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Q5:
In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Solution :
Given: The positions of four friends seated in a classroom are represented by coordinates on a Cartesian plane. Based on the provided figure, the coordinates are:
$A = (3, 4)$
$B = (6, 7)$
$C = (9, 4)$
$D = (6, 1)$
To Find: Determine whether the quadrilateral $ABCD$ is a square by calculating the lengths of its sides and diagonals using the distance formula.
Visual Representation:
Step 1: The Distance Formula
The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Step 2: Calculating Side Lengths
We calculate the lengths of the four sides $AB, BC, CD,$ and $DA$.
Length of $AB = \sqrt{(6 - 3)^2 + (7 - 4)^2} = \sqrt{(3)^2 + (3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.
Length of $BC = \sqrt{(9 - 6)^2 + (4 - 7)^2} = \sqrt{(3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.
Length of $CD = \sqrt{(6 - 9)^2 + (1 - 4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.
Length of $DA = \sqrt{(3 - 6)^2 + (4 - 1)^2} = \sqrt{(-3)^2 + (3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.
[Since all sides are equal ($AB = BC = CD = DA = 3\sqrt{2}$), the quadrilateral is at least a rhombus.]
Step 3: Calculating Diagonal Lengths
For a rhombus to be a square, its diagonals must also be equal in length.
Length of diagonal $AC = \sqrt{(9 - 3)^2 + (4 - 4)^2} = \sqrt{(6)^2 + (0)^2} = \sqrt{36} = 6$ units.
Length of diagonal $BD = \sqrt{(6 - 6)^2 + (1 - 7)^2} = \sqrt{(0)^2 + (-6)^2} = \sqrt{36} = 6$ units.
Step 4: Conclusion
Since all four sides are equal ($3\sqrt{2}$ units) and both diagonals are equal ($6$ units), the quadrilateral $ABCD$ satisfies the geometric properties of a square.
Final Answer: Champa is correct; ABCD is a square.
More Questions from Class 10 Mathematics Coordinate geometry EXERCISE 7.1
- Q1(i): Find the distance between the following pairs of points : (i) (2, 3), (4, 1)
- Q1(ii): Find the distance between the following pairs of points : (ii) (– 5, 7), (– 1, 3)
- Q1(iii): Find the distance between the following pairs of points : (iii) ($a$, $b$), (– $a$, – $b$)
- Q10: Find a relation between $x$ and $y$ such that the point ($x$, $y$) is equidistant from the point (3, 6) and (– 3, 4).
- Q2: Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B.
- Q3: Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.
- Q4: Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
- Q6(i): Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)
- Q6(ii): Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)
- Q6(iii): Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (iii) (4, 5), (7, 6), (4, 3), (1, 2)
- Q7: Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
- Q8: Find the values of $y$ for which the distance between the points P(2, – 3) and Q(10, $y$) is 10 units.
- Q9: If Q(0, 1) is equidistant from P(5, –3) and R($x$, 6), find the values of $x$. Also find the distances QR and PR.
CBSE Solutions for Class 10 Mathematics Coordinate geometry
Chapters in CBSE - Class 10 Mathematics
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