Ch 1 End of Chapter Exercise Class 9 Ganita Manjari
Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the Ch 1 End of Chapter Exercise Class 9 Ganita Manjari. Designed to reflect how top-scoring students write their answers in school examinations, this review walkthrough covers coordinate geometry, distance formulas, midpoint properties, and Cartesian plotting entirely.
🔑 Chapter Overview & Key Formulas:
- Cartesian Axes & Origin: The x-axis and y-axis intersect at the origin \((0, 0)\).
- Distance Formula: The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
- Midpoint Formula: The midpoint of a line segment with endpoints \((x_1, y_1)\) and \((x_2, y_2)\) is \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).
- Cyclic & Geometric Properties: Using right-angled triangles, Pythagorean relations, and collinearity checks.
NCERT Class 9 Maths Ganita Manjari Chapter 1 End of Chapter Exercise Solutions
Question 1
What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
Solution:
The x-axis and y-axis intersect at the origin.
Therefore:
• x-coordinate \(= 0\)
• y-coordinate \(= 0\)
So, the point of intersection is \(\mathbf{(0, 0)}\).
Question 2
Point W has x-coordinate equal to –5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?
Solution:
If point W has x-coordinate \(-5\), then any point on the line through W parallel to the y-axis will also have x-coordinate \(-5\).
Therefore, the coordinates of H will be of the form: \(\mathbf{H = (-5, y)}\), where \(y\) can be any real number.
• If \(y > 0\), H lies in Quadrant II.
• If \(y < 0\), H lies in Quadrant III.
• If \(y = 0\), H lies on the x-axis.
So, H can lie in Quadrant II, Quadrant III, or on the x-axis.
Question 3
Consider the points R (3, 0), A (0, –2), M (–5, –2) and P (–5, 2). If they are joined in the same order, predict:
Solution:
(i) Two sides of RAMP that are perpendicular to each other.
AM joins \(A(0, -2)\) to \(M(-5, -2)\), so it is horizontal.
MP joins \(M(-5, -2)\) to \(P(-5, 2)\), so it is vertical.
Since horizontal and vertical lines are perpendicular, the two perpendicular sides are \(\mathbf{AM}\) and \(\mathbf{MP}\) (\(\mathbf{AM \perp MP}\)).
(ii) One side of RAMP that is parallel to one of the axes.
• \(\mathbf{AM}\) is parallel to the x-axis (both points have y-coordinate \(-2\)).
• \(\mathbf{MP}\) is parallel to the y-axis (both points have x-coordinate \(-5\)).
(iii) Two points that are mirror images of each other in one axis. Which axis will this be?
Comparing \(M(-5, -2)\) and \(P(-5, 2)\), they have the same x-coordinate and y-coordinates equal in magnitude but opposite in sign.
Therefore, the points \(\mathbf{M}\) and \(\mathbf{P}\) are mirror images of each other in the \(\mathbf{x\text{-axis}}\).
Question 4
Plot point Z (5, –6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides.
Solution:
Take vertices: \(I = (5, 0)\), \(Z = (5, -6)\), and \(N = (0, -6)\).
\(\triangle IZN\) is right-angled at \((5, -6)\) because \(IZ\) is vertical and \(ZN\) is horizontal.
Lengths of the three sides:
• \(\mathbf{IZ} = 6\) units
• \(\mathbf{ZN} = 5\) units
• \(\mathbf{IN} = \sqrt{(5 - 0)^2 + (0 - (-6))^2} = \sqrt{25 + 36} = \mathbf{\sqrt{61}}\) units.
Question 5
What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?
Solution:
If we lacked negative numbers, coordinates could only be zero or positive, restricting usage strictly to Quadrant I.
Quadrants II, III, and IV could not be represented.
Answer: No, such a system would not allow us to locate all points across a complete 2-D plane.
Question 6
Are the points M (–3, –4), A (0, 0) and G (6, 8) on the same straight line? Suggest a method to check this without plotting and joining the points.
Solution:
Using the distance formula to find segment lengths:
• \(MA = \sqrt{(0 + 3)^2 + (0 + 4)^2} = \sqrt{9 + 16} = 5\)
• \(AG = \sqrt{(6 - 0)^2 + (8 - 0)^2} = \sqrt{36 + 64} = 10\)
• \(MG = \sqrt{(6 + 3)^2 + (8 + 4)^2} = \sqrt{81 + 144} = 15\)
Since \(MA + AG = 5 + 10 = 15 = MG\), the sum of two smaller segments equals the third segment.
Answer: Yes, points M, A, and G lie on the same straight line (collinear).
Question 7
Use your method from Problem 6 to check if the points R (–5, –1), B (–2, –5) and C (4, –12) are on the same straight line.
Solution:
Using distance formula:
• \(RB = \sqrt{(-2 + 5)^2 + (-5 + 1)^2} = \sqrt{9 + 16} = 5\)
• \(BC = \sqrt{(4 + 2)^2 + (-12 + 5)^2} = \sqrt{36 + 49} = \sqrt{85}\)
• \(RC = \sqrt{(4 + 5)^2 + (-12 + 1)^2} = \sqrt{81 + 121} = \sqrt{202}\)
Since \(RB + BC \neq RC\) (\(5 + \sqrt{85} \neq \sqrt{202}\)),
Answer: No, the points R, B, and C do not lie on the same straight line.
Question 8
Find coordinates/examples for:
Solution:
(i) A right-angled isosceles triangle
Vertices: \(O = (0, 0)\), \(A = (4, 0)\), and \(B = (0, 4)\).
