Class 10 Areas Related to Circles Case Study Questions

Class 10 Areas Related to Circles Case Study Questions

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

Class 10 Areas Related to Circles Case Study Questions

In mathematics, Case Study Class 10 Maths Areas Related to Circles primarily focuses on real-life applications of circular figures. Students often solve Case Study Questions on Areas Related to Circles Class 10 in order to practice using formulas for sectors, segments, as well as shaded regions. Moreover, these questions not only enhance logical reasoning but also gradually prepare learners for CBSE exams. Consequently, regular exposure to such problems makes the learning process smoother and more effective.

Importance of Class 10 Areas Related to Circles Case Study Questions

By practicing Class 10 Areas Related to Circles Case Study Questions, students clearly understand how geometry applies to daily life situations. Additionally, Cbse Class 10 Case Study Questions on Circles steadily build accuracy in using π and related formulas. Therefore, solving such problems not only improves problem-solving skills but also strengthens conceptual clarity. In fact, these questions serve as a bridge between theoretical knowledge and practical application, thereby enriching mathematical learning.

Preparation Strategy for Success

Students should consistently and regularly attempt Case Study Questions Maths Class 10 Areas Related to Circles from NCERT books and sample papers. Furthermore, referring to Ncert Case Study Questions Class 10 Maths Circles ensures a thorough as well as systematic revision of the chapter. Thus, working with solved Areas Related to Circles Case Study Class 10 examples gradually helps in scoring better in exams. Finally, with persistent effort, consistent practice, and timely revisions, students can undoubtedly guarantee both confidence and speed.

Case Study 3: Circular Amphitheatre Design

Case Study 3: Circular Amphitheatre Design

A city council plans a circular open-air amphitheatre for community events. The outer boundary of the amphitheatre is a perfect circle of radius $28\ \text{m}$. At the centre there is a decorative circular pond of radius $8\ \text{m}$ used as a focal feature. The seating area occupies the annular region between the pond and the outer boundary. For an annual festival the council reserves a VIP zone which is a $135^{\circ}$ sector of the outer circle (seating included) and marks it with special tiles. A service walkway runs across the seating area and is designed as a straight chord of the outer circle at a perpendicular distance $20\ \text{m}$ from the centre; this chord cuts off a smaller circular segment that will be fenced for equipment. Near the outer rim, a decorative mosaic will occupy a ring-sector between radii $24\ \text{m}$ and $28\ \text{m}$ subtending an angle of $60^{\circ}$. The engineering team must compute exact and simplified areas and lengths (in terms of $\pi$ where appropriate) for materials, fencing and tiles. All numerical answers must be exact symbolic expressions or simplified fractions involving $\pi$ and radicals as required.

MCQ Questions:

1. What is the circumference of the amphitheatre’s outer boundary?

  • (a) $56\pi\ \text{m}$
  • (b) $28\pi\ \text{m}$
  • (c) $112\pi\ \text{m}$
  • (d) $784\pi\ \text{m}$
Answer: (a) $56\pi\ \text{m}$
Solution: Circumference $=2\pi R$ with $R=28$. Compute step by step: \[ 2\times 28=56,\qquad \text{so } 2\pi R=56\pi. \]

2. What is the exact area of the seating annulus (area between the outer boundary and the pond)?

  • (a) $720\pi\ \text{m}^2$
  • (b) $784\pi\ \text{m}^2$
  • (c) $640\pi\ \text{m}^2$
  • (d) $576\pi\ \text{m}^2$
Answer: (a) $720\pi\ \text{m}^2$
Solution: Area of annulus $=\pi(R^2-r^2)$ with $R=28$, $r=8$. Compute $R^2=28^2=784$, $r^2=8^2=64$. Then \[ R^2-r^2=784-64=720, \] hence area $=720\pi\ \text{m}^2$.

3. The VIP zone is a $135^{\circ}$ sector of the outer circle. What is its area?

  • (a) $294\pi\ \text{m}^2$
  • (b) $196\pi\ \text{m}^2$
  • (c) $392\pi\ \text{m}^2$
  • (d) $245\pi\ \text{m}^2$
Answer: (a) $294\pi\ \text{m}^2$
Solution: Sector area $=\dfrac{\theta}{360^{\circ}}\pi R^2$ with $\theta=135^{\circ}$, $R=28$. Compute: \[ \frac{135}{360}=\frac{3}{8},\qquad R^2=784, \] thus sector area $=\dfrac{3}{8}\cdot 784\pi$. Compute $784\div 8=98$, then $98\times 3=294$, so area $=294\pi\ \text{m}^2$.

