Q1: Give an example of a monomial and a binomial having degrees of 82 and 99, respectively. Sol:An example of a monomial having a degree of 82 = x82 An example of a binomial having a degree of 99 = x99 + x
Q2: Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1. Sol:Let the polynomial be f(x) = 5x – 4x2 + 3 Now, for x = 2, f(2) = 5(2) – 4(2)2 + 3 ⇒ f(2) = 10 – 16 + 3 = –3 Or, the value of the polynomial 5x – 4x2 + 3 at x = 2 is -3. Similarly, for x = –1, f(–1) = 5(–1) – 4(–1)2 + 3 ⇒ f(–1) = –5 –4 + 3 = -6 The value of the polynomial 5x – 4x2 + 3 at x = -1 is -6.
Q3:Factorise the quadratic polynomial by splitting the middle term: x2 + 14x + 45 Sol: x2 + 14x + 45
here a = 1, b = 14, c = 45
ac = 45 = 9 x 5 , b = 14 = 9 + 5
x2 + 14x + 45
= x2 + 9x + 5x + 45
= x( x + 9 ) + 5(x + 9)
= (x + 9) (x + 5)
Q4: Check whether 3 and -1 are zeros of the polynomial x+4.
Sol: Let p(x)=x+4.
Now, check for each value:
p(3) =3+4 =7
p(−1)=−1+4 =3
Therefore, 3 and -1 are not zeros of the polynomial x+4.
Q5: Check whether (7 + 3x) is a factor of (3x3 + 7x). Sol:Let p(x) = 3x3 + 7x and g(x) = 7 + 3x. Now g(x) = 0 ⇒ x = –7/3. By the remainder theorem, we know that when p(x) is divided by g(x) then the remainder is p(–7/3). Now, p(–7/3) = 3(–7/3)3 + 7(–7/3) = –490/9 ≠ 0. ∴ g(x) is not a factor of p(x).
Q6: Verify whether 1 and -2 are zeros of the polynomial x2+3x. Sol: Let p(x)=x2+3x.
Now, check for each value:
p(1) =12+3(1)=1+3 =4
p(−2) =(−2)2+3(−2)=4−6 =−2
Therefore, 1 and -2 are not zeros of the polynomial x2+3x.
Q7: Calculate the perimeter of a rectangle whose area is 25x2 – 35x + 12. Sol:Given, Area of rectangle = 25x2 – 35x + 12 We know, area of rectangle = length × breadth So, by factoring 25x2 – 35x + 12, the length and breadth can be obtained. 25x2 – 35x + 12 = 25x2 – 15x – 20x + 12 ⇒ 25x2 – 35x + 12 = 5x(5x – 3) – 4(5x – 3) ⇒ 25x2 – 35x + 12 = (5x – 3)(5x – 4) So, the length and breadth are (5x – 3)(5x – 4). Now, perimeter = 2(length + breadth) So, perimeter of the rectangle = 2[(5x – 3)+(5x – 4)] = 2(5x – 3 + 5x – 4) = 2(10x – 7) = 20x – 14 So, the perimeter = 20x – 14
Q8: Find the value of k, if x+2 is a factor of 3x3+5x2−2x+k. Sol: As x+2 is a factor of p(x)=3x3+5x2−2x+k, we know that p(−2)=0.
Now, calculate p(−2) :
p(−2) =3(−2)3+5(−2)2−2(−2)+k
p(−2) =3(−8)+5(4)+4+k = −24+20+4+k
0 =−24+20+4+k
0 = 0+k
k =0.
Q9: Find the values of a and b so that (2x3 + ax2 + x + b) has (x + 2) and (2x – 1) as factors. Sol:Let p(x) = 2x3 + ax2 + x + b. Then, p( –2) = and p(½) = 0. p(2) = 2(2)3 + a(2)2 + 2 + b = 0 ⇒ –16 + 4a – 2 + b = 0 ⇒ 4a + b = 18 ….(i) p(½) = 2(½)3 + a(½)2 + (½) + b = 0 ⇒ a + 4b = –3 ….(ii) On solving (i) and (ii), we get a = 5 and b = –2. Hence, a = 5 and b = –2.
Q1: Find five rational numbers between 1 and 2. Sol: We have to find five rational numbers between 1 and 2. So, let us write the numbers with denominator 5 + 1 = 6 Thus, 6/6 = 1, 12/6 = 2 From this, we can write the five rational numbers between 6/6 and 12/6 as: 7/6, 8/6, 9/6, 10/6, 11/6
Q2: Locate √3 on the number line. Sol:
Construct BD of unit length perpendicular to OB (here, OA = AB = 1 unit) as shown in the figure. By Pythagoras theorem, OD = √(2 + 1) = √3 Taking O as the centre and OD as radius, draw an arc which intersects the number line at the point Q using a compass. Therefore, Q corresponds to the value of √3 on the number line.
Q3: Find the decimal expansions of 10/3, 7/8 and 1/7. Sol: Therefore, 10/3 = 3.3333… 7/8 = 0.875 1/7 = 0.1428571…
Q4: What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer. Sol: Thus, 1/17 = 0.0588235294117647…. Therefore, 1/17 has 16 digits in the repeating block of digits in the decimal expansion.
Q5: Visualise 3.765 on the number line, using successive magnification. Sol: Visualisation of 3.765 on the number line, using successive magnification is given below: Q6: Simplify: (√3+√7) (√3-√7). Sol: (√3 + √7)(√3 – √7) Using the identity (a + b)(a – b) = a2 – b2, (√3 + √7)(√3 – √7) = (√3)2 – (√7)2 = 3 – 7 = -4
Q7: Find five rational numbers between 3/5 and 4/5. Sol: We have to find five rational numbers between 3/5 and 4/5. So, let us write the given numbers by multiplying with 6/6, (here 6 = 5 + 1) Now, 3/5 = (3/5) × (6/6) = 18/30 4/5 = (4/5) × (6/6) = 24/30 Thus, the required five rational numbers will be: 19/30, 20/30, 21/30, 22/30, 23/30
Q8: Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number. Sol: No, since the square root of a positive integer 16 is equal to 4. Here, 4 is a rational number.
