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These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.3

Question 1.
Construct a ADEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90 .
Answer:
Steps of Construction

(i) First we draw a rough sketch with measures marked on it.
(ii) Draw a line segment DE = 5 cm.
(iii) At D, construct ∠EDX = 90
(iv) From the ray DX, cut off DF = 3 cm.
(v) Join EF.
Then, ∆DEF is the required triangle.

Question 2.
Construct an isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110.
Answer:
Steps of Construction

(i) First we draw a rough sketch with measures marked on it.
(ii) Draw a line segment AB = 6.5 cm.
(iii) At B, using protractor construct ∠ABX = 110°.
(iv) From BX, cut off BC = 6.5 cm.
(v) Join AC.
Thus, ∆ABC is the required isosceles triangle.

Question 3.
Construct ∆ABC with BC = 7.5 cm AC = 5 cm and m∠C – 60
Answer:
Steps of Construction
(i) First we draw a rough sketch with measures marked on it.
(ii) Draw a line segment BC = 7.5 cm.
(iii) At ‘C’ using protractor construct ∠BCX = 60°.

(iv) From the ray CX, cut off CA = 5 cm.
(v) Join AB.
Thus, ∆ABC is the required triangle.

These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.2

Question 1.
Construct ΔXYZ in which XY = 4.5 cm YZ = 5 cm and ZX = 6 cm.
Answer:
Steps of Construction
(i) First we draw a rough sketch with given measure.

(ii) Draw a line segment YZ = 5 cm
(iii) With centre Y and radius 4.5 cm, draw an arc.
(iv) With centre Z and radius 6 cm, draw another arc to cut the previous arc at X.
(v) Join XY and XZ.
Thus, ΔXYZ is the required triangle.

Question 2.
Construct an equilateral triangle of side 5.5 cm.
Answer:
Steps of Construction

(i) First we draw a rough sketch with given measure.
(ii) Draw a line segment, BC = 5.5 cm.
(iii) With B as centre, 5.5 cm as radius draw an arc.
(iv) With C as centre, 5.5 cm as radius draw an arc to cut the previous arc at A.
(v) Join AB and AC.
Thus, ΔABC is the required equilateral triangle.

Question 3.
Draw ΔPQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this ?
Answer:
Steps of Construction
(i) First we draw a rough sketch with given measure.
(ii) Draw a line segment QR = 3.5 cm.

(iii) With Q as centre, 4 cm as radius, draw an arc.
(iv) With R as centre, 4 cm as radius, draw an arc to cut the previous arc at P.
(v) Join PQ and PR.
Thus, ΔPQR is the required triangle since PQ = PR = 4 cm.
.’. ΔPQR is an isosceles triangle.

Question 4.
Construct AABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
Answer:
Steps of Construction

(i) First we draw a rough sketch with the given measure.
(ii) Draw a line segment BC = 6 cm.
(iii) With B as centre and 2.5 cm as radius, draw an arc.
(iv) With C as centre and 6.5 cm as radius, draw an arc to cut the previous arc at A.
(v) Join AB and AC.
Thus, ΔABC is the required triangle. On measuring, we find that ∠B = 90°.

These NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.1

Question 1.
Draw a line say AB, take a point ‘C’ outside it. Through C, draw a line parallel to AB using ruler and compass only.
Answer:
Steps of Construction

(i) Draw a line AB.
(ii) Take a point C outside it.
(iii) Take any point ‘D’ on AB.
(iv) Join C to D.
(v) With D as centre and a convenient radius draw an arc cutting AB at F and CD at E.
(vi) Now with C as centre and the same radius as in step 5, draw an arc GH cutting CD at I.
(vii) Place the pointed tip of the compass at F and adjust the opening so that the pencil tip is at E.
(viii) With the same opening as in step 7 and with I as centre, draw an arc cutting the arc GH at J.
(ix) Now Join CJ to draw a line ‘KL’ Then KL is the required line.