4. The service chord is at perpendicular distance $d=20\ \text{m}$ from the centre of the outer circle ($R=28\ \text{m}$). The chord cuts off the smaller circular segment. Which expression equals the exact area of that smaller segment?

  • (a) $784\arccos\!\bigl(\tfrac{5}{7}\bigr)-160\sqrt{6}\ \text{m}^2$
  • (b) $392\arccos\!\bigl(\tfrac{5}{7}\bigr)-80\sqrt{6}\ \text{m}^2$
  • (c) $784\arccos\!\bigl(\tfrac{20}{28}\bigr)-80\sqrt{6}\ \text{m}^2$
  • (d) $392\arccos\!\bigl(\tfrac{20}{28}\bigr)-160\sqrt{6}\ \text{m}^2$
Answer: (a) $784\arccos\!\bigl(\tfrac{5}{7}\bigr)-160\sqrt{6}\ \text{m}^2$
Solution: Let $R=28$, $d=20$. Set $\theta=\arccos\!\bigl(\tfrac{d}{R}\bigr)=\arccos\!\bigl(\tfrac{20}{28}\bigr)=\arccos\!\bigl(\tfrac{5}{7}\bigr)$. The area of the smaller segment is \[ \text{segment area}=R^{2}\theta – d\sqrt{R^{2}-d^{2}}. \] Compute $R^{2}=28^{2}=784$. Next $R^{2}-d^{2}=784-400=384$. Factor: $384=64\cdot 6$, so $\sqrt{384}=8\sqrt{6}$. Then \[ d\sqrt{R^{2}-d^{2}}=20\cdot 8\sqrt{6}=160\sqrt{6}. \] Therefore the area equals \[ 784\arccos\!\bigl(\tfrac{5}{7}\bigr)-160\sqrt{6}\ \text{m}^2, \] which matches option (a). (Options (c) and (d) use equivalent arccos arguments but their coefficients on the radical term are incorrect; options (b) and (d) have incorrect overall factors.)

5. The decorative mosaic occupies the ring-sector between radii $24\ \text{m}$ and $28\ \text{m}$ subtending an angle $60^{\circ}$. What is its exact area?

  • (a) $\dfrac{104}{3}\pi\ \text{m}^2$
  • (b) $\dfrac{208}{3}\pi\ \text{m}^2$
  • (c) $48\pi\ \text{m}^2$
  • (d) $\dfrac{52}{3}\pi\ \text{m}^2$
Answer: (a) $\dfrac{104}{3}\pi\ \text{m}^2$
Solution: Ring-sector area $=\dfrac{\theta}{360^{\circ}}\pi\bigl(R^{2}-r^{2}\bigr)$ with $\theta=60^{\circ}$, $R=28$, $r=24$. Compute $R^{2}-r^{2}=28^{2}-24^{2}=784-576=208$. Next $\dfrac{60}{360}=\dfrac{1}{6}$. Thus area $=\dfrac{1}{6}\cdot 208\pi=\dfrac{208}{6}\pi$. Simplify $208\div 2=104$, $6\div 2=3$, hence area $=\dfrac{104}{3}\pi\ \text{m}^2$.

Your Results:

Correct Answers: 0

Incorrect Answers: 0

Percentage Score: 0%

Educational Resources Footer

Free Educational Resources

Free Study Material English Vocabulary Tests Grammar for Hearing Impaired Math for Hearing Impaired English Vocabulary Case Study Questions Math GRE Vocabulary Tests LaTeX Tutorial R Programming Tests Online English Games Online Math Games Learn Computer Basics JEE Math Tests CAT Quant Tests Increase Calculation Speed Class 10 Math Tests Prompt Engineering SAT Math Tutorial Logic Quizzes Grade 1 Logic Quizzes Grade 2

© 2023 Udgam Welfare Foundation. All rights reserved.