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Arithmetic Progressions
EXERCISE 5.3
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take = 22 7 ) 70 Fig. 5.4 [Hint : Length of successive semicircles is l1, l2, l3, l4, . . . with centres at A, B, A, B, . . ., respectively.]
Step-by-Step Solution
Key Concept: The radii of the successive semicircles form an arithmetic progression (AP) with first term a = 0.5 cm and common difference d = 0.5 cm. The length of a semicircle of radius r is (π r). Hence the total length = π × (sum of the radii). Use the formula for the sum of the first n terms of an AP and substitute π = 22/7.
1. Identify the AP of radii\
The radii are 0.5 cm, 1.0 cm, 1.5 cm, … .
\[ r_k = 0.5 + (k-1)\times0.5 = 0.5k \] for k = 1,2,…,13.
2. Sum of the radii\
Sum of the first n terms of an AP: \[ S_n = \frac{n}{2}\big[2a+(n-1)d\big] \]
Here, n = 13, a = 0.5, d = 0.5.\
\[ S_{13} = \frac{13}{2}\big[2(0.5)+(13-1)(0.5)\big] = \frac{13}{2}\big[1+6\big] = \frac{13}{2}\times7 = 45.5 \text{ cm} \]
(Alternatively, \(S_{13}=0.5\sum_{k=1}^{13}k =0.5\times\frac{13\times14}{2}=45.5\) cm.)
3. Length of each semicircle\
Length of a semicircle of radius r is \(\ell = \pi r\).
Therefore total length \(L\) of 13 semicircles is \[ L = \pi \times S_{13} = \pi \times 45.5 = \frac{91\pi}{2} \] cm.
4. Insert the given value of \(\pi\)\
Given \(\pi = \frac{22}{7}\).\
\[ L = \frac{91}{2}\times\frac{22}{7}=\frac{91\times22}{14}=\frac{2002}{14}=143 \text{ cm} \]
5. Result\
The total length of the spiral consisting of thirteen consecutive semicircles is 143 cm.
The radii are 0.5 cm, 1.0 cm, 1.5 cm, … .
\[ r_k = 0.5 + (k-1)\times0.5 = 0.5k \] for k = 1,2,…,13.
2. Sum of the radii\
Sum of the first n terms of an AP: \[ S_n = \frac{n}{2}\big[2a+(n-1)d\big] \]
Here, n = 13, a = 0.5, d = 0.5.\
\[ S_{13} = \frac{13}{2}\big[2(0.5)+(13-1)(0.5)\big] = \frac{13}{2}\big[1+6\big] = \frac{13}{2}\times7 = 45.5 \text{ cm} \]
(Alternatively, \(S_{13}=0.5\sum_{k=1}^{13}k =0.5\times\frac{13\times14}{2}=45.5\) cm.)
3. Length of each semicircle\
Length of a semicircle of radius r is \(\ell = \pi r\).
Therefore total length \(L\) of 13 semicircles is \[ L = \pi \times S_{13} = \pi \times 45.5 = \frac{91\pi}{2} \] cm.
4. Insert the given value of \(\pi\)\
Given \(\pi = \frac{22}{7}\).\
\[ L = \frac{91}{2}\times\frac{22}{7}=\frac{91\times22}{14}=\frac{2002}{14}=143 \text{ cm} \]
5. Result\
The total length of the spiral consisting of thirteen consecutive semicircles is 143 cm.
Correct Answer: 143 cm
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