Home > Areas Related to Circles > CBSE_NCERT_EXEMPLAR_CH10 > 0d0bbb73-8a82-5339-9503-8245632d3616
Not the exact question you were looking for?

Paste your question to our Mathbee AI Mentor below to get an instant step-by-step solution.

Areas Related to Circles
NCERT Exemplar Ch 10
CBSE_NCERT_EXEMPLAR_CH10
Grade 10

Question:

A brooch is made with silver wire in the form of a circle with diameter $35\text{ mm}$. The wire is also used in making $5$ diameters which divide the circle into $10$ equal sectors. Find:
(i) total length of silver wire required
(ii) area of each sector of the brooch.

Step-by-Step Solution

Key Concept: Wire length $= \pi d + 5d$. Sector angle $\theta = 360^\circ / 10 = 36^\circ$.
Stepwise Solution:

(i) Wire length $= \pi d + 5 d = \dfrac{22}{7}(35) + 5(35) = 110 + 175 = 285\text{ mm}$. [1.5 Marks]

(ii) $\theta = 36^\circ, r = 35/2\text{ mm}$. Sector area $= \dfrac{36}{360} \times \dfrac{22}{7} \times \dfrac{35}{2} \times \dfrac{35}{2} = \dfrac{1}{10} \times \dfrac{1925}{2} = \dfrac{385}{4}\text{ mm}^2 = 96.25\text{ mm}^2$. [1.5 Marks]

Marking Scheme:

• Part (i) Total wire length $= 285\text{ mm}$: 1.5 Marks
• Part (ii) Sector area $= 385/4\text{ mm}^2 = 96.25\text{ mm}^2$: 1.5 Marks

Correct Answer:
Mathbee AI Mentor (Free Demo)

Confused by the solution? Ask the AI to explain a specific step, tell you where you went wrong, or break down the key trap in this question.

Master Areas Related to Circles with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free