\nSince coordinate of a given point is the distance of point from x-axis.
\n
![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-7-Coordinate-Geometry-1a.png?resize=189%2C165&ssl=1\")
<\/p>\nQuestion 1.<\/p>\n
(A) 2
\n(B) 3
\n(C) 1
\n(D) 5
\nAnswer:
\n(B) 3<\/p>\n
Explanation:
\nSince coordinate of a given point is the distance of point from x-axis.
\n<\/p>\n
![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions.png?resize=159%2C13&ssl=1\")
<\/p>\n
Question 2.<\/p>\n
(A) 6
\n(B) 8
\n(C) 4
\n(D) 2
\nAnswer:
\n(B) 8<\/p>\n
Explanation: Question 3.<\/p>\n (A) 8 Explanation: Question 4.<\/p>\n (A) 5 Explanation: Question 5.<\/p>\n (A) 5 Explanation: Question 6.<\/p>\n (A) 5 (A) right triangle Explanation: Question 8.<\/p>\n Explanation: Assertion and Reason Based MCQs<\/span><\/p>\n Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. Question 1.<\/p>\n Answer: Explanation: Question 2.<\/p>\n Answer: Explanation: Question 3.<\/p>\n Answer: Explanation: Case – Based MCQs<\/span><\/p>\n Attempt any four sub-parts from each question. Each sub-part carries 1 mark. Question 1.<\/p>\n (A) \\(\\frac{33}{2}, \\frac{15}{2}\\) Explanation: Question 2.<\/p>\n (A) 4 Explanation: Question 3.<\/p>\n (A) 4 Explanation: Question 4.<\/p>\n (A) (8.5,2.0) Explanation: Question 5.<\/p>\n (A) x + y = 13 Explanation: II. Read the following text and answer the below questions: Question 1.<\/p>\n (A) 5 km Explanation: Question 2.<\/p>\n (A) 5 km Explanation: Question 3.<\/p>\n (A) 24.6 km Explanation: Question 4.<\/p>\n (A) 5 km Explanation: Question 5.<\/p>\n (A) 2 km Explanation: III. Read the following text and answer the below questions: Question 1.<\/p>\n (A) (2,25) Question 2.<\/p>\n (A) (8,0) Question 3.<\/p>\n (A) \u221a41 Explanation: Question 4.<\/p>\n (A) (5,22.5) Explanation: Question 5.<\/p>\n (A) (3.5,24) Explanation: IV. Read the following text and answer the below questions: Question 1.<\/p>\n (A) (4,6) Question 2.<\/p>\n (A) (8,6) Question 3.<\/p>\n (A) (6,13) Question 4.<\/p>\n (A) 4.5 Explanation: Question 5.<\/p>\n (A) 8 Coordinate Geometry Class 10 MCQ Questions with Answers Question 1. The distance of the point P (2, 3) from the x-axis is: (A) 2 (B) 3 (C) 1 (D) 5 Answer: (B) 3 Explanation: Since coordinate of a given point is the distance of point from x-axis. Question 2. The distance between the points A …<\/p>\n
\nThe distance between two points (x1<\/sub>, y1<\/sub>) and (x2<\/sub>, y2<\/sub> ) is given as,
\nd = \\(\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}\\)
\nWhere, x1, = 0, y1 = 6 and x2 = 0, y2 = -2 So, distance between A (0, 6) and B (0, -2):
\nAB = \\(\\sqrt{(0-0)^{2}+(-2-6)^{2}}\\)
\n= \\(\\sqrt{0+(-8)^{2}}\/latex]
\n= [latex]\\sqrt{8^{2}}\\)
\n= 8<\/p>\nThe distance of the point P (-6,8) from the origin is:<\/h2>\n
\n(B) 2\u221a7
