Carom board is a very popular game. The board is a square of side length 65 cm. It has circular pockets in each corner. Ansh strikes a disc, kept at position P with a striker. The disc, hits the boundary of the board at R and goes straight to pocket at corner C. It is given that \(PS = 9\) cm, \(PQ = 35\) cm, \(BR = x\), \(\angle PRQ = \alpha\) and \(\angle CRB = \theta\). Based on the above information, answer the following questions:
Question: 1
Using law of reflection i.e. \(\angle PRT = \angle CRT\), prove that \(\theta = \alpha\).
Show Hint
Angles formed with the surface are equal if the angles formed with the normal are equal. This is a common property used in physics and geometry.
Step 1: Understanding the Concept:
The law of reflection states that the angle of incidence equals the angle of reflection, usually measured from a normal (perpendicular) line. Step 3: Detailed Explanation:
In the diagram, let line \(RT\) be the normal to the boundary \(AB\) at point \(R\).
By the law of reflection: \(\angle PRT = \angle CRT\).
The line \(RT\) is perpendicular to the side of the carom board, so \(\angle QRT = \angle BRT = 90^{\circ}\).
Now, \(\angle PRQ = \alpha = 90^{\circ} - \angle PRT\).
And \(\angle CRB = \theta = 90^{\circ} - \angle CRT\).
Since \(\angle PRT = \angle CRT\), their complements must also be equal:
\[ 90^{\circ} - \angle PRT = 90^{\circ} - \angle CRT \]
\[ \alpha = \theta \] Step 4: Final Answer:
Hence, \(\theta = \alpha\).
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Question: 2
Prove that \(\triangle PQR \sim \triangle CBR\) given that \(PQ\) is perpendicular to \(AB\).
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Similarity is the bridge between angle properties and side length ratios. Once proven, you can equate ratios of corresponding sides.
Step 1: Understanding the Concept:
Two triangles are similar if two of their corresponding angles are equal (AA Similarity Criterion). Step 3: Detailed Explanation:
In \(\triangle PQR\) and \(\triangle CBR\):
1. \(\angle PQR = \angle CBR = 90^{\circ}\) (Given \(PQ \perp AB\) and the corner of the square carom board is \(90^{\circ}\)).
2. \(\angle PRQ = \angle CRB\) (Proved in part (i) as \(\alpha = \theta\)).
Therefore, by AA Similarity Criterion:
\[ \triangle PQR \sim \triangle CBR \] Step 4: Final Answer:
Hence proved.
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Question: 3
Find the value of \(x\) using similarity of triangles.
Show Hint
Label the total side lengths and partial segments carefully to ensure the expressions for \(QR\) and \(BR\) are correct relative to the corners.
Step 1: Understanding the Concept:
When triangles are similar, the ratios of their corresponding sides are equal. Step 2: Key Formula or Approach:
From \(\triangle PQR \sim \triangle CBR\):
\[ \frac{PQ}{CB} = \frac{QR}{BR} \] Step 3: Detailed Explanation:
Side of square board = 65 cm. Thus, \(CB = 65\).
Given \(PQ = 35\).
From diagram, \(S\) is on \(AD\) and \(PQ\) is perpendicular to \(AB\). \(PS = 9\) cm represents the distance of \(Q\) from corner \(A\).
So, \(AQ = 9\).
Since \(AB = 65\), the length \(QB = 65 - 9 = 56\).
We are given \(BR = x\). Since \(R\) is on the segment \(QB\), \(QR = QB - BR = 56 - x\).
Now substitute into the similarity ratio:
\[ \frac{35}{65} = \frac{56 - x}{x} \]
\[ \frac{7}{13} = \frac{56 - x}{x} \]
Cross-multiply:
\[ 7x = 13(56 - x) \]
\[ 7x = 728 - 13x \]
\[ 20x = 728 \]
\[ x = \frac{728}{20} = 36.4 \text{ cm} \] Step 4: Final Answer:
The value of \(x\) is 36.4 cm.
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Question: 4
If \(\frac{\text{Area } \triangle PQR}{\text{Area } \triangle CBR} = \frac{PQ^2}{CB^2}\), then find the value of \(x\).
Show Hint
The relationship between area ratios and side ratios confirms that the triangles are similar. You can solve for the unknown side using the simpler linear side ratio.
Step 1: Understanding the Concept:
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Step 3: Detailed Explanation:
We have already proven \(\triangle PQR \sim \triangle CBR\).
The property given in the question (\(\frac{\text{Area } PQR}{\text{Area } CBR} = \frac{PQ^2}{CB^2}\)) is always true for similar triangles.
To find \(x\), we still use the ratio of sides derived from similarity:
\[ \frac{PQ}{CB} = \frac{QR}{BR} \]
As solved in the previous part:
\[ \frac{35}{65} = \frac{56 - x}{x} \implies x = 36.4 \text{ cm} \] Step 4: Final Answer:
The value of \(x\) is 36.4 cm.