Question 2.
Draw a line 1. Draw a perpendicular to T at any point on T. On this perpendicular Choose a point X, 4 cm away from l. Through X, draw a line ‘m’ parallel to T
Answer:
Steps of Construction

(i) Draw a line ‘l’.
(ii) Take any point A on line l.
(iii) Construct an angle of 90° at point ‘A’ on line ‘l’ and draw a line ‘AL’ perpendicular to line ‘l’.
(iv) Make a point ‘X’ on AL such that AX = 4 cm.
(v) At X, construct an angle of 90 and draw a line XC perpendicular to line AL.
(vi) Then line XC (line m) is the required line through X such that m || l.

Question 3.
Let ‘l’ be a line and ‘P’ be point not on ‘l’. Through P, draw a line ‘m’ parallel to ‘l’. Now Join ‘P’ to any point ‘Q’ on ‘l’. Choose any other point ‘R’ on ‘m’ Through ‘R’ draw a line parallel to PQ. Let this meet ‘l’ at ‘S’. What shape do the two sets of parallel lines enclose?
Answer:
Steps of Construction
(i) Draw a line ‘l’ and take a point ‘P’ not on it.
(ii) Take any point ‘Q’ on ‘l’.
(iii) Join ‘Q’ to ‘P’.
(iv) Draw a line ‘m’ parallel to the line T as shown in figure.
Then line ‘m’ || line ‘l’.
(v) Join ‘P’ to any point ‘Q’ on ‘l’.
(vi) Choose any point ‘R’ on ‘m’.

(vii) Join R to Q
(viii) Through R, draw a line ‘n’ parallel to the line PQ.
(ix) Let the line ‘n’ meet the line ‘l’ at ‘S’.
(x) Then the shape enclosed by the two sets of parallel lines is a ‘Parallelogram’.

These NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Exercise 5.2

Question 1.
State the property that is used in each of the following statements?
(i) If a || b, then ∠1 = ∠5.
(ii) If ∠4 = ∠6, then a || b.
(iii) If ∠4 + ∠5 = 180°, then a || b.

Answer:
(i) If two parallel lines are intersected by a transversal, then corresponding angles are equal.
(ii) If two given lines are intersected by a transversal such that alternate interior angles are equal, then the lines are parallel.
(iii) If two parallel lines are cut by a transversal such that the pairs of a interior angles on the same side of the transversal are supplementary, then the lines are parallel.

Question 2.
In the figure given below, identify
(i) the pairs of corresponding angles.
(ii) the pairs of alternate interior angles.
(iii) the pairs of interior angles on the same side of the transversal.
(iv) the vertically opposite angles.

Answer:
(i) The pairs of corresponding angles
are (∠1, ∠5); (∠2, ∠6); (∠3, ∠7); and (∠4, ∠8)
(ii) The pairs of alternate interior angles are (∠2, ∠8) and (∠3, ∠5)
(iii) The pairs of interior angles on the same side of the transversal are
(∠2, ∠5) and (∠3, ∠8)
(iv) The vertically opposite angles are (∠1, ∠3); (∠2, ∠4); (∠5, ∠7); and (∠6, ∠8)

Question 3.
In the figure given below, p || q. Find the unknown angles.
Answer:
Since, ∠e + 125° = 180°… (linear pair)
∠e = 180° – 125°
∠e = 55°
∠e = ∠f
(vertically opposite angles)
∴ ∠f = 55°
∴ ∠a = ∠e
(corresponding angles)
∠a = 55°
Again ∠b = 125°
(alternate exterior angles)
since, ∠b and ∠c form a linear pair
∴ ∠b+ ∠c= 180°
or 125° + ∠c = 180°
∠c = 180° – 125°
= 55°

Now, ∠b and ∠d are vertically opposite angles.
∴ ∠d = ∠b = 125°
[∵ ∠b = 125°]
Thus, the required measures are as follows: ∠a = 55°, ∠b = 125°, ∠c = 55°,∠d = 125°, ∠e = 55°, ∠f = 55°

Question 4.
Find the value of x in each of the following figures if 1 || m.