Q9: Find three different irrational numbers between the rational numbers 5/7 and 9/11. Sol: The given two rational numbers are 5/7 and 9/11. 5/7 = 0.714285714….. 9/11 = 0.81818181…… Hence, the three irrational numbers between 5/7 and 9/11 can be: 0.720720072000… 0.730730073000… 0.808008000…
Q1: For a particular year, following is the distribution of ages (in years) of primary school teachers in a district :
(i) Write the lower limit of first class interval. (ii) Determine the class limits of the fourth class interval. (iii) Find the class mark of the class 45 – 50. (iv) Determine the class size. Ans: (i) First class interval is 15 – 20 and its lower limit is 15. (ii) Fourth class interval is 30 – 35. Lower limit is 30 and upper limit is 35. (iii) Class mark of the class 45 – 50 = (45 + 50) / 2 = 95 / 2 = 47.5. (iv) Class size = Upper limit of each class interval – Lower limit of each class interval. ∴ Here, class size = 20 – 15 = 5. Q2: Find the mean of the following distribution :
Ans:
Now,
Explanation: Use the formula mean = Σ(f × x) / Σf, where f is frequency and x is class mark. The required values of Σ(f × x) and Σf are shown in the calculation above. Substituting these values gives the mean = 19.
Q3: In figure, there is a histogram depicting daily wages of workers in a factory. Construct the frequency distribution table.
Ans:
Explanation: The frequency distribution table is obtained by reading the class intervals and the heights of the bars from the histogram. The table constructed from the given histogram is shown above.
Q4: Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. The median of the data is 24. Find the value of x. Ans:
Here, the arranged data is 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43.
Total number of observations = 10.
For an even number (10) of observations, the median is the average of the 5th and 6th observations.
5th observation = x + 1 6th observation = 2x – 13
Therefore, (x + 1 + 2x – 13) / 2 = 24
⇒ (3x – 12) / 2 = 24
⇒ 3x – 12 = 48
⇒ 3x = 60
⇒ x = 20.
∴ The value of x = 20.
Q5: Draw a histogram for the given data :
Ans:
Let us represent class-intervals along x-axis and corresponding frequencies along y-axis on
a suitable scale. For each class interval draw a rectangle whose base is the class width and whose height is the frequency (or frequency density if class widths differ). The required histogram is as under :
Q6: Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 Find the mean score and median score of the data. Ans: Ascending order of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25.
Mean score: Sum of scores = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 = 158. Number of students = 9. Mean = Sum / Number = 158 / 9 = 17.555… ≈ 17.56 marks.
Median score: Since there are 9 (odd) observations, the median is the 5th value in the ordered list. 5th value = 19. ∴ Median = 19 marks.
Q7: Draw a histogram of the weekly pocket expenses of 125 students of a school given below :
Ans: Here, the class sizes are different, so calculate the frequency density (adjusted frequency) for each class by using the formula:
frequency density = frequency/class width.
Alternatively, if you wish to scale densities to the smallest class width (here minimum class width = 10), you may use:
adjusted frequency = frequency × (Minimum class size/class width).
Here, the minimum class size = 10 – 0 = 10.
Let us represent weekly pocket money along x-axis and corresponding frequency densities (or adjusted frequencies) along y-axis on a suitable scale. The required histogram using these densities is given below :
Q8: The weight in grams of 35 mangoes picked at random from a consignment are as follows: 131, 113, 82, 75, 204, 81, 84, 118, 104, 110, 80, 107, 111, 141, 136, 123, 90, 78, 90, 115, 110, 98, 106, 99, 107, 84, 76, 186, 82, 100, 109, 128, 115, 107, 115 From the grouped frequency table by dividing the variable range into interval of equal width of 20 grams, such that the mid-value of the first class interval is 70 g. Also, draw a histogram. Ans:
It is given that the size of each class interval = 20 and the mid-value of the first class interval is 70.
Let the lower limit of the first class interval be a, then its upper limit = a + 20.
Mid-value of first interval = (a + a + 20) / 2 = a + 10 = 70 …………..
⇒ a = 70 – 10 = 60.
Thus, the first class interval is 60 – 80 and the other class intervals are 80 – 100, 100 – 120, 120 – 140, 140 – 160, 160 – 180, 180 – 200 and 200 – 220.
So, the grouped frequency table formed by counting the number of mango weights falling in each interval is as under :
Let us represent weight (in g) along x-axis and corresponding frequencies along y-axis on a suitable scale. The required histogram is as under :
Q9: Find the mean salary of 60 workers of a factory from the following table :
Ans:
Explanation: Use the formula mean = Σ(f × x) / Σf, where x is the class mark and f the frequency. The table above shows the class marks, f, and f × x values. Summing these gives Σf = 60 and Σ(f × x) = 304,999.8 (as computed in the table). Therefore mean salary = Σ(f × x) / Σf = 304,999.8 / 60 = ₹5,083.33.
Q10: In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon.
Ans:
∴ Lower limit of first class interval is 305 – 10/2 = 300.
Upper limit of first class interval is 305 + 10/2 = 310.
Thus, first class interval is 300 – 310.
Required histogram and frequency polygon are drawn on graph paper as shown below:
Q11: The following two tables gives the distribution of students of two sections according to the marks obtained by them :
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections. Ans:
The class marks are as under :
Let us take class marks on X-axis and frequencies on Y-axis. To plot frequency polygon of Section-A, plot the points (5, 3), (15, 9), (25, 17), (35, 12), (45, 9) and join these points by straight line segments. To plot frequency polygon of Section-B, plot the points (5, 5), (15, 19), (25, 15), (35, 10), (45, 1) on the same scale and join these points by dotted line segments.
From the two polygons, Section A’s polygon lies above Section B’s polygon for most of the middle and higher class marks. This indicates that Section A has more students scoring in the middle and higher ranges compared with Section B; hence, overall performance of Section A is better.
Q1: A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm, find the volume of the spherical ball. [use π = 3.14] Ans: Since curved surface of half of the spherical ball = 56.57 cm2 2πr2 = 56.57
= 113.04 cm3
Q2: Find the capacity in litres of a conical vessel having height 8 cm and slant height 10 cm. Ans: Height of conical vessel (h) = 8 cm Slant height of conical vessel (l) = 10 cm ∴ r2 + h2 = l2 ⇒ r2 + 82 = 102 ⇒ r2 = 100 – 64 = 36 ⇒ r = 6 cm Now, volume of conical vessel = 1/3πr2h = 1/3 × 227 × 6 × 8 = 301.71 cm3 = 0.30171 litre
Q3: Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. Ans: Here, radius of hemispherical dome (r) = 14 m Surface area of dome = 2πr2 = 2 × 22/7 × 14 × 14 = 1232 m2 Hence, total surface area to be whitewashed from outside is 1232 m2.
