Case Study Class 10 Maths Areas Related to Circles

Case Study Class 10 Maths Areas Related to Circles

Case Study Class 10 Maths Areas Related to Circles | CBSE & NCERT Questions

Introduction to Circles in Case Studies

In mathematics, Case Study Class 10 Maths Areas Related to Circles focuses on real-life applications of circular figures. Students often solve Case Study Questions on Areas Related to Circles Class 10 to practice using formulas for sectors, segments, and shaded regions. Moreover, these questions enhance logical reasoning and prepare learners for CBSE exams.

Importance of Circle-Based Case Studies

By practicing Class 10 Areas Related to Circles Case Study Questions, students understand how geometry applies to daily life situations. Additionally, Cbse Class 10 Case Study Questions on Circles build accuracy in using π and related formulas. Therefore, solving such problems improves both problem-solving skills and conceptual clarity.

Preparation Strategy for Success

Students should regularly attempt Case Study Questions Maths Class 10 Areas Related to Circles from NCERT books and sample papers. Referring to Ncert Case Study Questions Class 10 Maths Circles ensures thorough revision of the chapter. Thus, working with solved Areas Related to Circles Case Study Class 10 examples helps in scoring better in exams. Finally, consistent practice guarantees confidence and speed.

Case Study 2: Circular Lake and Surrounds

Case Study 2: Circular Lake and Surrounds

A horticulture club is redesigning a recreational circular lake and its surrounds to make the area educational for Grade 10 students learning about areas related to circles. The original lake is a perfect circle of radius $21\ \text{m}$. Around the lake there is a concentric circular walking belt (an annular strip) used for jogging and exhibitions. At one sector of the outer edge (a $120^\circ$ sector) the club intends to lay decorative stones forming a fan-shaped pattern (a sector of the outer circle). On the opposite side of the lake, at a radius of $14\ \text{m}$ from the centre, a narrow service path (a straight chord) runs and cuts off a small circular segment of the inner garden; the perpendicular distance from the centre to the chord is $10\ \text{m}$. Additionally, the club will convert a ring-sector between radii $18\ \text{m}$ and $21\ \text{m}$ subtending an angle of $150^\circ$ into a flowerbed. The students must compute circumferences, areas of circles, annuli, sectors, segments and ring-sectors to prepare accurate material quantities and labels for the information boards. The paragraph above describes the exact geometric layout; all questions that follow are based only on this description and require exact or simplified symbolic answers where appropriate.

MCQ Questions:

1. What is the circumference of the lake (in metres)?

  • (a) $21\pi$
  • (b) $42\pi$
  • (c) $84\pi$
  • (d) $441\pi$
Answer: (b) $42\pi$
Solution: Circumference $=2\pi r=2\pi(21)=42\pi\ \text{m}$.

2. Find the area of the annular walking belt that lies between the lake (radius $21\ \text{m}$) and a concentric inner ornamental moat of radius $7\ \text{m}$ which the club plans to keep water-free.

  • (a) $392\pi\ \text{m}^2$
  • (b) $350\pi\ \text{m}^2$
  • (c) $1225\pi\ \text{m}^2$
  • (d) $98\pi\ \text{m}^2$
Answer: (a) $392\pi\ \text{m}^2$
Solution: Area of annulus $=\pi(R^2-r^2)=\pi(21^2-7^2)=\pi(441-49)=392\pi\ \text{m}^2$.

3. The decorative fan-shaped stone pattern occupies a $120^\circ$ sector of the outer circle of radius $21\ \text{m}$. What is the area of that sector?

  • (a) $147\pi\ \text{m}^2$
  • (b) $77\pi\ \text{m}^2$
  • (c) $63\pi\ \text{m}^2$
  • (d) $441\pi\ \text{m}^2$
Answer: (a) $147\pi\ \text{m}^2$
Solution: Sector area $=\dfrac{\theta}{360^\circ}\pi r^2$ with $\theta=120^\circ$, $r=21$. Thus area $=\dfrac{120}{360}\pi(21^2)=\dfrac{1}{3}\pi\cdot441=147\pi\ \text{m}^2$.

4. On the inner garden (radius $14\ \text{m}$) the service chord is at perpendicular distance $10\ \text{m}$ from the centre and cuts off the smaller circular segment. Express the area of this smaller segment exactly (in simplest symbolic form) and choose the correct option.

  • (a) $196\arccos\!\bigl(\tfrac{5}{7}\bigr)-40\sqrt{6}\ \text{m}^2$
  • (b) $196\arccos\!\bigl(\tfrac{5}{7}\bigr)-20\sqrt{6}\ \text{m}^2$
  • (c) $98\arccos\!\bigl(\tfrac{5}{7}\bigr)-40\sqrt{6}\ \text{m}^2$
  • (d) $196\arccos\!\bigl(\tfrac{10}{14}\bigr)-40\sqrt{6}\ \text{m}^2$
Answer: (a) $196\arccos\!\bigl(\tfrac{5}{7}\bigr)-40\sqrt{6}\ \text{m}^2$
Solution: Let the radius be $r=14$ and let the perpendicular distance from centre to chord be $d=10$. Denote $\alpha=\arccos\!\bigl(\tfrac{d}{r}\bigr)=\arccos\!\bigl(\tfrac{5}{7}\bigr)$. The central angle corresponding to the smaller segment is $2\alpha$. The area of a segment with central angle $2\alpha$ is \[ \text{segment area}=\tfrac{1}{2}r^2(2\alpha-\sin 2\alpha) =r^2\bigl(\alpha-\sin\alpha\cos\alpha\bigr). \] Compute $\cos\alpha=\tfrac{5}{7}$ and $\sin\alpha=\sqrt{1-\bigl(\tfrac{5}{7}\bigr)^2}=\tfrac{2\sqrt{6}}{7}$. Hence \[ \sin\alpha\cos\alpha=\frac{5}{7}\cdot\frac{2\sqrt{6}}{7}=\frac{10\sqrt{6}}{49}. \] Therefore the segment area equals \[ r^2\Bigl(\alpha-\frac{10\sqrt{6}}{49}\Bigr) =14^2\alpha-14^2\cdot\frac{10\sqrt{6}}{49} =196\arccos\!\bigl(\tfrac{5}{7}\bigr)-40\sqrt{6}\ \text{m}^2, \] which matches option (a). Note option (d) is equivalent to (a) in the $\arccos$ argument but (a) is the simplified correct form.

5. The flowerbed occupies the ring-sector between radii $18\ \text{m}$ and $21\ \text{m}$ subtending an angle of $150^\circ$. What is the exact area of this ring-sector?

  • (a) $\dfrac{195}{4}\pi\ \text{m}^2$
  • (b) $48\pi\ \text{m}^2$
  • (c) $\dfrac{117}{6}\pi\ \text{m}^2$
  • (d) $50\pi\ \text{m}^2$
Answer: (a) $\dfrac{195}{4}\pi\ \text{m}^2$
Solution: Area of a ring-sector $=\dfrac{\theta}{360^\circ}\pi(R^2-r^2)$ with $\theta=150^\circ$, $R=21$, $r=18$. Compute $R^2-r^2=441-324=117$. Thus area $=\dfrac{150}{360}\pi\cdot117=\dfrac{5}{12}\cdot117\pi=\dfrac{585}{12}\pi=\dfrac{195}{4}\pi\ \text{m}^2$.

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