\n(C) 10
\n(D) 6
\nAnswer:
\n(C) 10<\/p>\n
\nDistance between two points (x1<\/sub>, y1<\/sub>) and (x2<\/sub>, y2<\/sub>) is given as,
\nd = \\(\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}\\)
\nWhere, x1<\/sub>= -6, y1<\/sub> =8 and x2<\/sub> = 0, y2<\/sub> = 0 So, distance between P (-6,8) and origin O (0,0) is given by,
\nPO = \\(\\sqrt{[0-(-6)]^{2}+[(0-8)]^{2}}\\)
\n= \\(\\sqrt{(6)^{2}+(-8)^{2}}\\)
\n= \\(\\sqrt{36+64}\\)
\n= \\(\\sqrt{100}\\)
\n= 10<\/p>\nThe distance between the points (0, 5) and (-5, 0) is:<\/h2>\n
\n(B) 5\u221a2
\n(C) 2\u221a5
\n(D) 10
\nAnswer:
\n(B) 5\u221a2<\/p>\n
\nDistance between two points (x1<\/sub>,y1<\/sub>) and (x2<\/sub>,y2<\/sub>) is given as,
\n= \\(\\sqrt{(-5-0)^{2}+(0-5)^{2}} \\)
\nWhere, x1<\/sub>= 0, y1<\/sub>= 5 and x2<\/sub> = -5, y2<\/sub> = 0 So that, distance between the point (0, 5) and I (-5,0),
\n= \\(\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}\\)
\n= \\(\\sqrt{(-5-0)^{2}+(0-5)^{2}}\\)
\n= \\(\\sqrt{25+25}\\)
\n= \\(\\sqrt{50}\\)
\n= 5\u221a2<\/p>\nAOBC is a rectangle whose three vertices are vertices A (0, 3), O (0, 0) and B (5, 0). The length of its diagonal is:<\/h2>\n
\n(B) 3
\n(C) \u221a34
\n(D) 4
\nAnswer:
\n(C) \u221a34<\/p>\n
\nAccording to the question, a triangle can be represented as:
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-7-Coordinate-Geometry-2a.png?resize=186%2C121&ssl=1\")
\n\u2234 Distance between the points A (0,3) and B (5, 0) is
\nAB = \\(\\sqrt{(5-0)^{2}+(0-3)^{2}}\\)
\n= \\(\\sqrt{25+9}\\)
\n= \\(\\sqrt{34}\\)
\nHence, the required length of diagonal is \u221a34.<\/p>\n<\/p>\nThe perimeter of a triangle with vertices (0, 4), (0,0) and (3,0) is:<\/h2>\n
\n(B) 12
\n(C) 11
\n(D)7+\u221a5
\nAnswer:
\n(B) 12
\nExplanation:
\nSince, Perimeter of triangle = OA + AB + OB
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-7-Coordinate-Geometry-2ab.png?resize=201%2C174&ssl=1\")
\nHere OA = 4 units, AB = 5 units (using pythagoras theorem in triangle AOB) and OB =3 units.
\nTherefore, Perimeter of triangle = 4 + 5+3
\n= 12 units.
\nQuestion 7.<\/p>\nThe points (-4,0), (4,0) and (0,3) are the vertices of a:<\/h2>\n
\n(B) isosceles triangle
\n(C) equilateral triangle
\n(D) scalene triangle E
\nAnswer:
\n(B) isosceles triangle<\/p>\n
\nLet X (-4, 0), Y (4, 0) and Z (0, 3) are the given vertices.
\nNow, distance between X ( -4,0) and Y (4, 0),
\nXY = \\(\\sqrt{[4-(-4)]^{2}+(0-0)^{2}}\\)
\n= \\(\\sqrt{(4+4)^{2}}\\)
\n= \\(\\sqrt{8^{2}}\\)
\n= 8
\nDistance between Y(4, 0) and Z (0, 3),
\nYZ = \\(\\sqrt{(0-4)^{2}+(3-0)^{2}}\\)
\n= \\(\\sqrt{16+9}\\)
\n= \\(\\sqrt{25}\\)
\n= 5
\nDistance between Y (4,0) and Z (0, 3),
\nZX = \\(\\sqrt{[0-(-4)]^{2}+(3-0)^{2}}\\)
\n= \\(\\sqrt{16+9}\\)
\n= \\(\\sqrt{25}\\)
\n= 5
\nAs, YZ = ZX, i.c., two sidps of the triangle ane H equal.
