(iii) (a) What is the total length of silver wire required ?

Q: (iii) (a) What is the total length of silver wire required ?

Total length of silver wire required = **285 cm**, obtained by adding the circumference of the rim (110 cm) and the length of the 5 diameters (175 cm) that form the internal spokes of the brooch.

Figure shows a circle of diameter 35 cm with 5 diameters drawn through the centre, dividing the circle into 10 equal sec — Fig. for Q55

Areas Related to Circles

Circles — Circumference and Perimeter of Composite Shape

Official Answer

Total length of silver wire required = 285 cm, obtained by adding the circumference of the rim (110 cm) and the length of the 5 diameters (175 cm) that form the internal spokes of the brooch.

circumferencediametersilver wireπ = 22/7broochperimeter of composite figure285 cm

Marking Scheme

1½ mark: correctly identifying that total wire = circumference + 5 diameters.
2½ mark: computing circumference = $\pi d = \left(\frac{22}{7}\right) \times 35 = 110 \text{ cm}$ .
3½ mark: computing length of 5 diameters = $5 \times 35 = 175 \text{ cm}$ .
4½ mark: correct final answer, total wire = $110 + 175 = 285 \text{ cm}$ (with unit cm).

Hint

Total wire = circumference of the circle $(\pi d)$ + length of the 5 diameters $(5 \times d)$ . Use $\pi = \frac{22}{7}$ since $d = 35$ is a multiple of 7.

Quick Oral Answer

Total wire equals the circumference of the circle plus the length of the 5 diameters: $\pi d + 5d = 110 \text{ cm} + 175 \text{ cm} = 285 \text{ cm}$ .

Analysis & Explanation

Tests whether the wire used is only the circumference or also the internal diameters of the brooch design.

Concept

The brooch's wire outline = circular rim (circumference) + 5 straight diameters through the centre.
Diameter $d = 35 \text{ cm}$ is given so that $\pi = \frac{22}{7}$ gives a clean value.

Key points

$\text{Circumference} = \pi d = \left(\frac{22}{7}\right) \times 35 = 110 \text{ cm}$ .
Length of 5 diameters = $5 \times 35 = 175 \text{ cm}$ .
Total wire = $110 + 175 = 285 \text{ cm}$ .

Common mistakes

Computing only the circumference and forgetting the 5 diameters.
Confusing 5 diameters with 10 radii counted twice with wrong values.

Real-world

Mirrors how a jeweller/wire-craft designer budgets material for both the outer frame and internal decorative lines.

Common Mistakes

1Calculating only the circumference (110 cm) and forgetting to add the wire used for the 5 internal diameters.
2Adding 10 radii as $10 \times 17.5 = 175 \text{ cm}$ using the wrong count of diameters versus number of sectors, though the numeric result coincides here — confusion often leads to errors with different sector counts.
3Using $\pi = 3.14$ instead of $\frac{22}{7}$ , leading to a non-exact, messier value when the diameter is a multiple of 7.

Interesting Facts

Case-study/brooch-style circle mensuration questions have appeared in CBSE Class 10 board papers regularly since 2020 as part of the 'competency-based' question format introduced in that year.

Real jewellery designers use similar wire-length calculations (circumference + internal partitions) to estimate the exact grams of precious metal wire needed, since silver/gold wire is costed by weight per unit length.

Using a diameter that is a multiple of 7 (like 35 cm here) is a deliberate CBSE design choice so that π = 22/7 yields whole-number answers, making the question gradeable without calculators.

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Frequently Asked Questions

Why is $\pi = \frac{22}{7}$ used instead of 3.14 in this problem?

Because the diameter 35 cm is a multiple of 7, using $\pi = \frac{22}{7}$ gives a clean, exact answer (110 cm) without decimals, which is the intended approach for such CBSE case-study numbers.

Why do we add 5 diameters and not 10 radii separately?

Each of the 5 straight wires passes fully through the centre from one side of the circle to the other, so each one is a complete diameter (35 cm long), not two separate radii — though mathematically 5 diameters = 10 radii in total length, the wire is physically laid as 5 straight pieces.