Question 552 marksShort AnswerSection E

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

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.

Answer keywords:circumferencediametersilver wireπ = 22/7broochperimeter of composite figure285 cm

Marking scheme

  1. 1.½ mark: correctly identifying that total wire = circumference + 5 diameters.
  2. 2.½ mark: computing circumference = πd=(227)×35=110 cm\pi d = \left(\frac{22}{7}\right) \times 35 = 110 \text{ cm}.
  3. 3.½ mark: computing length of 5 diameters = 5×35=175 cm5 \times 35 = 175 \text{ cm}.
  4. 4.½ mark: correct final answer, total wire = 110+175=285 cm110 + 175 = 285 \text{ cm} (with unit cm).

Hint

Total wire = circumference of the circle (πd)(\pi d) + length of the 5 diameters (5×d)(5 \times d). Use π=227\pi = \frac{22}{7} since d=35d = 35 is a multiple of 7.

Quick oral answer

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

Common mistakes

  1. 1.Calculating only the circumference (110 cm) and forgetting to add the wire used for the 5 internal diameters.
  2. 2.Adding 10 radii as 10×17.5=175 cm10 \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.
  3. 3.Using π=3.14\pi = 3.14 instead of 227\frac{22}{7}, leading to a non-exact, messier value when the diameter is a multiple of 7.

Analysis and 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 cmd = 35 \text{ cm} is given so that π=227\pi = \frac{22}{7} gives a clean value.

Key points

  • Circumference=πd=(227)×35=110 cm\text{Circumference} = \pi d = \left(\frac{22}{7}\right) \times 35 = 110 \text{ cm}.
  • Length of 5 diameters = 5×35=175 cm5 \times 35 = 175 \text{ cm}.
  • Total wire = 110+175=285 cm110 + 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.

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.

Question details

Q55 of 56 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Standard)
Year
2026
Section
Section E
Marks
2
Type
Short Answer
Chapter
Areas Related to Circles
Topic
Circles — Circumference and Perimeter of Composite Shape
Bloom's level
apply
Syllabus topic
Unit 5: Mensuration — Areas Related to Circles: Circumference of a Circle and Perimeter of Combined Figures
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is π=227\pi = \frac{22}{7} used instead of 3.14 in this problem?

Because the diameter 35 cm is a multiple of 7, using π=227\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.