\nSo that, the AXYZ is an isosceles triangle.<\/p>\nThe coordinates of the point which is equidistant from the three vertices of the \u2206AOB as shown in the figure is:<\/h2>\n
![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-7-Coordinate-Geometry-3.png?resize=184%2C173&ssl=1\")
\n(A) (x, y)
\n(B) (y,x)
\n(C) \\(\\left(\\frac{x}{2}, \\frac{y}{2}\\right)\\)
\n(D) \\(\\left(\\frac{y}{2}, \\frac{x}{2}\\right)\\)
\nAnswer:
\n(A) (x, y)<\/p>\n
\nFrom the given figure, coordinates of three vertices of a triangle arc: O (0, 0), A (0,2y) and B (2x, 0). Suppose the required point be P whose coordinates are (h, k) Now, P is equidistant trom the three vertices of \u2206AOB, therefore,
\nPO = PA = PB
\nor (PO)2<\/sup> = (PA)2<\/sup> = (PB)2<\/sup>
\nBy using distance formula, we get:
\n\\(\\left(\\sqrt{(h-0)^{2}+(k-0)^{2}}\\right)^{2}\\) = \\(\\left(\\sqrt{(h-0)^{2}+(k-2 y)^{2}}\\right)^{2}\\)
\n= \\(\\left(\\sqrt{(h-2 x)^{2}+(5-0)^{2}}\\right)^{2}\\)
\n\u21d2 h2<\/sup> + k2<\/sup> = h2<\/sup> +(k – 2y)2 = (h – 2x)2<\/sup> + k2<\/sup> ………(i)
\n\u21d2 k2<\/sup> = k2<\/sup> + 4y2<\/sup> – 4yk
\n\u21d2 0 = 4y2<\/sup> – 4yk
\n\u21d2 4y (y – k) = 0
\n\u21d2 y = k
\nTaking first and second parts of EQuestion (i),
\nh2<\/sup> + k2<\/sup> = (h- 2x)2<\/sup> + k2<\/sup>
\n\u21d2 h2<\/sup> = h2<\/sup> + 4x2<\/sup> – 4xh
\n\u21d2 4x (x – h) = 0
\n\u21d2 h = x
\nRequired coordinates of the point are (h, k) = (x, y)<\/p>\n
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\nAssertion (A): If the distance between the point (4, p) and (1,0) is 5, then the value of p is 4.
\nReason (R): The point which divides the line segment joining the points (7, – 6) and (3,4) in ratio 1: 2 internally lies in the fourth quadrant.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIn case of assertion:
\nDistance between two points (x1<\/sub>,y1<\/sub>) and (x2<\/sup>, y2<\/sup>) is given as,
\nd = \\(\\sqrt{\\left(x_{2}+x_{1}\\right)^{2}+\\left(y_{2}+y_{1}\\right)^{2}}\\)
\nwhere,
\n(x1<\/sub>,y1<\/sub>) = (4, p)
\n(x2<\/sub>,y2<\/sub>) = (1, 0)
\nAnd, d = 5
\nPut the values, we have
\n52<\/sup> = (1 \u2014 4)2<\/sup> + (0 – p)2<\/sup>
\n25 = (-3)2<\/sup> + (-p)2<\/sup>
\n25-9 = p2<\/sup>
\n16 = p2<\/sup>
\n+4, -4 = p
\n\u2234 Assertion is incorrect.
\nIn case of reason:
\nLet (x, y) be the point
\nThen, x = \\(\\frac{m x_{2}+n x_{1}}{m+n}\\) and y = \\(\\frac{m y_{2}+n y_{1}}{m+n}\\) ………..(i)
\nHere, x1<\/sub> = 7, y1<\/sub>= -6, x1<\/sub> = 3, y1<\/sub> = 4, m = 1 and n = 2
\n\u2234 x = \\(\\frac{1(3)+2(7)}{1+2}\\)
\n\u21d2 x = \\(\\frac{3+14}{3}=\\frac{17}{3}\\)
\nAnd, y = \\(\\frac{1(4)+2(-6)}{1+2}\\)
\n\u21d2 y = \\(\\frac{4-12}{3}=-\\frac{8}{3}\\)
\nSo, the required point (x,y) = \\(\\left(\\frac{17}{3},-\\frac{8}{3}\\right)\\) lies in IV th<\/sup> quadrant.