Q4: A rectangular piece of paper is 22 cm long and 10 cm wide. A cylinder is formed by rolling the paper along its length. Find the volume of the cylinder. Ans: Since rectangular piece of paper is rolled along its length. ∴ 2πr = 22 r = 22 / (2π) = 22 / (2 × 22/7) = 22 × 7 / 44 = 3.5 cm Height of cylinder (h) = 10 cm ∴ Volume of cylinder = πr2h = 22/7 × 3.5 × 3.5 × 10 = 385 cm3
Q5: A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find it volume. If 1m3 wheat cost is ₹10, then find total cost. Ans: Diameter of cone = 10.5 m Radius of cone (r) = 5.25 m Height of cone (h) = 3 m Volume of cone = 13πr2h = 13 × 22/7 × 5.25 × 5.25 × 3 = 86.625 m3 Cost of 1m3 of wheat = ₹10 Cost of 86.625 m3 of wheat = ₹10 × 86.625 = ₹866.25
Q6: A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and1cm3 of water weighs 1 g, find the depth of water. Ans: Since 1 cm3 of water weighs 1 g. ∴ Volume of cylindrical vessel = 154 cm3 πr2h = 154 22/7 × 3.5 × 3.5 × h = 154 h = 4cm Hence, the depth of water is 4 cm.
Q7: A wall of length 10 m is to be built across an open ground. The height of the wall is 5 m and thickness of the wall is 42 cm. If this wall is to be built with brick of dimensions 42 cm × 12 cm × 10 cm, then how many bricks would be required? Ans: Here, length of the wall (L) = 10 m = 1000 cm Breadth of the wall (B) = 42 cm Height of the wall (H) = 5 m = 500 cm ∴ Volume of the wall = L × B × H = 1000 × 42 × 500 cm3 Volume of each brick = 42 × 12 × 10 cm3 = 4167 Hence, the required number of bricks is 4167.
Q8: The volume of cylindrical pipe is 748 cm. Its length is 0.14 m and its internal radius is 0.09 m. Find thickness of pipe. Ans: Internal radius (r) of cylindrical pipe = 0.09 m = 9 cm Length (height) of cylindrical pipe (h) = 0.14 m = 14 cm Let external radius of the cylindrical pipe be R cm. Volume of cylindrical pipe = 748 cm3 ⇒ π(R2 – r2)h = 748 ⇒ 22/7 (R2 – 92)14 = 748 ⇒ R2 = 81 + 17 = 98 ⇒ R = √98 = 7√2 cm = 9.9 cm Thus, thickness of the pipe = 9.9 -9 = 0.9 cm
Q9: The curved surface area of a cylinder is 154 cm. The total surface area of the cylinder is three times its curved surface area. Find the volume of the cylinder. Ans: Since curved surface area of cylinder = 154 cm2 (given] Total surface area of cylinder = 3 × curved surface area 3 × CSA = 3 × 154 = 462 cm2 So 2πr2 = 462 – 154 = 308
Q10: A right-angled ∆ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. Find the volume of the solid generated. Also, find the total surface area of the solid. Ans: When right angled ∆ABC is revolved about AB = 4 cm, it forms a right circular cone of radius 3 cm and height 4 cm. Slant height of the cone is 5 cm.
Q11: A semicircular sheet of metal of radius 14 cm is bent to form an open conical cup. Find the capacity of the cup. Ans: Radius of semicircular sheet (r) = 14 cm ∴ Slant height (1) = 14 cm Circumference of base = Circumference of semicircular sheet
Q12: It costs ₹3300 to paint the inner curved surface of a 10 m deep well. If the rate cost of painting is of ₹30 per m2, find : (a) inner curved surface area (b) diameter of the well (c) capacity of the well. Ans: Depth of well (h) = 10 m Cost of painting inner curved surface is ₹30 per m2 and total cost is ₹3300. Hence, inner curved surface area is 110 m2, diameter of the well is 2 × 1.75 i.e., 3.5 m and capacity of the well is 96.25 m3.
Q13: Using clay, Anant made a right circular cone of height 48 cm and base radius 12 cm. Versha reshapes it in the form of a sphere. Find the radius and curved surface area of the sphere so formed. Ans: Height of cone (h) = 48 cm Radius of the base of cone = 12 cm Let R be the radius of sphere so formed ∴ Volume of sphere = Volume of cone 4/3πR3 = 1/3πr2h 4R3 = 12 × 12 × 48 R3 = 12 × 12 × 12 R = 12 cm Now, curved surface area of sphere = 4πR2 = 4 × 22/7 × 12 × 12 = 1810.29 cm
Q14: A dome of a building is in the form of a hemisphere. From inside, it was whitewashed at the cost of ₹498.96. If the rate of whitewashing is ₹4 per square metre, find the : (i) Inside surface area of the dome (ii) Volume of the air inside the dome. Ans: Here, dome of building is a hemisphere. Total cost of whitewashing inside the dome = ₹498.96 Rate of whitewashing = ₹4 per m2
Q15: A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm. Find the volume of the solid so obtained. If it is now revolved about the side 12 cm, then what would be the ratio of the volumes of the two solids obtained in two cases ? Ans: Here, right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm. ∴ Radius of the base of cone = 12 cm Height of the cone = 5 cm = 12 : 5
Q16: A right triangle of hypotenuse 13 cm and one of its sides 12 cm is made to revolve taking side 12 cm as its axis. Find the volume and curved surface area of the solid so formed. Ans: Here, hypotenuse and one side of a right triangle are 13 cm and 12 cm respectively. Now, given triangle is revolved, taking 12 cm as its axis ∴ Radius of the cone (r) = 5 cm Height of the cone (h) = 12 cm Slant height of the cone (1) = 13 cm ∴ Curved surface area = πrl = π(5)(13) = 65π cm2 Volume of the cone = 1/2πr2h = 1/2π × 5 × 5 × 12 = 100π cm3 Hence, the volume and curved surface area of the solid so formed are 100 π cm3 and 65 π cm2 respectively.
Q1: Find the area of a triangle whose sides are 11 m, 60 m and 61 m.
Ans: Let a = 11 m, b = 60 m and c = 61 m : Firstly, to calculate semi perimeter (s) ∴ a + b + c2 = 11 + 60 + 612 = 1322 = 66 m Now, s − a = 66 − 11 = 55 m s − b = 66 − 60 = 6 m s − c = 66 − 61 = 5 m ∴ Area = √[s(s − a)(s − b)(s − c)] = √[66(55)(6)(5)] = √108900 = 330 m2
Q2: Suman has a piece of land, which is in the shape of a rhombus. She wants her two sons to work on the land and produce different crops. She divides the land in two equal parts by drawing a diagonal. If its perimeter is 400 m and one of the diagonals is of length 120 m, how much area each of them will get for his crops ?
