
= 2 × 3 × 375
= 2 × 3 × 3 × 125
= 2 × 3 × 3 × 5 × 25
= 2 × 3 × 3 × 5 × 5 × 5
= 2¹ × 3² × 5³

(iv) 3600 = 2 × 1800
= 2 × 2 × 900
=2 × 2 × 2 × 450
=2 × 2 × 2 × 2 × 225
= ( 2 × 2 × 2 × 2 ) × 3 × 75
= (2 × 2 × 2 × 2) × (3 × 3) × 25
= (2 × 2 × 2 × 2) × (3 × 3) × (5 × 5)
= 2⁴ × 3² × 5²

(v) 2400
2400 = 2 × 1200
= 2 × 2 × 600
= 2 × 2 × 2 × 300
= 2 × 2 × 2 × 2 × 150
= 2 × 2 × 2 × 2 × 2 × 75
= (2 × 2 × 2 × 2 × 2) × 3 × 25
= (2 × 2 × 2 × 2 × 2) × 3 × (5 × 5)
= 2⁵ × 3¹

Question 4.
Identify the greater number in each of the following pairs.
(i) 2³ or 3²
(ii) 5³ or 3⁵
(iii) 2⁸ or 8²
Solution:
(i) 2³ or 3² = 2³ = 2 × 2 × 2 = 8 and 3² = 3 × 3 = 9
∴ 2³ < 3² or 3² > 2³

(ii) 5³ or 3⁵ = 5³ = 5 × 5 × 5 = 125 and 3⁵ = 3 × 3 × 3 × 3 × 3 = 243
∴ 3⁵ > 5³

(iii) 2⁸ or 8² = 2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
8² = 8 × 8 = 64
∴ 2⁸ > 8²

Question 5.
If a = 3, b = 2 find the value of
(i) a^b + b^a
(ii) a^a + b^b
(iii) (a + b)a^b
(iv)(a – b)^a
Solution:
(i) a^b + b^a = 3² + 2³ = 3 × 3 + 2 × 2 × 2 = 9 + 8 =17
(ii) a^a + b^b = 3³ + 2² = 3 × 3 × 3 + 2 × 2 = 27 + 4 = 31
(iii) (a + b)^b = (3 + 2)² = 5² = 5 × 5 = 25
(iv)(a – b)^a = (3 – 2)²= 1² = 1 × 1 = 1 .

Students can go through AP Board 7th Class Maths Notes Chapter 8 Congruency of Triangles to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 8 Congruency of Triangles

→ Two figures are said to be identical if their shapes are same.
Eg: Any two squares, circles or equilateral triangles.

→ Two figures are said to be congruent if they are identical in shape and equal in size.

→ Two line segments are congruent if they have same lengths.

→ Two triangles are congruent if the corresponding angles are equal.

→ We establish the congruency of the triangles by following criteria.

→ S.S.S. criterion: If three sides of a triangle are equal to the corresponding three sides of another triangle, then the triangles are congruent.

FA = TI; AN = IN; FN = TN then △FAN ≅ △TIN

→ S.A.S. criterion: If two sides and the angle included between the two sides of a triangle are equal to the corresponding two sides and the included angle of another triangle, then the two triangles are congruent.

CA = PI; ∠C = ∠P; CT = PG
then, △CAT ≅ △PIG

→ A.S.A. criterion : If two angles and the included side of a triangle are equal to the corresponding two angles and included side of another triangle then the triangles are congruent.

∠A = ∠I; AT = IM; ∠T = ∠M then △MAT ≅ △DIM.

→ R.H.S. criterion: In two right angled triangles, if the hypotenuse and one corresponding side are equal then they are congruent.

∠B = ∠E = 90°
BC = EF
AC = DF
then △ABC ≅ △DEF

→ If by any criterion two triangles are congruent then all the corresponding parts are equal.