Class 7

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 13th Lesson Area and Perimeter Exercise 3

Question 1.
Find the area of each of the following triangles.

Solution:
î) Area = \(\frac{1}{2}\) = \(\frac{1}{2}\) x 5 x 8 = 20 cm²
ii) Area = \(\frac{1}{2}\) = \(\frac{1}{2}\) x 6 x 4 = 12 cm²
iii) Area = \(\frac{1}{2}\) = \(\frac{1}{2}\) x 5.4 x 7.5 = 20.25 cm²
iv)Area = \(\frac{1}{2}\) = \(\frac{1}{2}\) x 6 x 4 = 12 cm²

Question 2.
In ΔPQR, PQ = 4 cm, PR =8 cm and RT = 6cm. Find (i) the area of ΔPQR (ii) the length of QS.

Solution:
Given PQ 4cm, PR = 8cm, RT = 6cm
1) Area of ΔPQR = \(\frac { 1 }{ 2 }\) × base × height = \(\frac { 1 }{ 2 }\) × PQ × RT
= \(\frac { 1 }{ 2 }\) × 4 × 6 = 12cm²

ii) Also area of ΔPQR = \(\frac { 1 }{ 2 }\) × base x height = \(\frac { 1 }{ 2 }\) × PR × QS
12 = \(\frac { 1 }{ 2 }\) × 8 × QS
QR = \(\frac{12 \times 2}{8}\) = 3cm

Question 3.
ΔABC is right-angled at A. AD is perpendicular to BC, AB =5 cm, BC = 13 cm and AC = 12 cm. Find the area of ΔABC. Also, find the length of AD.

Solution:
Given.
In ΔABC, ∠A = 90°
AB = 5cm; AC = 12cm; AD ⊥ BC; BC = 13cm.
Now area of ΔABC = \(\frac { 1 }{ 2 }\) × base × height
= \(\frac { 1 }{ 2 }\) x AB x AC = \(\frac { 1 }{ 2 }\) × 5 × 12 = 30cm²
Also area of MBC = \(\frac { 1 }{ 2 }\) × BC × AD
30 = \(\frac { 1 }{ 2 }\) × 13 × AD
∴ \(\frac{30 \times 2}{13}=\frac{60}{13}=4 \frac{8}{13}\) cm

Question 4.
ΔPQR is isosceles with PQ = PR = 7.5 cm and QR = 9 cm. The height PS from P to QR, is 6 cm. Find the area of ΔPQR. What will be the height from R to PQ i.e. RT?

Solution:
Given: PQ = PR = 7.5cm, QR = 9cm PS = 6cm
Area of ΔPQR = \(\frac { 1 }{ 2 }\) bh = latex]\frac { 1 }{ 2 }[/latex] x 9 x 6 = 27cm²
Also area of ΔPQR = \(\frac { 1 }{ 2 }\) × PQ × RT
27 = \(\frac { 1 }{ 2 }\) × 7.5 × RT [PQ = PR]
∴ RT = \(\frac{2 \times 27}{7.5}=\frac{2 \times 27 \times 2}{15}=\frac{36}{5}\) = 7.2 cm

Question 5.
ABCD rectangle with AB =8 cm, BC = 16 cm and AE = 4 cm. Find the area of ABCE. Is the area of ΔBEC equal to the sum of the area of ΔBAE and ΔCDE. Why?

Solution:
Given : In rectangle ABCD,
AB = 8cm, BC = 16 cm, AE = 4cm
Now Area of ΔBAE = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) × \(\overline{\mathrm{BC}} \times \overline{\mathrm{AB}}\)
= \(\frac { 1 }{ 2 }\) × 16 x 8 = 64cm²

Area of ΔCDE = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) × \(\overline{\mathrm{ED}} \times \overline{\mathrm{DC}}\)
= \(\frac { 1 }{ 2 }\) × 12 × 8 = 48cm²
(∵ \(\overline{\mathrm{ED}}\) = AD – AE = 16 – 4 = 12cm and \(\overline{\mathrm{DC}}=\overline{\mathrm{AB}}\))
Now ΔBAE + ΔCDE = 16 + 48 = 64 = ΔBCE.

Question 6.
Ramu says that the area of ΔPQR is, A = \(\frac{1}{2}\) × 7 × 5 cm².
Gopi says that it is, A = \(\frac{1}{2}\) × 8 × 5 cm². Who is correct’? Why?

Solution:
Ramu is correct.
Since area of a triangle = \(\frac { 1 }{ 2 }\) × base × corresponding height
= \(\frac { 1 }{ 2 }\) × 7 × 5cm²
But height 5cm is not corresponding to side &m.
∴ Gopi is not correct.

Question 7.
Find the base of a triangle whose area is 220 cm² and height is 11cm.
Solution:
Given Area = 220 cm², h = 11 cm
Area of a triangle = \(\frac { 1 }{ 2 }\)b.h
= \(\frac { 1 }{ 2 }\) × b × 11 = 220cm² (given)
b × 11 = 220 × 2
b = \(\frac{220 \times 2}{11}\) = 40
∴ base of the triangle = 40 cm.

