A Spiral Is Made Up of Successive Semicircles With Centres Alternately at A and B, of Radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … What Is the Total Length of Such a Spiral Made Up of Thirteen Consecutive Semicircles

NCERT Class 10 Mathematics — Arithmetic Progressions, Chapter 5, Question 18.

Problem

A spiral is made up of successive semicircles, with centres alternately at $A$ and $B$, starting with centre at $A$, of radii $0.5\text{ cm}, 1.0\text{ cm}, 1.5\text{ cm}, 2.0\text{ cm}, \ldots$. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $\pi=\dfrac{22}7$.)

Key insight. Each semicircle’s arc length is $\pi r$ for its own radius, and those 13 radii themselves form an arithmetic progression — so the total spiral length is $\pi$ times the sum of an AP, not 13 separate circumference calculations.

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Approach

Express the length of the $k$-th semicircle in terms of the base radius $r=0.5$ cm, using the fact that a semicircle’s arc length is half the full circumference, $\pi r$. Since the radii themselves increase by $0.5$ cm each time — $r, 2r, 3r, \ldots, 13r$ — the total length becomes $\pi r$ times the sum of the first 13 natural numbers, which is an ordinary AP sum. Only at the very end is the numerical value of $r$ substituted in, to avoid decimal arithmetic until it’s unavoidable.

Solution

Step 1 — Express each semicircle’s radius in terms of r

Let $r=0.5$ cm be the smallest radius. The given radii $0.5, 1.0, 1.5, 2.0, \ldots$ are then $r, 2r, 3r, 4r, \ldots$, up to the 13th semicircle with radius $13r$.

Step 2 — Write the arc length of each semicircle

The circumference of a full circle of radius $\rho$ is $2\pi\rho$; a semicircle’s arc is half of that, $\pi\rho$. So the lengths of the successive semicircles are:

$$L_1 = \pi r, \quad L_2 = \pi(2r) = 2\pi r, \quad L_3 = \pi(3r) = 3\pi r, \quad \ldots, \quad L_{13} = 13\pi r$$

Step 3 — Sum all 13 semicircle lengths

$$S = L_1+L_2+\cdots+L_{13} = \pi r(1+2+3+\cdots+13)$$

The sum $1+2+\cdots+13$ is the sum of the first 13 natural numbers — an AP with first term 1, common difference 1, and 13 terms:

$$1+2+\cdots+13 = \frac{13}2\left[2(1)+(13-1)(1)\right] = \frac{13}2\times14 = 13\times7 = 91$$

Step 4 — Combine

$$S = \pi r\times91$$

Step 5 — Substitute the numerical values

With $r=0.5$ cm $=\dfrac12$ cm and $\pi=\dfrac{22}7$:

$$S = \frac{22}7\times\frac12\times91$$

Since $91=7\times13$, the factor of 7 cancels with the denominator:

$$S = \frac{22\times13}2 = \frac{286}2 = 143$$

Answer

Common mistakes

• Using the full-circle circumference formula instead of the semicircle arc length. Each piece of the spiral is a semicircle, not a full circle, so the length is $\pi r$, not $2\pi r$ — using the wrong one doubles the final answer.
• Substituting the numerical value of $r=0.5$ too early. Working with $r$ symbolically until the very last step, as in Step 3–4, keeps the arithmetic to a single clean multiplication; substituting $0.5$ at every intermediate step invites decimal-rounding slips.
• Miscounting the number of terms in the AP sum. With “13 consecutive semicircles,” the sum $1+2+\cdots+13$ has exactly 13 terms — it’s easy to accidentally sum only to 12 or extend to 14 when writing out the formula.

Practise next

• A similar spiral is made of 10 consecutive semicircles with the same starting radius of $0.5$ cm, increasing by $0.5$ cm each time. Find its total length, using the same $\pi r \times (\text{AP sum})$ approach.