CBSE Board Exam 2026 | Q.P. Code 30/8/1 (Set-1) | 1 Mark MCQ
In the given figure, there is a circle with centre 'O' having three tangents at points P, T and S. If $QR=12\text{ cm}$ and radius of a circle is $7\text{ cm}$, then perimeter of polygon PQRSO is:
A. $26\text{ cm}$
B. $31\text{ cm}$
C. $38\text{ cm}$
D. $33\text{ cm}$
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1
Theorem 10.2: The lengths of tangents drawn from an external point to a circle are perfectly equal. This applies to both point $Q$ ($QP = QT$) and point $R$ ($RS = RT$).
2
Perimeter Definition: The perimeter of a polygon is the total length of its continuous outer boundary. For polygon PQRSO, the boundary consists of sides $OP$, $PQ$, $QR$, $RS$, and $SO$.
1
Write out the perimeter formula for the requested polygon PQRSO: Perimeter = $OP + PQ + QR + RS + SO$.
2
You are given three of these values directly: Base $QR = 12$, Radius $OP = 7$, and Radius $SO = 7$.
3
You still need $PQ$ and $RS$. Look at the tangency point $T$ lying on the base $QR$. Notice that $QR$ is made of two segments: $QR = QT + TR$.
4
By Theorem 10.2, the tangent $QP$ is equal to $QT$, and the tangent $RS$ is equal to $TR$.
5
Therefore, $PQ + RS = QT + TR = QR = 12$. Substitute this sum back into your perimeter equation!
!
Assuming Midpoints: A common mistake is assuming that $T$ is exactly in the middle of $QR$, meaning $QT=6$ and $TR=6$, which leads to $QP=6$ and $RS=6$. While this gives the correct final sum, it's mathematically flawed. The sum $PQ + RS$ will ALWAYS be 12 regardless of where $T$ touches, even if the shape is entirely asymmetric!
🚀
The "Tangent Fold" Trick: Whenever a straight tangent line ($QR$) touches a circle and you need the perimeter of the polygon bounded by its corresponding external tangents and radii (the "hat" shape), the formula is ALWAYS simply:
Perimeter = $2 \times (\text{Tangent Base}) + 2 \times (\text{Radius})$
Here, Base = 12 and Radius = 7.
Perimeter = $(2 \times 12) + (2 \times 7) = 24 + 14 = 38$. Done in 3 seconds!
Perimeter = $2 \times (\text{Tangent Base}) + 2 \times (\text{Radius})$
Here, Base = 12 and Radius = 7.
Perimeter = $(2 \times 12) + (2 \times 7) = 24 + 14 = 38$. Done in 3 seconds!
Topper's Answer Sheet
Given: $QR = 12\text{ cm}$. Radius $OP = OS = 7\text{ cm}$.
The perimeter of polygon PQRSO is the sum of its outer boundaries: Perimeter = $OP + PQ + QR + RS + SO$ [½ Mark]
We know that lengths of tangents drawn from an external point to a circle are equal. From point Q: $QP = QT$ From point R: $RS = RT$
Now, observe the line segment $QR$: $QR = QT + TR$ Substituting $QT=QP$ and $TR=RS$: $QR = QP + RS$
Since $QR = 12\text{ cm}$, it means the sum of the other two tangents is also 12: $QP + RS = 12\text{ cm}$ [½ Mark]
Now substitute all known values into the perimeter equation: Perimeter = $OP + SO + QR + (PQ + RS)$ Perimeter = $7 + 7 + 12 + 12$ Perimeter = $14 + 24$
$\therefore$ Perimeter = $38\text{ cm}$
Ans: Option (C) $38\text{ cm}$
The perimeter of polygon PQRSO is the sum of its outer boundaries: Perimeter = $OP + PQ + QR + RS + SO$ [½ Mark]
We know that lengths of tangents drawn from an external point to a circle are equal. From point Q: $QP = QT$ From point R: $RS = RT$
Now, observe the line segment $QR$: $QR = QT + TR$ Substituting $QT=QP$ and $TR=RS$: $QR = QP + RS$
Since $QR = 12\text{ cm}$, it means the sum of the other two tangents is also 12: $QP + RS = 12\text{ cm}$ [½ Mark]
Now substitute all known values into the perimeter equation: Perimeter = $OP + SO + QR + (PQ + RS)$ Perimeter = $7 + 7 + 12 + 12$ Perimeter = $14 + 24$
$\therefore$ Perimeter = $38\text{ cm}$
Ans: Option (C) $38\text{ cm}$
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