Ans: Here, perimeter of the rhombus is 400 m. ∴ Side of the rhombus = 400/4 = 100 m Let diagonal BD = 120 m and this diagonal divides the rhombus ABCD into two equal parts.
∴ s = 100 + 120 + 1002 = 3202 = 160 ∴Area of ΔABD = √[s(s − a)(s − b)(s − c)] = √[160(160 − 100)(160 − 100)(160 − 120)] = √[160 × 60 × 60 × 40] = 80 × 60 = 4800 m2 Hence, area of land allotted to two sons for their crops is 4800 m2 each.
Q3: The perimeter of a triangular field is 144 m and its sides are in the ratio 3:4:5. Find the length of the perpendicular from the opposite vertex to the side whose length is 60 m.
Ans: Let the sides of the triangle be 3x, 4x and 5x ∴ The perimeter of the triangular field = 144 m ⇒ 3x + 4x + 5x = 144m ⇒ 12x = 144m
⇒ x = 14412 = 12 m Sides of the triangle are: (3 × 12 = 36) m, (4 × 12 = 48) m, (5 × 12 = 60) m ∴ s = a + b + c2 = 36 + 48 + 602 = 1442 = 72 m Area of ΔABC = √[s(s − a)(s − b)(s − c)] (using Heron’s formula) = √[72(72 − 36)(72 − 48)(72 − 60)] = √[72 × 36 × 24 × 12] = √746496 = 864 m2 ∴ Also, Area (ΔABC) = 12 × AD × BC (Using the formula for area) = 1 2 × AD × 60 = 30 x AD ∴ 30 x AD = 864 AD = 86430 = 28.8 m
Q4:Find the area of the triangle whose perimeter is 180 cm and two of its sides are of lengths 80 cm and 18 cm. Also, calculate the altitude of the triangle corresponding to the shortest side.
Ans:
Perimeter of given triangle = 180 cm Two sides are 18 cm and 80 cm ∴ Third side = 180 – 18 – 80 = 82 cm s = 1802 = 90 cm Area of triangle using Heron’s Formula: Area = √[s(s − a)(s − b)(s − c)] = √[90(90 − 18)(90 − 80)(90 − 82)] = √[90 × 72 × 10 × 8] = √518400 = 720 cm2 Use area formula to calculate height (h): 12 × base × height = Area 12 × 18 × h = 720 9 × h = 720 h = 7209 = 80 cm
Hence, area of triangle is 720 cm2and altitude of the triangle corresponding to the shortest side is 80 cm.
Q5: Calculate the area of the shaded region.
Ans:
Area of ΔAOB:
Area = 12 × OA × OB
= 12 × 12 × 5 = 30 cm2
Calculate AB:
AB2 = OA2 + OB2
= 122 + 52 = 144 + 25 = 169
AB = √169= 13 cm
Area of ΔABC: a = BC = 14 cm, b = CA = 15 cm, c = AB = 13 cm s = a + b + c2 = 14 + 15 + 132 = 422 = 21 cm
Using Heron’s Formula: Area = √[s(s − a)(s − b)(s − c)] = √[21(21 − 14)(21 − 15)(21 − 13)] = √[21 × 7 × 6 × 8] = √[3 × 7 × 7 × 2 × 3 × 2 × 2 × 2] = 2 × 2 × 3 × 7 = 84 cm2 Area of shaded region = Area of ∆ABC – Area of ∆AOB = 84 cm2– 30 cm2= 54 cm2
Q6: The sides of a triangular park are 8 m, 10 m and 6 m respectively. A small circular area of diameter 2 m is to be left out and the remaining area is to be used for growing roses. How much area is used for growing roses ? (use n = 3.14)
Ans: The sides of the triangular park are 8 m, 10 m and 6 m. ∴ s = a + b + c2 = 8 + 10 + 62 = 242 = 12 m Area of the park using Heron’s Formula: Area = √[s(s − a)(s − b)(s − c)] = √[12(12 − 8)(12 − 10)(12 − 6)] = √[12 × 4 × 2 × 6] = √[2 × 2 × 3 × 2 × 2 × 2 × 3] = 2 × 2 × 2 × 3 = 24 m2 Radius of the circle = 2/2 = 1 m Area of the circle = πr2 = 3.14 × 1 × 1 = 3.14 m2 ∴ Area to be used for growing roses = Area of the park – area of the circle ⇒ 24 – 3.14 = 20.86 m2
Q7: The Triangular garden has sides 40 m, 60 m, and 80 m. A landscaping company needs to install a pond in the center of the garden. How much area will the company need to clear for the pond?
Ans:
Given, The sides of the triangular garden are 40 m, 60 m, and 80 m.
Q8: The sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540 cm. Find its area.
Ans:
Given, Ratio of the sides of the triangle is 12: 17: 25
Let the sides of triangle be 12x, 17x and 25x
Given, perimeter of the triangle = 540 cm
12x + 17x + 25x = 540 cm
⇒ 54x = 540cm
So, x = 10
Thus, the sides of the triangle are:
12 x 10 = 120 cm
17 x 10 = 170 cm
25 x 10 = 250 cm
Semi perimeter, s = 5402 = 270 cm
Using Heron’s formula,
Area of the triangle = √[s (s-a) (s-b) (s-c)]
= √[270 (270 – 120) (270 – 170) (270 – 250)]
= √(270 x 150 x 100 x 20) = 9000 cm2
Q9: Find the area of a triangle whose two sides are 14 cm and 12 cm, respectively, and the perimeter is 38 cm.
Ans: Given Two sides of the triangle are 14 cm and 12 cm, and the perimeter is 38 cm. Let the third side be c. The perimeter of the triangle is given as 38 cm: a + b + c = 38 14 + 12 + c = 38 c = 38 – 26 = 12 cm The semi-perimeter s is calculated as: s = 382 = 19 cm Using Heron’s formula: Area = √[s(s – a)(s – b)(s – c)] Substitute the values: Area = √[19(19 – 14)(19 – 12)(19 – 12)] Area = √[19 × 5 × 7 × 7] Area = √[4655] The area of the triangle is approximately 68.22 cm²
Q10: The sides of a quadrilateral, taken in order as 5, 12, 14, and 15 meters, respectively, and the angle contained by the first two sides is a right angle. Find its area.
Ans:
Here, AB = 5 m, BC = 12 m, CD =14 m and DA = 15 m
Join the diagonal AC.
Now, the area of △ABC = 1/2 ×AB×BC
= 1/2×5×12 = 30
The area of △ABC is 30 m2
In △ABC, (right triangle).