Since \(OA = OB = 4\) units and \(OA \perp OB\), \(\triangle OAB\) is a right-angled isosceles triangle.
(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV
Vertices: \(O = (0, 0)\), \(P = (-3, -4)\) [in Quad III], and \(Q = (3, -4)\) [in Quad IV].
Since \(OP = OQ = 5\) units, \(\triangle OPQ\) is an isosceles triangle meeting the criteria.
Question 9
State whether M is the midpoint of segment ST in each case.
Solution:
• (i) S(−3, 0), M(0, 0), T(3, 0): \(SM = 3\), \(MT = 3\). Since \(SM = MT\), M is the midpoint.
• (ii) S(2, 3), M(3, 4), T(4, 5): \(SM = \sqrt{2}\), \(MT = \sqrt{2}\). Since \(SM = MT\), M is the midpoint.
• (iii) S(0, 0), M(0, 5), T(0, −10): \(SM = 5\), \(MT = 15\). Since \(SM \neq MT\), M is NOT the midpoint.
• (iv) S(−8, 7), M(0, −2), T(6, −3): \(SM = \sqrt{145}\), \(MT = \sqrt{37}\). Since \(SM \neq MT\), M is NOT the midpoint.
Question 10
Find the coordinates of B given that M (–7, 1) is the midpoint of A (3, –4) and B (x, y).
Solution & Calculation:
Using midpoint formula: \(\left(\frac{3 + x}{2}, \frac{-4 + y}{2}\right) = (-7, 1)\)
• \(\frac{3 + x}{2} = -7 \implies 3 + x = -14 \implies x = -17\)
• \(\frac{-4 + y}{2} = 1 \implies -4 + y = 2 \implies y = 6\)
Answer: The coordinates of B are \(\mathbf{(-17, 6)}\).
Question 11
Let P and Q be points of trisection of AB, with P closer to A and Q closer to B. Find the coordinates of P and Q for A (4, 7) and B (16, –2).
Solution & Calculation:
Since P and Q trisect AB, P divides AB in ratio \(1:2\) and Q divides AB in ratio \(2:1\).
• Coordinates of P \(= \mathbf{(8, 4)}\)
• Coordinates of Q \(= \mathbf{(12, 1)}\)
Question 12
Circle Properties & Point Locations
Solution:
(i) Show that A (1, –8), B (–4, 7) and C (–7, –4) lie on a circle centered at the origin.
Find distances from origin \((0, 0)\):
• \(OA = \sqrt{1^2 + (-8)^2} = \sqrt{65}\)
• \(OB = \sqrt{(-4)^2 + 7^2} = \sqrt{65}\)
• \(OC = \sqrt{(-7)^2 + (-4)^2} = \sqrt{65}\)
Since all distances are equal (\(\sqrt{65}\)), the points lie on a circle with radius \(\mathbf{\sqrt{65}}\).
(ii) Check whether D (–5, 6) and E (0, 9) lie inside, on, or outside the circle.
• \(OD = \sqrt{(-5)^2 + 6^2} = \sqrt{61}\). Since \(\sqrt{61} < \sqrt{65}\), point D lies inside the circle.
• \(OE = \sqrt{0^2 + 9^2} = 9\). Since \(9 > \sqrt{65}\), point E lies outside the circle.
Question 13
The midpoints of the sides of triangle ABC are D(5, 1), E(6, 5), and F(0, 3). Find the coordinates of A, B and C.
Solution & Calculation:
Using midpoint relations for triangle vertices:
• \(\mathbf{A = (1, 7)}\)
• \(\mathbf{B = (-1, -1)}\)
• \(\mathbf{C = (11, 3)}\)
Question 14
City Model Grid & Intersections
Solution:
• (i) Model: Sketch 10 vertical and 10 horizontal parallel lines spaced 1 cm apart (\(1\text{ cm} = 200\text{ m}\) scale).
• (ii)(a) Intersections for (4, 3): Exactly 1 intersection point.
• (ii)(b) Intersections for (3, 4): Exactly 1 unique intersection point.
Question 15
A computer graphics screen is 800 pixels wide and 600 pixels high. Two circular icons are drawn. Determine whether any part lies outside the screen and whether the circles intersect.
Solution:
Circle A: Center \((100, 150)\), Radius \(80\)
Circle B: Center \((250, 230)\), Radius \(100\)
• (i) Screen Bounds: Both circles lie completely inside the \(800 \times 600\) pixel screen.
• (ii) Intersection: Yes, the two circular icons intersect each other.
Question 16
Plot the points A (2, 1), B (–1, 2), C (–2, –1), and D (1, –2). Is ABCD a square? Find its area.
Solution & Calculation:
• Side lengths: \(AB = BC = CD = DA = \sqrt{10}\) units.
• Diagonals: \(AC = BD = \sqrt{20}\) units (Diagonals are equal and sides are equal).
Therefore, ABCD is a square.
• \(\text{Area} = \text{side}^2 = (\sqrt{10})^2 = \mathbf{10}\) square units.
Frequently Asked Questions (FAQs)
Q1: Are these the latest solutions for the 2026-27 Ganita Manjari textbook?
A: Yes. These solutions are strictly based on the latest NCERT Ganita Manjari Class 9 Maths textbook released for the current 2026-27 academic session.
Q2: Where can I find the End of Chapter Exercise for Class 9 Maths Chapter 1?
A: The End of Chapter Exercise is located at the very end of Chapter 1 in the new Ganita Manjari book, covering all core review concepts.
Q3: Is it important to solve the End of Chapter Exercise for exams?
A: Absolutely. These EOC questions often test higher-level analytical thinking frequently featured in school and board assessments.
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