Answer:
(i) ∠x = ∠P
(alternate angles are equal)
But ∠P + 110° = 180° (linear pair)
∠P = 180°- 110°
= 70°
∴ ∠x = 70°

(ii) 1 and m are parallel and ‘a’ is a transversal.
∴ ∠x = 100°
(corresponding angles are equal)

Question 5.
In the given figure, the arms of two angles are parallel.
If ∠ABC = 70°, then find
(i) ∠DGC
(ii) ∠DEF

Answer:
We have AB || ED and BC || EF
(i) BC is a transversal
∴ ∠DGC – ∠ABC
(corresponding angles)
But∠ABC – 70°
∴ ∠DGC = 70°

(ii) ED is a transversal to BC || EF
∴ ∠DEF = ∠DGC
(corresponding angles)
But∠DGC = 70°
∴ ∠DEF = 70°

Question 6.
In the given figures below, decide whether 1 is parallel to m.

Answer:
(i) 44° + 126° = 170° ≠ 180°
Sum of the interior angles on the same side of the transversal is not 180°.
∴ 1 and m are not parallel.

(ii) n is a transversal to 1 and m and ∠P = 75°(vertically opposite angles)
also ∠P + 75°= 75° + 75° = 150° ≠ 180°
∴ 1 and m are not parallel as sum of interior angles on same side of transversal is not 180°.

(iii) ∠P + 123°= 180° (linear pair)
∠P = 180° – 123° = 57°
∠P = 57°(linear pair)
i.e., Corresponding angles are equal.
∴ 1 and m are parallel (corresponding angles are not equal)

(iv) ∠1 + ∠2 = 180°
∠1 + 98° = 180°
∠1 = 180°- 98°= 82°
∠3 = 72°
∠1 ≠∠3
∴1 and m are not parallel.

These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.3

Question 1.
What could be the possible ‘ones’ digits of the square root of each of the following numbers?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
Solution:
(i) The unit’s digit of the square root of the number 9801 could be 1 or 9 [1 × 1 = 1, 9 × 9 = 81]
(ii) The units digit of the square root of the number 99856 could be 4 or 6 [4 × 4 = 16 or 6 × 6 = 36]
(iii) The unit’s digit of the square root of the number 998001 could be 1 or 9 [1 × 1 = 1 or 9 × 9 = 81]
(iv) The units digit of the square root of the number 657666025 could be 5 [5 × 5 = 25]

Question 2.
Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153
(ii) 257
(iii) 408
(iv) 441
Solution:
We know that the ending digit of a perfect square is 0, 1, 4, 5, 6 and 9
A number ending in 2, 3, 7, or 8 can never be a perfect square
(i) 153 cannot be a perfect square.
(ii) 257 cannot be a perfect square.
(iii) 408 cannot be a perfect square.
(iv) 441 can be a perfect square.

Question 3.
Find the square roots of 100 and 169 by the method of repeated subtraction.
Solution:
(i) √100
100 – 1 = 99
99 – 3 = 96
96 – 5 = 91
91 – 7 = 84
84 – 9 = 75
75 – 11 = 64
64 – 13 = 51
51 – 15 = 36
36 – 17 = 19
19 – 19 = 0
∴ We reach at 0 by successive subtraction of 10 odd numbers
∴ √100 = 10

(ii) √169
169 – 1 = 168
168 – 3 = 165
165 – 5 = 160
160 – 7 = 153
153 – 9 = 144
144 – 11 = 133
133 – 13 = 120
120 – 15 = 105
105 – 17 = 88
88 – 19 = 69
69 – 21 = 48
48 – 23 = 25
25 – 25 = 0
∴ We reach at 0 by successive subtraction of 13 odd numbers
∴ √169 = 13

Question 4.
Find the square roots of the following numbers by the Prime Factorisation Method.
(i) 729
(ii) 400
(iii) 1764
(iv) 4096
(v) 7744
(vi) 9604
(vii) 5929
(viii) 9216
(ix) 529
(x) 8100
Solution:
(i) √729
= \(\sqrt{3^{2} \times 3^{2} \times 3^{2}}\)
= 3 × 3 × 3
= 27
∴ √729 = 27

(ii) √400
= \(\sqrt{2^{4} \times 5^{2}}\)
= 2² × 5
= 20
∴ √400 = 20

(iii) √1764
= \(\sqrt{2^{2} \times 3^{2} \times 7^{2}}\)
= 2 × 3 × 7
= 42
∴ √1764 = 42