From Pythagoras theorem,
AC2 = AB2 + BC2
AC2 = 52 + 122
AC2 = 25 + 144 = 169
or AC = 13 m
Now in △ADC,
All sides are known, apply Heron’s Formula:
Area of triangle = √[s(s − a)(s − b)(s − c)] Semi-Perimeter, s = a + b + c2 Where: a, b, and c are the sides of the triangle. Perimeter of △ADC = 2s = AD + DC + AC 2s = 15 m +14 m +13 m s = 21 m Area of triangle ADC = √[21 × (21 − 13) × (21 − 14) × (21 − 15)] Area of triangle ADC = √[21 × 8 × 7 × 6] Area of △ADC = 84 m2 Area of quadrilateral ABCD = Area of △ABC + Area of △ADC = (30 + 84) m2 = 114 m2
Question 1. Look at the adjoining figure. If O is the center of the circle. PQ=12cm and ST = 3 cm, then find the radius of the circle when RS ⊥ PQ. Solution: Let us join O and P such that OP = r. ∵ RS ⊥ PQ
∴ T is the mid-point of PQ .⇒ PT = (1/2) PQ⇒ PT = (1/2) x 12 cm = 6 cm [∵ PQ = 12 cm (Given)] And ∠ OTP = 90° Also OS = r and TS = 3 cm ∴ OT = OS – TS = (r – 3) cm Now, in right ΔOTP, we have OP2 = PT2 + OT2 ⇒ r2 = 62 + (r – 3)2 ⇒ r2 = 36 + r2 + 9 – 6r ⇒ 6r = 45
⇒ r =(45/6) = (15/2) = 7.5 cm Thus, the radius of the circle is 7.5 cm.
Question 2. An equilateral triangle ABCABC is inscribed in a circle. Each side of the triangle is 99cm. Find the radius of the circle. Solution: Let ABC be an equilateral triangle such that AB = BC = AC = 9 cm (each) Let us draw a median AD corresponding to BC. ∴ BD =(1/2) BC ⇒ BD = (1/2) x 9 cm = (9/2)cm Also, AD ⊥ BC [∵ O is the centre of the circle] Now, in right ΔADB,
AD2 = AB2 – BD2
Since, in an equilateral triangle, the centroid and circumcentre coincide.
Q1: In the figure, P, Q and R are the mid-points of the sides BC, AC and AB of ΔABC. If BQ and PR intersect at X and CR and PQ intersect at Y, then show that XY = 1/4 BC.
Ans: R and Q are mid-points of AB and AC respectively. ∴ By the mid-point theorem, RQ || BC and RQ = 1/2 BC. Also P is the mid-point of BC, so BP = PC = 1/2 BC. Thus RQ ∥ BP and RQ = BP. Consider quadrilateral BPQR with vertices B-P-Q-R. Since one pair of opposite sides (BP and RQ) are equal and parallel, BPQR is a parallelogram. Therefore its diagonals PR and BQ bisect each other, so X (the intersection of PR and BQ) is the mid-point of PR. Similarly, in quadrilateral PCQR, the opposite sides PC and RQ are equal and parallel, so PCQR is a parallelogram. Hence its diagonals PQ and CR bisect each other, so Y (the intersection of PQ and CR) is the mid-point of PQ. Now, in ΔPQR, X and Y are mid-points of PR and PQ respectively. Therefore XY is parallel to RQ and XY = 1/2 · RQ (mid-point theorem applied in ΔPQR). But RQ = 1/2 BC, so XY = 1/2 · (1/2 BC) = 1/4 BC. Thus XY = 1/4 BC.
Q2: In ΔABC, AB = 8 cm, BC = 9 cm and AC = 10 cm. X, Y and Z are mid-points of AO, BO and CO respectively as shown in the figure. Find the lengths of the sides of ΔXYZ.
Ans: Given AB = 8 cm, BC = 9 cm and AC = 10 cm. In ΔAOB, X and Y are mid-points of AO and BO respectively. ∴ By the mid-point theorem, XY = 1/2 · AB = 1/2 × 8 cm = 4 cm. In ΔBOC, Y and Z are mid-points of BO and CO respectively. ∴ YZ = 1/2 · BC = 1/2 × 9 cm = 4.5 cm. In ΔCOA, Z and X are mid-points of CO and AO respectively. ∴ ZX = 1/2 · AC = 1/2 × 10 cm = 5 cm. Hence the sides of ΔXYZ are XY = 4 cm, YZ = 4.5 cm and ZX = 5 cm.
Q3: ABCD is a parallelogram. If the bisectors DP and CP of angles D and C meet at P on side AB, then show that P is the mid-point of side AB. Ans:
Ans: Let DP and CP bisect ∠D and ∠C respectively. So the two halves of ∠D are equal and the two halves of ∠C are equal. Since AB ∥ DC and CP is a transversal, an angle formed at P with AB is equal to the corresponding half of ∠C. Hence an angle at B in triangle BCP equals the corresponding half of ∠D. Therefore, in triangle BCP two angles are equal, which implies BC = BP. (angles opposite equal sides are equal and conversely) Similarly, since AB ∥ DC and DP is a transversal, an angle at A in triangle ADP equals the corresponding half of ∠D and so in triangle ADP two angles are equal; hence DA = AP. But in a parallelogram opposite sides are equal, so BC = AD. From BC = BP, DA = AP and BC = AD, we get BP = AP. Hence P is the mid-point of AB.
Q4: In the figure, ΔBCD is a trapezium in which AB || DC. E and F are the mid-points of AD and BC respectively. DF and AB are produced to meet at G. Also, AC and EF intersect at the point O. Show that : (i) EO || AB (ii) AO = CO
Ans: F is the mid-point of BC, so BF = FC. Lines DF and AB are produced to meet at G, so points D, F, G are collinear. Consider triangles BFG and CFD: BF = FC, ∠BFG = ∠CFD (vertical angles), and ∠BGF = ∠CDF (alternate interior angles, as AB ∥ DC). Thus ΔBFG ≅ ΔCFD by AAS, and so DF = FG. Since D, F, G are collinear and DF = FG, F is the mid-point of DG. Now, in ΔAGD, E is the mid-point of AD and F is the mid-point of DG. By the mid-point theorem, EF ∥ AG. But G lies on AB, so AG is a segment of AB; therefore EF ∥ AB and, in particular, EO (part of EF) ∥ AB. This proves (i). Because AB ∥ DC, EF ∥ AB implies EF ∥ DC as well. In ΔADC, E is the mid-point of AD and EO ∥ DC, so the line through the mid-point E parallel to DC bisects AC. Hence O is the mid-point of AC and AO = CO. This proves (ii).