\n\u2234 Reason is correct.
\nHence, assertion is incorrect but reason is correct.<\/p>\n<\/p>\nAssertion (A): AABC with vertices A(-2, 0), B(2, 0) and C(0,2) is similar to ADEF with vertices D(-4,0), E(4,0) and F(0,4).
\nReason (R): A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6,8) lies outside the circle.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn case of assertion:
\nBy using Distance Formula,
\n\u2234Distance between A (-2,0) and B (2,0),
\nAB = \\(\\sqrt{[2-(-2)]^{2}+(0-0)^{2}}\\)=4
\n[\u2234distance between the points (x1, y1) and (x2,y2)
\nD = \\(\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}\\)
\nSimilarly, distance between B (2,0) and C (0, 2),
\nBC = \\(\\sqrt{(0-2)^{2}+(2-0)^{2}}\\) = \\(\\sqrt{4+4}\\) = 2\u221a2
\nIn \u2206ABC, distance between C (0, 2) and A (-2,0),
\nCA = \\(\\sqrt{\\left[0-(-2)^{2}\\right]+(2-0)^{2}}\\) = \\(\\sqrt{4+4}\\) = 2\u221a2
\n\u2234Distance between F (0,4) and D (-4, 0),
\nFD = \\(\\sqrt{(0+4)^{2}+(0-4)^{2}}\\) = \\(\\sqrt{4^{2}+(4)^{2}}\\) = 4\u221a2
\n\u2234Distance between F (o,4) and E (4,0) and D (-4, 0),
\nFE = \\(\\sqrt{(0-4)^{2}+(4-0)^{2}}\\) = \\(\\sqrt{4^{2}+4^{2}}\\) = 4\u221a2
\nand distance between E (4,0) and D (-4, 0),
\nED = \\(\\sqrt{[4-(-4)]^{2}+(0)^{2}}\\) = \\(\\sqrt{8^{2}}\\) = 4\u221a2
\nNow, \\(\\frac{A B}{O E}\\) = \\(\\frac{4}{9}\\) = \\(\\frac{1}{2}\\),
\n\\(\\frac{A C}{D E}\\) = \\(\\frac{2 \\sqrt{2}}{4 \\sqrt{2}}\\) = \\(\\frac{1}{2}\\)
\n\\(\\frac{B C}{E F}\\) = \\(\\frac{2 \\sqrt{2}}{4 \\sqrt{2}}\\)= \\(\\frac{1}{2}\\)
\n\u2234 \\(\\frac{A B}{D E}\\) = \\(\\frac{A C}{D F}\\) = \\(\\frac{B C}{E F}\\)
\nHere, we see that sides of AABC and AFDE are proportional.
\nTherefore, by SSS similarity rule, both the triangles are similar.
\n\u2234 Assertion is correct.
\nIn case of reason:
\nFaint Q (6, 8) will lie outside the circle if its distance from the centre of circle is greater than the radius of the circle.
\nDistance between centre O (0,0) and P (5,0),
\nOP = \\(\\sqrt{(5-0)^{2}+(0-0)^{2}}\\)
\n= \\(\\sqrt{5^{2}+0^{2}}\\)
\n= 5
\nAs, point P lies at the circle, therefore, OP = Radius of circle.
\nDistance between centre O (0,0) and Q (6,8),
\nOQ = \\(\\sqrt{(6-0)^{2}+(8-0)^{2}}\\)
\n= \\(\\sqrt{6^{2}+8^{2}}\\)
\n= \\(\\sqrt{36+64}\\)
\n= \\(\\sqrt{100}\\)
\n= 10
\nAs OQ > OP, therefore, point Q (6, 8) lies outside of the circle.
\n\u2234 Reason is correct.
\nHence, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\nAssertion (A): The ordinate of a point A on y-axis is 5 and B has coordinates (-3,1). Then the length of AB is 5 units.