(iv) √4096
= \(\sqrt{2^{12}}\)
= 2⁶
= 2 × 2 × 2 × 2 × 2 × 2
= 64
∴ √4096 = 64

(v) √7744
= \(\sqrt{2^{6} \times 11^{2}}\)
= 2³ × 11
= 88
∴ √7744 = 88

(vi) √9604
= \(\sqrt{2^{2} \times 7^{4}}\)
= 2 × 7²
= 2 × 49
= 98
∴ √9604 = 98

(vii) √5929
= \(\sqrt{7^{2} \times 11^{2}}\)
= 7 × 11
= 77
∴ √5929 = 77

(viii) √9216
= \(\sqrt{2^{5} \times 3^{2}}\)
= 2⁵ × 3
= 2 × 2 × 2 × 2 × 2 × 3
= 32 × 3
= 96
∴ √9216 = 96

(ix) √529
= \(\sqrt{23^{2}}\)
= 23
∴ √529 = 23

(x) √8100
= \(\sqrt{2^{2} \times 3^{4} \times 5^{2}}\)
= 2 × 3² × 5
= 2 × 9 × 5
= 90
∴ √8100 = 90

Question 5.
For each of the following numbers, find the smallest whole number by which it should be multiple so as to get a perfect square number. Also, find the square root of the square number so obtained.
(i) 252
(ii) 180
(iii) 1008
(iv) 2028
(v) 1458
(vi) 768
Solution:
(i) 252 = 2² × 3² × 7
The prime factor 7 has no pair
∴ The number 252 should be multiply by 7 to get a perfect square number 252 × 7 = 1764
√1764
= \(\sqrt{2^{2} \times 3^{2} \times 7^{2}}\)
= 2 × 3 × 7
= 42

(ii) 180 = 2² × 3² × 5
The prime factor 5 has no pair
∴ The number 180 should be multiplied by 5 to get a perfect square number 180 × 5 = 900
√900
= \(\sqrt{2^{2} \times 3^{2} \times 5^{2}}\)
= 2 × 3 × 5
= 30

(iii) 1008 = 2⁴ × 3² × 7
The prime factor 7 has no pair
∴ The number 1008 should be multiplied by 7 to get a perfect square number 1008 × 7 = 7056
√7056
= \(\sqrt{2^{4} \times 3^{2} \times 7^{2}}\)
= 4 × 3 × 7
= 84

(iv) 2028 = 2² × 3³ × 13²
The Prime factor 3 has no pair
∴ The number 2028 should be multiplied by 3 to get a perfect square number 2028 × 3 = 6084
√6084
= \(\sqrt{2^{2} \times 3^{2} \times 13^{2}}\)
= 2 × 3 × 13
= 78

(v) 1458 = 2 × 3⁶
The prime factor 2 has no pair
∴ The number 1458 should be multiplied by 2 to get perfect square number 1458 × 2 = 2916
√2916
= \(\sqrt{2^{2} \times 3^{6}}\)
= 2 × 3³
= 2 × 3 × 3 × 3
= 54

(vi) 768 = 2⁸ × 3
∴ The prime factor 3 has no pair
∴ The number 768 should be multiply be 3 to make it a perfect square.
768 × 3 = 2304
√2304
= \(\sqrt{2^{8} \times 3^{2}}\)
= 2⁴ × 3
= 2 × 2 × 2 × 2 × 3
= 48

Question 6.
For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also, find the square root of the square number so obtained
(i) 252
(ii) 2925
(iii) 396
(iv) 2645
(v) 2800
(vi) 1620
Solution:
(i) 252 = 2² × 3² × 7
The prime factor 7 has no pair
∴ The given number should be divided by 7 to get a perfect square
\(\frac{252}{7}=\frac{2^{2} \times 3^{2} \times 7}{7}\) = 36
√36
= \(\sqrt{2^{2} \times 3^{2}}\)
= 2 × 3
= 6