Q5: PQRS is a square and ∠ABC = 90° as shown in the figure. If AP = BQ = CR, then prove that ∠BAC = 45°
Ans: In square PQRS, all sides are equal, so PQ = QR. (1) Given AP = BQ = CR. Subtracting the equal lengths BQ and CR from PQ and QR respectively gives: PQ – BQ = QR – CR ⇒ PB = QC. (2) In ΔAPB and ΔBQC: AP = BQ (given), ∠APB = ∠BQC = 90° (angles in a square), and PB = QC (from (2)). So ΔAPB ≅ ΔBQC by SAS congruence. Hence AB = BC. In ΔABC, if AB = BC, the base angles opposite those equal sides are equal; let each be x°. Then ∠B + ∠ACB + ∠BAC = 180° ⇒ 90° + x + x = 180° ⇒ 2x = 90° ⇒ x = 45°. Therefore ∠BAC = 45°.
Q6: In the given figure, AE = DE and BC || AD. Prove that the points A, B, C and D are concyclic. Also, prove that the diagonals of the quadrilateral ABCD are equal.
Ans: Since AE = DE ∠D = ∠A …. (i) [∵ ∠s opp. to equal sides of a Δ] Again, BC || AD ∠EBC = ∠A …. (ii) (corresponding ∠s] From (i) and (ii), we have ∠D = ∠EBC …. (iii) But ∠EBC + ∠ABC = 180° (a linear pair] ∠D + ∠ABC = 180° (using (iii)] Now, a pair of opposite angles of quadrilateral ABCD is supplementary Thus, ABCD is a cyclic quadrilateral i.e., A, B, C and D’are concyclic. In ΔABD and ΔDCA ∠ABD = ∠ACD [∠s in the same segment for cyclic quad. ABCD] ∠BAD = ∠CDA [using (i)] AD = AD (common] So, by using AAS congruence axiom, we have ΔABD ≅ ΔDCA Hence, BD = CA [c.p.c.t.]
Q1: In the given figure, AP and DP are bisectors of two adjacent angles A and D of quadrilateral ABCD. Prove that 2 ∠APD = ∠B + 2C. Ans: Here, AP and DP are angle bisectors of ∠A and ∠D ∴ ∠DAP = 1/2∠DAB and ∠ADP = 1/2∠ADC ……(i) In ∆APD, ∠APD + ∠DAP + ∠ADP = 180° ⇒ ∠APD + 1/2 ∠DAB + 1/2∠ADC = 180° ⇒ ∠APD = 180° – 1/2(∠DAB + ∠ADC) ⇒ 2∠APD = 360° – (∠DAB + ∠ADC) ……(ii) Also, ∠A + ∠B + C + ∠D = 360° ∠B + ∠C = 360° – (∠A + ∠D) ∠B + ∠C = 360° – (∠DAB + ∠ADC) ……(iii) From (ii) and (iii), we obtain 2∠APD = ∠B + ∠C
Q2: In right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see fig.). Show that : (i) ∆AMC ≅ ∆BMD (ii) ∠DBC = 90° (ii) ∆DBC ≅ ∆ACB (iv) CM = 1/2AB Ans: Given : ∆ACB in which 4C = 90° and M is the mid-point of AB. To Prove : (i) ∆AMC ≅ ∆BMD (ii) ∠DBC = 90° (iii) ∆DBC ≅ ∆ACB (iv) CM = 1/2AB Proof : Consider ∆AMC and ∆BMD, we have AM = BM [given] CM = DM [given – construction] ∠AMC = ∠BMD [vertically opposite angles] ∴ ∆AMC ≅ ∆BMD [by SAS congruence axiom] ⇒ AC = DB …(i) [by c.p.c.t.] and ∠1 = ∠2 [by c.p.c.t.] But ∠1 and ∠2 are alternate angles.
⇒ BD || CA Now, BD || CA and BC is transversal. ∴ ∠ACB + ∠CBD = 180° ⇒ 90° + CBD = 180° ⇒ ∠CBD = 90° In ∆DBC and ∆ACB, we have CB = BC [common] DB = AC [using (i)] ∠CBD = ∠BCA ∴ ∆DBC ≅ ∆ACB ⇒ DC = AB ⇒ 1/2AB = 1/2DC ⇒ 1/2AB = CM or CM = 1/2AB (∵ CM = 1/2DC)
Q3: Prove that two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle. Ans:
Given : Two As ABC and DEF in which ∠B = ∠E, ∠C = ∠F and BC = EF To Prove : ∆ABC = ∆DEF Proof : We have three possibilities Case I. If AB = DE, we have AB = DE, ∠B = ∠E and BC = EF. So, by SAS congruence axiom, we have ∆ABC ≅ ∆DEF
Case II. If AB < ED, then take a point Mon ED such that EM = AB. Join MF. Now, in ∆ABC and ∆MEF, we have AB = ME, ∠B = ∠E and BC = EF. So, by SAS congruence axiom, we have ΔΑΒC ≅ ΔΜEF ⇒ ∠ACB = ∠MFE But ∠ACB = ∠DFE ∴ ∠MFE = ∠DFE
Which is possible only when FM coincides with B FD i.e., M coincides with D. Thus, AB = DE ∴ In ∆ABC and ∆DEF, we have AB = DE, ∠B = ∠E and BC = EF So, by SAS congruence axiom, we have ∆ABC ≅ ∆DEF Case III. When AB > ED Take a point M on ED produced such that EM = AB. Join MF Proceeding as in Case II, we can prove that ∆ABC = ∆DEF Hence, in all cases, we have ∆ABC = ∆DEF.
Q4: In the given figure, side QR is produced to the point S. If the bisectors of ∠PQR and ∠PRS meet at T, prove that ∠QTR = 1/2 ∠QPR.
In triangle QTR, ∠TRS is an exterior angle.
:. ∠QTR + ∠TQR = ∠TRS
∠QTR = ∠TRS – ∠TQR ——(1)
For triangle PQR, ∠PRS is an external angle.
:. ∠QPR + ∠PQR = ∠PRS
∠QPR + 2∠TQR = 2∠TRS (As QT and RT are angle bisectors)
∠QPR = 2(∠TRS – ∠TQR)
∠QPR = 2∠QTR [By using equation (1)]
LQTR= 1/2 ∠QPR
Q5: In figure, ABC is an isosceles triangle with AB = AC. D is a point in the interior of ∆ABC such that ∠BCD = ∠CBD. Prove that AD bisects ∠BAC of ∆ABC.