\nReason (R): The point A(2, 7) lies on the perpendicular bisector of line segment joining the points P(6,5) and Q(0, -4).<\/h2>\n
\n(C) A is true but R is false<\/p>\n
\nIn case of assertion:
\nHere, A \u2192 (0,5) and B \u2192 (-3,1)
\nAB = \\(\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}\\)
\n= \\(\\sqrt{(-3-0)^{2}+(1-5)^{2}}\\)
\n= \\(\\sqrt{9+16}\\)
\n= \\(\\sqrt{25}\\)
\n= 5 units
\n\u2234 Assertion is correct.
\nIn case of reason:
\nIf A (2, 7) lies on perpendicular bisector of P (6,5) and Q (0, -4), then
\nAP = AQ
\n\u2234 By using Distance Formula,
\nAP = \\(\\sqrt{(6-2)^{2}+(5-7)^{2}}\\)
\n= \\(\\sqrt{(4)^{2}+(-2)^{2}}\\)
\n= \\(\\sqrt{20}\\)
\nAnd AQ = \\(\\sqrt{(0-2)^{2}+(-4-7)^{2}}\\)
\n= \\(\\sqrt{(-2)^{2}+(-11)^{2}}\\)
\n= \\(\\sqrt{125}\\)
\nAs, AP AQ
\nThercfore, A does not lie on the perpendicular bisector of PQuestion
\n\u2234 Reason is incorrect.
\nHence, assertion is correct but reason is incorrect.<\/p>\n
\nI. Read the following text and answer the below questions:
\nThe diagram show the plans for a sun room. It will be built onto the wall of a house. The four walls of the sun room are square clear glass panels. The roof is made using,<\/p>\n\n
\n
![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-7-Coordinate-Geometry-3a.png?resize=386%2C274&ssl=1\")
<\/p>\nRefer to Top View, find the mid-point of the seg\u00acment joining the points J(6,17) and 1(9,16).<\/h2>\n
\n(B) \\(\\frac{3}{2}, \\frac{1}{2}\\)
\n(C) \\(\\frac{15}{2}, \\frac{33}{2}\\)
\n(D) \\(\\frac{1}{2}, \\frac{3}{2}\\)
\nAnswer:
\n(C) \\(\\frac{15}{2}, \\frac{33}{2}\\)<\/p>\n
\nMid-point of J(6,17) and 1(9,16) is
\nx = \\(\\frac{6+9}{2}\\) and y = \\(\\frac{17+16}{2}\\)
\nx = \\(\\frac{15}{2}\\) and y = \\(\\frac{33}{2}\\)<\/p>\nRefer to front View, the distance of the point P from the y-axis is:<\/h2>\n
\n(B) 15
\n(C) 19
\n(D) 25
\nAnswer:
\n(A) 4<\/p>\n
\nThe distance of the point P from the Y-axis = 4.<\/p>\n<\/p>\nRefer to front view, the distance between the points A and S is<\/h2>\n
\n(B) 8
\n(C) 14
\n(D) 20
\nAnswer:
\n(C) 14<\/p>\n
\nA’s coordinates = (1,8)
\nS’s coordinates = (15,8)
\nThen, AS = \\(\\left|\\sqrt{(15-1)^{2}+(8-8)^{2}}\\right|\\)
\n= \\(\\sqrt{(14)^{2}}\\)
\n= 14.<\/p>\nRefer to front view, find the co-ordinates of the point which divides the line segment joining the points A and B in the ratio 1: 3 internally.<\/h2>\n
\n(B) (2.0,9.5)
\n(C) (3.0,7.5)
\n(D) (2.0,8.5)
\nAnswer:
\n(D) (2.0,8.5)<\/p>\n
\nThe coordinates of A = (1,8) The coordinates of B = (4,10) Also, m = 1 and n = 3
\nThen, (x,y) = \\(\\left(\\frac{1 \\times 4+3 \\times 1}{1+3}, \\frac{1 \\times 10+3 \\times 8}{1+3}\\right)\\)
\n= \\(\\left(\\frac{7}{4}, \\frac{34}{4}\\right)\\)
\n= (1. 75,8.5)<\/p>\nRefer to front view, if a point (x, y) is equidistant from the Q(9,8) and S(17,8), then<\/h2>\n
\n(B) x – 13 = 0
\n(C) y – 13 = 0
\n(D) Y – y = 13
\nAnswer:
\nOption (B) is correct<\/p>\n
\nLet point be P(x, y)
\nPQ2<\/sup> = PS2<\/sup>
\nor, (x – 9)2<\/sup> + (y – 8)2<\/sup> = (x – 17)2<\/sup> + (y – 8)2<\/sup> or, x – 13 =0<\/p>\n
\nAyush Starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office.