(ii) 2925 = 3² × 5² × 13
The prime factor 13 has no pair
∴ The given number should be divided by 13 to get a perfect square
\(\frac{2925}{13}=\frac{3^{2} \times 5^{2} \times 13}{13}\)
= 9 × 25
= 225
√225
= \(\sqrt{3^{2} \times 5^{2}}\)
= 3 × 5
= 15

(iii) 396 = 2² × 3² × 11
The prime factor 11 has no pair.
∴ The given no. should be divided by 11 to get a perfect square
\(\frac{396}{11}=\frac{2^{2} \times 3^{2} \times 11}{11}\)
= 4 × 9
= 36
√36
= \(\sqrt{2^{2} \times 3^{2}}\)
= 2 × 3
= 6

(iv) 2645 = 5 × 23²
The prime factor 5 has no pair.
∴ The given number should be divided by 5 to get a perfect square
\(\frac{2645}{5}=\frac{5 \times 23^{2}}{5}\) = 23² = 529
√529 = \(\sqrt{23^{2}}\) = 23

(v) 2800 = 2⁴ × 5² × 7
The prime factor 7 has no pair
∴ The given number should be divided by 7 to get a perfect square
\(\frac{2800}{7}=\frac{2^{4} \times 5^{2} \times 7}{7}\)
= 2⁴ × 5²
= 16 × 25
= 400
√400 = 20

(vi) 1620 = 2² × 3⁴ × 5
The prime factor 5 has no pair.
The given number should be divided by 5 to get a perfect square
\(\frac{1620}{5}=\frac{2^{2} \times 3^{4} \times 5}{5}\)
= 2² × 3⁴
= 4 × 81
= 324
√324
= \(\sqrt{2^{2} \times 3^{4}}\)
= 2 × 3²
= 2 × 9
= 18

Question 7.
The students of class VIII of a school donated ₹ 2401 in all, for Prime Minister National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
Solution:
Let the number of students be ‘x’
Amount donated by each student = x
Total amount donated by the class = ₹ x × x = ₹ x²
Given, x² = 2401
x = √2401
= \(\sqrt{7^{2} \times 7^{2}}\)
= 7 × 7
= 49
Number of students in the class = 49.

Question 8.
2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solution:
Let the number of rows be x’
Number of plants in a row = x
Number of plants to be planted = x × x = x²
Given, x² = 2025
x = √2025
= \(\sqrt{3^{4} \times 5^{2}}\)
= 32 × 5
= 9 × 5
= 45
∴ The required number of rows = 45
Also, no. of plants in a row = 45

Question 9.
Find the smallest square number that is divisible by each of the numbers 4, 9, and 10.
Solution:
The least number divisible by each one of 4, 9, and 10 is their L.C.M.

∴ L.C.M = 2 × 2 × 3 × 3 × 5 = 180
Prime factors of 180 = 2² × 3² × 5
The Prime factor 5 is not in pair
180 is not a perfect square
In order to get a perfect square 180 should be multiplied by 5
The required square number = 180 × 5 = 900

Question 10.
Find the smallest square number that is divisible by each of the numbers 8, 15, and 20.
Solution:
The least number divisible by each one of 8, 15, and 20 is their L.C.M.

∴ L.C.M = 2 × 2 × 2 × 3 × 5 = 120
The prime factors 2, 3, and 5 are not in pairs.
∴ 120 is not a perfect square.
To get a perfect square 120 should be multiplied by 2, 3, and 5 i.e. 30.
∴ The required smallest square number = 120 × 30 = 3600

These NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area InText Questions

NCERT In-text Question Page No. 205

Question 1.
What would you need to find, area or
perimeter, to answer the following?
1. How much space does a blackboard occupy?
2. What is the length of a wire required to fence a rectangular flower bed?
3. What distance would you cover by taking two rounds of a triangular park?
4. How much plastic sheet do you need to cover a rectangular swimming pool?
Answer:
1. Area 2.Perimeter 3. Perimeter 4. Area

NCERT In-text Question Page No. 206

Question 1.
Experiment with several such shapes and cutouts. You might find it useful to draw these shapes on squared sheets and compute their areas and perimeters. You have seen that increase in perimeter does not mean that area will also increase.
Answer:
Please do this question yourself with the help of your subject teacher.