Ans: In ∆BDC, we have ∠DBC = ∠DCB (given). ⇒ CD = BD (sides opp. to equal ∠s of ∆DBC) Now, in ∆ABD and ∆ACD, we have AB = AC [given] BD = CD [proved above] AD = AD [common] ∴ By using SSS congruence axiom, we obtain ∆ABD ≅ ∆ACD ⇒ ∠BAD = ∠CAD [c.p.ç.t.] Hence, AD bisects ∠BAC of ∆ABC.
Q6: In figure, ABCD is a square and EF is parallel to diagonal BD and EM = FM. Prove that (i) DF = BE (ii) AM bisects ∠BAD.
Ans: (i) EF || BD = ∠1 = ∠2 and ∠3 = ∠4 [corresponding ∠s] Also, ∠2 = ∠4 ⇒ ∠1 = ∠3 ⇒ CE = CF (sides opp. to equals ∠s of a ∆] ∴ DF = BE [∵ BC – CE = CD – CF)
(ii) In ∆ADF and ∆ABE, we have AD = AB [sides of a square] DF = BE [proved above] ∠D = ∠B = 90° ⇒ ∆ADF ≅ ∆ABE [by SAS congruence axiom] ⇒ AF = AE and ∠5 = ∠6 … (i) [c.p.c.t.] In ∆AMF and ∆AME AF = AE [proved above] AM = AM [common] FM = EM (given) ∴ ∆AMF ≅ ∆AME [by SSS congruence axiom] ∴ ∠7 = ∠8 …(ii) [c.p.c.t.] Adding (i) and (ii), we have ∠5 + ∠7 = ∠6 + ∠8 ∠DAM = ∠BAM ∴ AM bisects ∠BAD.
Q1: In the adjoining figure, AB || CD. If ∠ APQ = 54° and ∠ PRD = 126°, then find x and y. Solution: ∵ AB || CD and PQ is a transversal, then interior alternate angles are equal.
⇒ ∠ APQ = ∠ PQR [alt. interior angles] ⇒ 54° = x [∵ ∠ APQ = 54° (Given)] Again, AB || CD and PR is a transversal, then ∠ APR = ∠ PRD [Interior alternate angles] But ∠ PRD = 126° [Given] ∴ ∠ APR = 126° Now, exterior∠ PRD + ∠ PRQ = 180° ⇒ 126°+∠ PRQ = 180° ⇒∠ PRQ= 180°-126° = 54°
Q2: In the adjoining figure AB || CD || EG, find the value of x. Solution: Let us draw FEG || AB || CD through E. Now, since FE || AB and BE are transversals, ∴ ∠ ABE + ∠ BEF = 180° [Interior opposite angles]
⇒ 127° + ∠ BEF = 180° [co int. angles] ⇒ ∠ BEF = 180° – 127° = 53° Again, EG || CD and CE is a transversal. ∴ ∠ DCE + ∠ CEG = 180° [Interior opposite angles] ⇒ 108° + ∠ CEG = 180° ⇒ ∠ CEG = 180° – 108° = 72° Since FEG is a straight line, then ⇒ ∠BEF + ∠BEC + ∠CEG = 180° [Sum of angles at a point on the same side of a line = 180°] ⇒ 53° + x + 72° = 180° ⇒ x = 180° – 53° – 72° = 55° Thus, the required measure of x = 55°.
Q3: AB and CD are parallel, and EF is a transversal. If ∠BEF = 70°, find ∠EFD and ∠CFE
Ans: Since AB || CD and EF is a transversal, by the Corresponding Angles Theorem:
∠BEF = ∠EFD
Thus, ∠EFD = 70°.
By the Linear Pair Axiom:
∠EFD + ∠CFE = 180°
70° + ∠CFE = 180°
∠CFE = 110°.
So, ∠EFD = 70° and ∠CFE = 110°.
Q4: If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of interior angles enclose a rectangle.
Ans:
Two parallel lines AB and CD are intersected by a transversal L at P and R
respectively
PQ, RQ, RS and PS are bisectors of ∠APR, ∠PRC, ∠PRD and ∠BPR respectively.
Since AB || CD and L is a transversal
∠APR = ∠PRD ( alt. interior angles)
∠APQ = 1/2(∠PRD) = ∠QPR = ∠PRS
these are alternate interior angles.
QP || RS. Similarly QR || PS.
PQRS is a parallelogram.
Also ray PR stands on AB
∠APR + ∠BPR = 180° ( linear pair)
∠QPR + (1/2) ∠BPR = 90°
∠QPR + ∠SPR = 90°
∠QPS = 90°
Therefore PQRS is a parallelograrn, one of whose angle is 90°.
Hence PQRS satisfies all the properties of being a rectangle.
Long Question Answer: Introduction To Euclid’s Geometry
Q1: Which of the following statements are true and which are false? Give reasons for your answers. (i) Only one line can pass through a single point. Ans: False Correct statement: Infinite many lines can pass through a single point. This is self-evident and can be seen visually by the student given below:
(ii) There are infinite number of lines which pass through two distinct points. Ans: False The given statement contradicts the postulate I of the Euclid that assures that there is a unique line that passes through two distinct points.
Through two points P and Q a unique line can be drawn. (iii) A terminated line can be produced indefinitely on both the sides. Ans: True
We need to consider Euclid’s Postulate 2: “A terminated line can be produced indefinitely. (iv) If two circles are equal, then their radii are equal. Ans: True Let us consider two circles with same radii. We can conclude that, when we make the two circles overlap with each other, we will get a superimposed figure of the two circles. Therefore, we can conclude that the radii of both the circles will also coincide and still be same.
(v) In Fig. if AB=PQ and PQ=XY, then AB=XY
Ans: True We are given that AB=PQ and PQ=XY. We need to consider the axiom: “Given two distinct points, there is a unique line that passes through them.” Therefore, we can conclude that AB,PQ and XY are the lines with same dimensions, and hence if AB=PQ and PQ=XY, then AB=XY.
Q2: Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them? (i) Parallel lines Ans: Two lines are said to be parallel, when the perpendicular distance between these lines is always constant or we can say that the lines that never intersect each other are called as parallel lines.
We need to define line first, in order to define parallel lines.
(ii) Perpendicular lines Ans: Two lines are said to be perpendicular lines, when angle between these two lines is 90∘.
We need to define line and angle, in order to define perpendicular lines.
(iii) Line segment Ans: A line of a fixed dimension between two given points is called as a line segment.
We need to define line and point, in order to define a line segment.
(iv) Radius of a circle Ans: The distance of any point lying on the boundary of a circle from the center of the circle is called as radius of a circle.
We need to define circle and center of a circle, in order to define radius of a circle.
(v) Square Ans: A quadrilateral with all four sides equal and all four angles of 90∘ is called as a square.