\n(Assume that all distances covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in km.<\/p>\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-7-Coordinate-Geometry-4.png?resize=397%2C188&ssl=1\")
<\/p>\nWhat is the distance between house and bank?<\/h2>\n
\n(B) 10 km
\n(C) 12 km
\n(D) 27 km
\nAnswer:
\n(A) 5 km<\/p>\n
\nSince,
\nDistance between two points (x1<\/sub>, y1<\/sub>) and (x2<\/sub>, y2<\/sub>).
\nd = \\(\\left|\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}\\right|\\)
\nNow, distance between house and bank,
\n= \\(\\left|\\sqrt{(5-2)^{2}+(8-4)^{2}}\\right|\\)
\n= \\(\\left|\\sqrt{(3)^{2}+(4)^{2}}\\right|\\)
\n= \\(|\\sqrt{9+16}|\\)
\n= \\(\\sqrt{25}\\)
\n= 5 km<\/p>\nWhat is the distance between Daughter’s School and bank ?<\/h2>\n
\n(B) 10 km
\n(C) 12 km
\n(D) 27 km
\nAnswer:
\n(B) 10 km<\/p>\n
\nDistance between bank and daughter’s school,
\n= \\(\\left|\\sqrt{(13-5)^{2}+(14-8)^{2}}\\right|\\)
\n= \\(\\left|\\sqrt{(8)^{2}+(6)^{2}}\\right|\\)
\n= \\(|\\sqrt{64+36}|\\)
\n= \\(|\\sqrt{100}|\\)
\n= 10 km<\/p>\n<\/p>\nWhat is the distance between house and office?<\/h2>\n
\n(B) 26.4 km
\n(C) 24 km
\n(D) 26 km
\nAnswer:
\n(B) 26.4 km<\/p>\n
\nDistance between house to office,
\n= \\(\\vec{a}\\)
\n= \\(\\vec{a}\\)
\n= \\(\\vec{a}\\)
\n= \\(\\vec{a}\\)
\n= 24. 59
\n= 24.6 km<\/p>\nWhat is the total distance travelled by Ayush to reach the office?<\/h2>\n
\n(B) 10 km
\n(C) 12 km
\n(D) 27 km
\nAnswer:
\n(D) 27 km<\/p>\n
\nDistance between daughter’s school and office.