Question 2.
Give two examples where the area increases as the perimeter increases.
Answer:
Please do this question yourself with the help of your subject teacher.

Question 3.
Give two examples where the area does not increase when perimeter increases.
Ans: Please do this question yourself with the help of your subject teacher.

NCERT In-text Question Page No. 210

Question 1.
Each of the following rectangles of length 6 cm and breadth 4 cm is composed of congruent polygons. Find the area of each polygon.

Answer:
∴ Length of the rectangle (l) = 6 cm
Breadth of the rectangle (b) = 4 cm
∴ Area of the rectangle = (l x b)
= 6 x 4 cm² = 24 cm²
(i) Here, number of congruent polygons = 6
∴ Area of each polygon = \(\frac{24}{6}\) cm² = 4 cm²

(ii) Here, number of congruent polygons = 4
Area of each polygon = \(\frac{24}{4}\)cm² = 6 cm²

(iii) Here, number of congruent polygons = 2
∴ Area of each polygon = \(\frac{24}{2}\)cm² = 12 cm²

(iv) Number of congruent polygons = 2
∴ Area of each polygon = \(\frac{24}{2}\)cm² = 12 cm²

(v) Number of congruent polygons = 8
∴ Area of each polygon = \(\frac{24}{8}\)cm² = 3 cm²

NCERT In-text Question Page No. 212

Question 1.
Find the area of the following parallelograms:

(iii) In a parallelogram ABCD, AB = 7.2 cm and the perpendicular from C on AB is 4.5 cm.
Answer:
(i) Base = 8 cm, Height = 3.5 cm
∴ Area of the parallelogram = Base x
Height = 8 cm x 3.5 cm = 28 cm²

(ii) Base = 8 cm, Height = 2.5 cm
Area of the parallelogram = Base x Height = 8 cm x 2.5 cm = 20 cm²

(iii) Base of parallelogram ABCD = (AB) = 7.2 cm
Height of parallelogram ABCD = 4.5 cm
Area of parallelogram ABCD
= Base x Height
= 7.2 cm x 4.5 cm = \(\frac { 72 }{ 10 }\)cm x \(\frac { 45 }{ 10 }\)cm
= \(\frac{3240}{100}\) cm² = 32.40 cm²

NCERT In-text Question Page No. 213

Question 1.
Try the activity given on page 213, NCERT Textbook with different types of triangles.
Answer:
Do it yourself.

Question 2.
Take different parallelograms. Divide each of the parallelograms into two triangles by cutting any of its diagonals. Are the triangles congruents.
Answer:
Do it yourself.

NCERT In-text Question Page No. 219

Question 1.
In the adjoining figure,
(a) Which square has the larger perimeter?
(b) Which is larger, perimeter of smaller square or the circumference of the circle?
Answer:
(a) The outer square has the larger perimeter.
(b) The circumference of the circle is larger than the perimeter of the smaller square.

NCERT In-text Question Page No. 222

Question 1.
Draw circles of different radii on a graph paper. Find the drea by counting the number of squares. Also find the area by using the formula. Compare the two answers.
Answer:
Do it yourself.

NCERT In-text Question Page No. 225

Question 1.
Convert the following:
(i) 50 cm² in mm²
(ii) 2 ha in m²
(iii) 10 m2 in cm²
(iv) 1000 cm² in m²
Answer:
(i) 50 cm² in mm²
∵ 1 cm² = 100 mm²
∴ 50 cm² = 50 x 100 mm² = 5000 mm²

(ii) 2 ha in m²
1 ha = 1000 m²
∴ 2 ha = 2 x 1000 m2 = 20000 m²

(iii) 10 m² in cm²
∵ 1 cm² = 10000 cm²
∴ 10 m²= 10 x 10000 cm² = 100000 cm²

(iv) 1000 cm² in m²
∵ 10000 cm² = 1 m²
∴ 1 cm² = \(\frac{1}{10000}\) m²
So, 1000 cm² = \(\frac{1}{10000}\) x 1000 m²
= \(\frac{1}{10}\) m² = 0.1m²