We need to define quadrilateral and angle, in order to define a square.
Q3: If a point C lies between two points A and B such that AC = BC, then prove that AC = 1/2AB⋅ Explain by drawing the figure. Ans: We are given that a point C lies between two points B and C, such that AC = BC. We need to prove that AC = 1/2AB⋅ Let us consider the given below figure.
We are given that AC = BC−…(i) An axiom of the Euclid says that “If equals are added to equals, the wholes are equal.” Let us add AC to both sides of equation (i). AC + AC = BC + AC. An axiom of the Euclid says that “Things which coincide with one another are equal to one another.” “ We can conclude that BC+AC coincide with AB, or AB = BC + AC.…(ii) An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.” From equations (i) and (ii), we can conclude that AC + AC = AB, or 2AC = AB An axiom of the Euclid says that “Things which are halves of the same things are equal to one another.” Therefore, we can conclude that AC = 1/2AB
Q4: In the following figure, if AC = BD, then prove that AB = CD.
Ans: We are given that AC = BD We need to prove that AB = CD in the figure given below.
From the figure, we can conclude that AC = AB + BC, and BD = CD + BC. An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.” AB + BC = CD + BC An axiom of the Euclid says that “when equals are subtracted from equals, the remainders are also equal.” We need to subtract BC from equation(i) to get AB + BC − BC = CD + BC− BC
AB = CD
Therefore, we can conclude that the desired result is proved.
Q5: Why is axiom 5, in the list of Euclid’s axioms, considered as a ‘universal truth’? (Note that the question if not about fifth postulate) Ans: We need to prove that Euclid’s fifth axiom is considered as a universal truth. Euclid’s fifth axiom states that “the whole is greater than the part.” The above given axiom is a universal truth. We can apply the fifth axiom not only mathematically but also universally in daily life. Mathematical proof: Let us consider a quantity z, which has different parts as a,b,x and y So, z = a + b + x + y. Therefore, we can conclude that z will always be greater than its corresponding parts a,b,x and y. Universal proof: We know that Mumbai is located in Maharashtra and Maharashtra is located in India. In other words, we can conclude that Mumbai is a part of Maharashtra and Maharashtra is a part of India. Therefore, we can conclude that whole India will be greater than Mumbai or Maharashtra or both. Therefore, we can conclude that Euclid’s fifth axiom is considered as a ‘Universal truth’. Q6: How would you rewrite Euclid’s fifth postulate so that it would be easier to understand? Ans: We need to rewrite Euclid’s fifth postulate so that it is easier to understand. We know that Euclid’s fifth postulate states that “No intersection of lines will take place when the sum of the measures of the interior angles on the same side of the falling line is exactly 180∘. We know that Playfair’s axiom states that “For every line l and for every point P not lying on 1, there exists a unique line m passing through P and parallel to Γ. The above mentioned Playfair’s axiom is easier to understand in comparison to the Euclid’s fifth postulate. Let us consider a line l that passes through a point p and another line m. Let these lines be at a same plane. Let us consider the perpendicular CD on l and FE on m.
From the above figure, we can conclude that CD = EF. Therefore, we can conclude that the perpendicular distance between lines m and l will b. constant throughout, and the lines m and l will never meet each other or in other words, w can say that the lines m and I are equidistant from each other.
Q7: Does Euclid’s fifth postulate imply the existence of parallel lines? Explain. Ans: We need to verify whether Euclid’s fifth postulate imply the existence of parallel lines or not. The answer to the above statement is Yes. Let us consider two lines m and l In the figure given below, we can conclude that the lines m and l will intersect further.
From the figure, we can conclude that ∠1 + ∠2 < 180∘ and ∠3 +∠4 > 180∘ We know that Euclid’s fifth postulate states that “No intersection of lines will take place when the sum of the measures of the interior angles on the same side of the falling line is exactly 180∘. Let us consider lines l and m.
From the above figure, we can conclude that lines I and m will never intersect from either side. Therefore, we can conclude that the lines l and m are parallel.
Q8: Consider the two ‘postulates’ given below: (i) Given any two distinct points A and B, there exists a third point C, which is between A and B. (ii) There exists at least three points that are not on the same line. Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain. Ans: Given any two distinct points A and B, there exists a third point C, which is between A and B. There exists at least three points that are not on the same line. The undefined terms in the given postulates are point and line. The two given postulates are consistent, as they do not refer to similar situations and they refer to two different situations. We can also conclude that, it is impossible to derive at any conclusion or any statement that contradicts any well-known axiom and postulate. The two given postulates do not follow from the postulates given by Euclid. The two given postulates can be observed following from the axiom, “Given two distinct points, there is a unique line that passes through them”
Q9: In the above question, point C is called a mid-point of line segment AB, prove that every line segment has one and only one mid-point. Ans: We need to prove that every line segment has one and only one mid-point. Let us consider the given below line segment AB and assume that C and D are the mid-points of the line segment AB.
If C is the mid-point of line segment AB, then AC = CB An axiom of the Euclid says that “If equals are added to equals, the wholes are equal.” AC + AC = CB + AC.(i) From the figure, we can conclude that CB + AC will coincide with AB. An axiom of the Euclid says that “Things which coincide with one another are equal to one another.” “ AC+AC = AB.(ii) An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.” “ Let us compare equations (i) and (ii), to get AC+AC = AB, or 2AC = AB.(iii) If D is the mid-point of line segment AB, then AD = DB An axiom of the Euclid says that “If equals are added to equals, the wholes are equal.” AD + AD = DB + AD. (iv) From the figure, we can conclude that DB+AD will coincide with AB. An axiom of the Euclid says that “Things which coincide with one another are equal to one another.” AD + AD = AB.(v) An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another:” Let us compare equations (iv) and (v), to get AD + AD = AB, or 2AD = AB.(vi) An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.” Let us compare equations (iii) and (vi), to get 2AC = 2AD An axiom of the Euclid says that “Things which are halves of the same things are equal to one another.” AC = AD Therefore, we can conclude that the assumption that we made previously is false and a line segment has one and only one mid-point.
If a point C lies between two points A and B such that AB = BC, then prove that AC = ½ AB
Find the number of dimensions a solid, surface, and point have.
Let x + y = 10, and x = z. Show that y + z = 10
There are two sales employees who received equal incentives during the month of July. In August, each sales employee received double incentives in comparison to the month of July. Compare the employees’ incentives for the month of August.
Prove that an equilateral triangle can be formed on any of the given line segments.
Thus, 1/17 = 0.0588235294117647….
∴ T is the mid-point of PQ