\n= \\(\\left|\\sqrt{(13-13)^{2}+(26-14)^{2}}\\right|\\)
\n= \\(\\left|\\sqrt{0+(12)^{2}}\\right|\\)
\n= 12 km
\nTotal distance (House + Bank + School + Office) travelled = 5 + 10 + 12 = 27 km<\/p>\nWhat is the extra distance travelled by Ayush?<\/h2>\n
\n(B) 2.2 km
\n(C) 2.4 km
\n(D) none of these
\nAnswer:
\n(C) 2.4 km<\/p>\n
\nExtra distance travelled by Ayush in reaching his office = 27 – 24.6 = 2.4 km.<\/p>\n
\nIn order to conduct Sports Day activities in your School, lines have been drawn with chalk powder at a distance of 1 m each, in a rectangular shaped ground ABCD, 100 flowerpots have been placed at a distance of 1 m from each other along AD, as shown in given figure below. Niharika runs 1\/4th<\/sup> the distance AD on the 2nd line and posts a green (G) flag runs 1\/5th<\/sup> distance AD on the eighth line and posts a red (R) flag.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-7-Coordinate-Geometry-5.png?resize=363%2C375&ssl=1\")
<\/p>\nFind the position of green flag<\/h2>\n
\n(B) (2,0.25)
\n(C) (25,2)
\n(D) (0,-25)
\nAnswer:
\nOption (A) is correct.<\/p>\nFind the position of red flag<\/h2>\n
\n(B) (20,8)
\n(C) (8,20)
\n(D) (8,0.2)
\nAnswer:
\n(C) (8,20)<\/p>\nWhat is the distance between both the flags?<\/h2>\n
\n(B) \u221a11
\n(C) \u221a61
\n(D) \u221a51
\nAnswer:
\n(C) \u221a61<\/p>\n
\nPosition of Green flag = (2,25) Position of Red flag = (8,20)
\nDistance between both the flags
\n\\(\\sqrt{(8-2)^{2}+(20-25)^{2}}\\) = \\(\\sqrt{6^{2}+(-5)^{2}}\\)
\n= \\(\\sqrt{36+25}\\)
\n= \u221a61<\/p>\n<\/p>\nIf Rashini has to post a blue llag exactly halfway between the line segment joining the two flags, where should she post her flag?<\/h2>\n
\n(B) (10,22)
\n(C) (2,8.5)
\n(D) (2.5,20)
\nAnswer:
\n(A) (5,22.5)<\/p>\n
\nPosition of blue flag Mid-point of line segment joining the green and red flags
\n= \\(\\left(\\frac{2+8}{2}, \\frac{25+20}{2}\\right)\\)
\n= (5, 22.5)<\/p>\nIf Joy has to post a flag at one-fourth distance from green flag, in the line segment joining the green and red flags, then where should he post his flag ?<\/h2>\n
\n(B) (0.5,12.5)
\n(C) (2.25,8.5)
\n(D) (25,20)
\nAnswer:
\n(A) (3.5,24)<\/p>\n
\nPosition of Joy\u2019s flag = Mid-point of line segment joining green and blue flags
\n= \\(\\left[\\frac{2+5}{2}, \\frac{25+22.5}{2}\\right]\\)
\n= [3.5,23.75] ~ [3.5,24]<\/p>\n
\nThe class X students school in krishnagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-7-Coordinate-Geometry-6.png?resize=325%2C181&ssl=1\")
<\/p>\nTaking A as origin, find the coordinates of P.<\/h2>\n
\n(B) (6,4)
\n(C) (0,6)
\n(D) (4,0)
\nAnswer:
\n(A) (4,6)<\/p>\nWhat will be the coordinates of R, if C is the origin?<\/h2>\n
\n(B) (3,10)
\n(C) (10,3)
\n(D) (0,6)
\nAnswer:
\n(C) (10,3)<\/p>\nWhat will be the coordinates of Q, if C is the origin?<\/h2>\n
\n(B) (-6,13)
\n(C) (-13,6)
\n(D) (13,6)
\nAnswer:
\n(D) (13,6)<\/p>\n<\/p>\nCalculate the area of the triangles if A is the origin.<\/h2>\n
\n(B) 6
\n(C) 8
\n(D) 6.25
\nAnswer:
\n(A) 4.5<\/p>\n
\nCoordinates of P = (4,6)
\nCoordinates of Q = (3,2)
\nCoordinates of R = (6,5)
\nArea of triangle PQR = [ x1<\/sub> (y2<\/sub> – y2<\/sub> + x2<\/sub> (y3<\/sub> – y1<\/sub>) + x3<\/sub>(y1<\/sub> – y2<\/sub>)]
\n= \\(\\frac{1}{2}\\) [4(-3) + 3(-1) + 6(4)]
\n= \\(\\frac{1}{2}\\) [- 12 + (-3) + 24]
\n= \\(\\frac{1}{2}\\)[- 12 + 21]
\n= \\(\\frac{1}{2}\\)[9]
\n= 4.5 sQuestion units.<\/p>\nCalculate the area of the triangle if C is the origin.<\/h2>\n
\n(B) 5
\n(C) 6.25
\n(D) 4.5
\nAnswer:
\n(D) 4.5<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"