Question 8.
In a triangle the height is double the base and the area is 400 cm². Find the length of the base and height.
Solution:
Area = 400 cm²
Let the base of the triangle be x.
height = 2x
Area of the triangle = \(\frac { 1 }{ 2 }\)b.h
\(\frac { 1 }{ 2 }\) × x × 2x = 400 (given)
x² = 400
x = \(\sqrt{400}\) = 20
∴ base of the triangle = 20 cm
height = 2(base)= 2(20) = 40 cm.

Question 9.
The area of triangle is equal to the area of a rectangle whose length and breadth are 20 cm and 15 cm respectively. Calculate the height of the triangle if its base measures 30 cm.
Solution:
Given :Length of the rectangle = 20 cm
Breadth of the rectangle = 15 cm
Base of the triangle = 30 cm
Area of the rectangle = Area of the triangle
Length x breadth = \(\frac { 1 }{ 2 }\) × base × height
20 × 15 = \(\frac { 1 }{ 2 }\) × 30 × height
∴ height = \(\frac{20 \times 15}{15}\) = 20cm

Question 10.
In Figure ABCD find the area of the shaded region.

Solution:
Given : Side of the square ABCD = 40cm
Area of the shaded rgion = (Area of the square) — (Area of the unshaded region)
= (side × side) – \(\frac { 1 }{ 2 }\) × base × height
= 40 × 40 – \(\frac { 1 }{ 2 }\) × 40 × 40
= 1600 – 800 = 800cm²

Question 11.
In Figure ABCD, find the area of the shaded region.

Solution:
Given: Side of the square = 20 cm
Area of the shade region
= (Area of the square) – (Area of unshaded region)
= (Square ABCD) – (ΔEAF + ΔFBC + ΔEDC)
= side × side – (\(\frac { 1 }{ 2 }\)AF × AE + \(\frac { 1 }{ 2 }\) FB × BC + \(\frac { 1 }{ 2 }\) DE × DC)
= 20 × 20 – (\(\frac { 1 }{ 2 }\) × 10 × 12 + \(\frac { 1 }{ 2 }\) × 10 × 20 + \(\frac { 1 }{ 2 }\) × 8 × 20)
= 400 – [60 + 100 + 80] = 400 – 240 = 160cm²

Question 12.
Find the area of a parallelogram PQRS, if PR =24 cm and QU = ST =8 cm.

Solution:
Given : In PQRS, PR = 24cm, ST = 8cm, QU =8cm
Area of PQRS = ΔPQR + ΔPRS
= \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\)base x height + x base x height
= \(\frac { 1 }{ 2 }\) × 24 x 8 + \(\frac { 1 }{ 2 }\) × 24 × 8 = 96 + 96 = 192 cm²

Question 13.
The base and height of the triangle are in the ratio 3:2 and its area is 108 cm². Find its base
and height.
Solution:
Given : Area = 108 cm2
Let the base of the triangle be 3x and height of the triangle be 2x.
Area of the triangle = \(\frac { 1 }{ 2 }\) × base × height = \(\frac { 1 }{ 2 }\) x (3x) (2x) = 108 cm² (given)
3x² = 108
x² = 108/3 = 36
x = \(\sqrt{36}\) = 6
∴ base 3x = 3(6) = 18 cm
height = 2x = 2(6) = 12cm

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 13th Lesson Area and Perimeter Exercise 2

Question 1.
Find the area of the each of the following parallelograms.

Solution:

Fig	Base	Height	A = Base x Height
(i)	7	4	7 x 4 = 28cm2
(ii)	5	3	5 x 3  = 15 cm2
(iii)	5.1	7.6	5.1 x 7.6 = 38.76 cm2
(iv)	4	6	4 x 6 = 24cm2

Question 2.
PQRS is a parallelogram. PM is the height from P to \(\overline{\mathrm{SR}}\) and PN is the height from P to \(\overline{\mathrm{QR}}\). If SR = 12cm and PM = 7.6 cm.
(i) Find the area of the parallelogram PQRS
(ii) Find PN, if QR = 8 cm.
Solution:

i) Given:SR = 12cm
PM = 7.6cm
∴ Area = base × height
= 12 × 7.6 = 91.2 cm²

ii) Also area = base × height
91.2 = QR × PN
91.2 = 8 × PN
∴ PN = \(\frac{91.2}{8}\) = 11.4cm

Question 3.
DF and BE are the height on sides AB and AD respectively in parallelogram ABCD. Ifthe area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BE and DF.

Solution:
Given : AB = 35 cm, AD = 49 cm
area of 🖾ABCD = 1470 cm²
Area of a parallelogram = base × height
AB × DF = area of 🖾ABCD
35 × DF = 1470
DF = \(\frac{1470}{35}\) = 42cm
Also AD × BE = 1470
49 × BE = 1470
∴ BE= \(\frac{1470}{49}\) = 30cm

Question 4.
The height of a parallelogram is one third of its base. If the area of the parallelogram is 192cm², find its height and base.
Solution:
Area of the parallelogram = 192 cm²
Area of parallelogram = base × height = 192 cm² (given)
Let the base of the parallelogram be x.
height = \(\frac{x}{3}\)
Area = base × height = x x \(\frac{x}{3}=\frac{x^{2}}{3}\) = 192
⇒ x² = 192 × 3 = 576
x = \(\sqrt{576}\) = 24
∴ base = 24 cm
height = \(\frac{x}{3}=\frac{24}{3}\) = 8cm.

