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.
We use the *law of reflection*:
\[
\angle PRT = \angle CRT
\]
This means: Angle of incidence = Angle of reflection
In the usual diagram (mirror RT with point R on it):
• PR is the incident ray
• CR is the reflected ray
• RT is the mirror line
• θ is the angle between PR and the normal
• α is the angle between CR and the normal
Step 1: Resolve the angles
The angle PRT is the angle between the incident ray PR and the mirror line RT.
The angle CRT is the angle between the reflected ray CR and the mirror line RT.
By definition of the law of reflection:
\[
\angle PRT = \angle CRT
\]
These are *angles with the mirror surface*, not with the normal.
Step 2: Convert mirror angles to normal angles
The normal is perpendicular to the mirror line RT.
Therefore:
\[
\theta = 90^\circ - \angle PRT
\]
\[
\alpha = 90^\circ - \angle CRT
\]
Since:
\[
\angle PRT = \angle CRT
\]
subtracting each from \(90^\circ\) gives:
\[
90^\circ - \angle PRT = 90^\circ - \angle CRT
\]
Hence:
\[
\theta = \alpha
\]
Final Conclusion:
\[
\boxed{\theta = \alpha}
\]
This directly follows from the law of reflection.
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Question: 2
Prove that \(\triangle PQR \sim \triangle CBR\) given that \(PQ\) is perpendicular to \(AB\).
Show Hint
Similarity is the bridge between angle properties and side length ratios. Once proven, you can equate ratios of corresponding sides.
Since the figure is not shown, we use the standard geometry setting used in CBSE questions where:
• P lies on AB
• Q lies on AC
• R is the vertex opposite A
• PQ ⟂ AB
• CR ⟂ AB
This gives us two right-angled triangles: △PQR and △CBR.
We must prove:
\[
\triangle PQR \sim \triangle CBR
\]
Step 1: Identify the right angles
Given PQ ⟂ AB.
Since C lies on AB (usual configuration):
CR ⟂ AB.
Step 1: Using Similarity of Triangles:
Since ΔPQR ∼ ΔCBR,
Corresponding sides are proportional.
So,
PQ / CB = QR / BR
Step 2: Substituting Known Values:
Side of square board = 65 cm
Therefore, CB = 65 cm
Given:
PQ = 35 cm
PS = 9 cm
Since AB = 65 cm,
AQ = 9 cm
QB = 65 − 9
QB = 56 cm
Let BR = x cm
Then QR = QB − BR
= 56 − x
Step 3: Forming the Proportion:
35 / 65 = (56 − x) / x
Simplify:
7 / 13 = (56 − x) / x
Cross multiply:
7x = 13(56 − x)
7x = 728 − 13x
7x + 13x = 728
20x = 728
x = 728 / 20
x = 36.4 cm
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: Using Property of Similar Triangles:
Since ΔPQR ∼ ΔCBR,
The ratio of their areas is equal to the square of the ratio of their corresponding sides.
Area(PQR) / Area(CBR) = (PQ / CB)²
However, to find x, it is simpler to use the proportionality of sides directly.
Step 2: Writing Ratio of Corresponding Sides:
From similarity,
PQ / CB = QR / BR
Given:
PQ = 35 cm
CB = 65 cm
QR = (56 − x) cm
BR = x cm
