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Mathematics · 2024 · 4 marks
CBSE 2024 · Region 5 · Set 1 · Q37
A garden is in the shape of a square. The gardener grew saplings of Ashoka tree on the boundary of the garden at the distance of 1 m from each other.…
A garden is in the shape of a square. The gardener grew saplings of Ashoka tree on the boundary of the garden at the distance of $\displaystyle 1$ m from each other. He wants to decorate the garden with rose plants. He chose a triangular region inside the garden to grow rose plants. In the above situation, the gardener took help from the students of class 10. They made a chart for it which looks like the given figure.
Based on the above, answer the following questions :
(i)
If A is taken as origin, what are the coordinates of the vertices of $\displaystyle \triangle \mathrm{PQR}$ ?
(ii)
(a)
Find distances PQ and QR .
OR
(b)
Find the coordinates of the point which divides the line segment joining points P and R in the ratio $\displaystyle 2$ : $\displaystyle 1$ internally.
(iii)
Find out if $\displaystyle \triangle \mathrm{PQR}$ is an isosceles triangle.
Marking-scheme solution
(i)
P $\displaystyle (4, 6)$, Q $\displaystyle (3, 2)$, R $\displaystyle (6, 5)$
(ii) (a) \(\displaystyle \mathrm{PQ}=\sqrt{(4-3)^{2}+(6-2)^{2}}=\sqrt{17}\)
\[\mathrm{QR}=\sqrt{(3-6)^{2}+(2-5)^{2}}=\sqrt{18}
\]
The coordinate of required point are \(\displaystyle \left(\frac{6 \times 2+1 \times 4}{3}, \frac{5 \times 2+1 \times 6}{3}\right)\)
i.e. \(\displaystyle \left(\frac{16}{3}, \frac{16}{3}\right)\)
(iii) \[\begin{aligned}
& \mathrm{PQ}=\sqrt{(4-3)^{2}+(6-2)^{2}}=\sqrt{17} \\
& \mathrm{QR}=\sqrt{(3-6)^{2}+(2-5)^{2}}=\sqrt{18} \\
& \mathrm{PR}=\sqrt{(4-6)^{2}+(6-5)^{2}}=\sqrt{5} \\
& \mathrm{PQ} \neq \mathrm{QR} \neq \mathrm{PR}
\end{aligned}
\]
\(\displaystyle \Delta \mathrm{PQR}\) is not isosceles
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CBSE Class 10 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.