MathsCBSE Class 10 · Basic · Board exam 2024, set 430/2/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2024: question 38

Set in the CBSE Class 10 board exam of 2024, paper 430/2/1, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedCoordinate Geometry

The top of a table is hexagonal in shape. The hexagonal top is shown as a 3D sketch, and its plan is drawn on a coordinate grid (x-axis 0 to 13, y-axis 0 to 13). The points are: A(1, 9), B(5, 13), C(9, 13), D(13, 9), E(13, 5), F(9, 1), G(5, 1), H(1, 5) forming the outer octagon; inner points M(5, 11), N(9, 11), O(11, 9), P(11, 5), Q(9, 3), R(5, 3), S(3, 5), T(3, 9) forming an inner octagon; the dashed lines from M, N, O, P, Q, R, S, T all pass through the centre (7, 7).

On the basis of the information given above, answer the following questions:

(i)
Write the coordinates of A and B.
1 mark
(ii)
Write the coordinates of the mid-point of line segment joining C and D.
1 mark
(iii) (a)
Find the distance between M and Q.
2 marks
OR
(iii) (b)
Find the coordinates of the point which divides the line segment joining M and N in the ratio 1:3 internally.
2 marks

Figure in the paper: Coordinate grid with x-axis (0 to 13) and y-axis (0 to 13). An outer hexagon-like octagon with vertices A(1,9), B(5,13), C(9,13), D(13,9), E(13,5), F(9,1), G(5,1), H(1,5), and an inner octagon with vertices M(5,11), N(9,11), O(11,9), P(11,5), Q(9,3), R(5,3), S(3,5), T(3,9). Dashed lines join M, N, O, P, Q, R, S, T to the centre at (7,7). A small 3D sketch of a hexagonal table top is shown above the grid. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 430/2/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Read each point by going across to its x-value first, then up to its y-value.

Hint 2

Mid-point is the average of the x-coordinates and the average of the y-coordinates; distance uses $(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .

Hint 3

For dividing in ratio $1 : 3$ , use $(\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})$ with M as the first point.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Reading from the grid: $A (1, 9)$ and $B (5, 13)$ .
1 mark
(ii) $C (9, 13)$ and $D (13, 9)$ .
Mid-point $= (\frac{9 + 13}{2}, \frac{13 + 9}{2}) = (11, 11)$
1 mark
(iii)(a) From the grid: $M (5, 11)$ and $Q (9, 3)$ .
1 mark
$M Q = (9 - 5)^{2} + (3 - 11)^{2} = 16 + 64 = 80 = 45$ units
1 mark
(iii)(b) From the grid: $M (5, 11)$ and $N (9, 11)$ ; ratio $m_{1} : m_{2} = 1 : 3$ .
1 mark
By the section formula, the point is $(\frac{1 \times 9 + 3 \times 5}{1 + 3}, \frac{1 \times 11 + 3 \times 11}{1 + 3}) = (\frac{24}{4}, \frac{44}{4}) = (6, 11)$
1 mark

Answer: (i) A(1, 9), B(5, 13); (ii) (11, 11); (iii)(a) MQ = √80 = 4√5 units; (iii)(b) (6, 11)

Where the marks are

(i) A(1, 9) and B(5, 13) (½ + ½ marks)
(ii) Reads C(9, 13), D(13, 9) (½ mark) and gets mid-point (11, 11) (½ mark)
(iii)(a) Coordinates M(5, 11), Q(9, 3) (1 mark)
(iii)(a) $M Q = 80 = 45$ (1 mark)
(iii)(b) Coordinates M(5, 11), N(9, 11) (1 mark)
(iii)(b) Section formula gives (6, 11) (1 mark)

Revise the chapter

Chapter: Coordinate Geometry (notes, formulas, where marks are lost and every past question)
Topic: Distance Formula

Practise this chapterBoard-style questions from Coordinate Geometry, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 430/2/1 (Basic): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Basic 2024: question 38

Hints

Step-by-step solution

Revise the chapter

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