Question 5.
In a parallelogram the base and height are is in the ratio of 5:2. If the area of the parallelogram is 360m², find its base and height.
Solution:
Let the base of the parallelogram be 5x and height of the parallelogram be 2x.
Area = height x base = 5x × 2x = 10x² = 360 cm² (given)
10x² = 360
x² = \(\frac{360}{10}\) = 36
x = √36 = 6
base = 5x = 5(6) = 30 cm
height = 2x = 2(6) = 12 cm

Question 6.
A square and a parallelogram have the same area. If a side of the square is 40m and the height of the parallelogram is 20m, find the base of the parallelogram.
Solution:
Given: Side of a square = 40m
Height of a parallelogram = 20m
Area of parallelogram = Area of the square
base × height side × side
base × 20 40 × 40
∴ base = \(\frac{40 \times 40}{20}\) = 80 m.

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

AP State Syllabus AP Board 7th Class Science Solutions Chapter 7 Electricity – Current and Its Effect Textbook Questions and Answers.

AP State Syllabus 7th Class Science Solutions 7th Lesson Electricity – Current and Its Effect

7th Class Science 7th Lesson Electricity – Current and Its Effect Textbook Questions and Answers

Improve Your Learning

Question 1.
Draw the symbols of the following electric components.
a) Cell
b) Battery
c) Switch
d) Electric bulb
Answer:
a) Symbol for cell: The longer line denotes the positive terminal, and the thicker, small line denotes the negative terminal.

b) Symbol for battery: Two or more cells joined together form a battery.

c) Symbol for switch: Switch is also called key. Switch is open.

d) Symbol for electric bulb: Electric bulb in on position.

Question 2.
Draw an electric circuit diagram consisting of a cell, a bulb and an electric switch.
Answer:
Electric Circuit diagram:

Question 3.
In a series connection of bulbs, if one bulb fails, why do all other bulbs go OFF?
Answer:

1. In a series electric circuit, electricity has only one path to flow through.
2. In series connection of bulbs, if one bulb fails, the circuit breaks and current do not flow in the circuit.
3. So other bulbs in the series connection of bulbs do not glow’and they go OFF.

Question 4.
Write the difference between series connection and parallel connection.
Answer:

Series Connection	Parallel Connection
Electricity has only one path to flow.	Electricity has more than one path to flow.
All the electrical components are connected in this path.	Each bulb in the circuit is connected in separate path through which electricity can flow.

Question 5.
What is the advantage of Miniature Circuit Breaker?
Answer:

1. The advantage miniature circuit breakers have over fuses is that they can reset (manually or automatically) to restore normal operation.
2. Fuses need to be replaced after every single operation.

Question 6.
Fill in the blanks.
a) Longer line in the symbol for a cell represents its ………… terminal.
b) Smaller line in the symbol for a cell represents its ………… terminal.
c) The combination of two or more cells is called a ………… .
d) Safety device used in electric circuit is ………… .
e) The device used to close or open an electric circuit is ………… .
Answer:
a) positive
b) negative
c) battery
d) fuse
e) switch

Question 7.
Mark T’ if the statement is true and F’ if it is false. Give reasons for choice of answer.
a) In series circuit the electricity has only one path. (T/F)
b) In parallel circuit the electricity has more than one path. (T/F)
c) To make a battery of two cells, the negative terminal of one cell is connected to the negative terminal of the other cell. (T/F)
d) When the electric current through the fuse exceeds a certain limit the fuse wire
melts and breaks. (T/F)
e) The switch is used to close or open an electric circuit. (T/F)
Answer:
a) T
b) T
c) F
d) T
e) T

Question 8.
Choose the correct answer.

i) Arun buys four bulbs of 15 W, 40 W, 60 W and 100 W respectively. Which one should be use in his room as a night bulb?
A) 15 W
B) 40 W
C) 60 W
D) 100 W
Answer:
A) 15 W

ii) Device used to close or open an electric circuit is ( C )
A) Electric bulb
B) Battery
C) Switch
D) Fuse
Answer:
C) Switch

iii) Which one of the following is used for light source? ( D )
A) Cassette player
B) Electric mixer
C) Rice Cooker
D) Table lamp
Answer:
D) Table lamp

iv) Safety device used in electric circuit is ( D )
A) Electric bulb
B) Battery
C) Switch
D) Fuse
Answer:
D) Fuse

Question 9.
Visit your classmates houses. Find out the meter readings of three months. Record your observations. Ask your parents about how electricity bill is paid?
Answer:

Question 10.
Draw the symbols of the following electric components.
Answer:

Question 11.
Draw the circuit diagram for the following series connection.

Answer:

Question 12.
Match the following.

Answer: