 # NCERT Exemplar Solution for Class 10 Maths: Areas Related to Circles (Part-IIIA)

NCERT Exemplar Problems and Solutions for CBSE Class 10 Mathematics chapter 10, Areas Related to Circles (Part-IIIA). This part includes solutions to Q. No. 1-8 from Exercise 11.3 of Class 10 Maths NCERT Exemplar Problems, chapter 11. It consists of only the Short Answer Type Questions. Here you get the CBSE Class 10 Mathematics chapter 11, Areas Related to Circles: NCERT Exemplar Problems and Solutions (Part-IIIA). This part of the chapter includes solutions to Question Number 1 to 8 from Exercise 11.3 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Areas Related to Circles (Part-IIIA). This exercise comprises only the Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

CBSE Class 10 Mathematics Syllabus 2017-2018

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Areas Related to Circles:

Exercise 11.3

Short Answer Type Questions (Q. No. 1-8)

Question. 1 Find the radius of a circle whose circumference is equal to the sum of the circumference of two circles

Solution.

According to the question,

Circumference of bigger circle = Circumference of first circle + Circumference of second circle Hence, required radius of circle is 33cm.

Question. 2 In figure, a square of diagonal 8 cm is inscribed in a circle. Find the area of the shaded region. Solution. To find the area of shaded region first we have to find the area of the circle and the inscribed square.

and the radius of circle be ‘r’. Let the side of the square be a cm.

Given that, length of diagonal of square, AC = 8cm

Thus, radius of circle, r = 8/2 = 4 cm

Here ∠ABC =  90o

By applying Pythagoras theorem in right angled ΔABC, we get: Thus, area of shaded region = 18.28 cm

Question. 3 Find the area of a sector of a circle of radius 28cm and central angle 45°.

Solution.

Given, radius of a circle, r = 28cm

And central angle, θ = 45° Question. 4 The wheel of a motor cycle is of radius 35cm. How many revolutions per minute must the wheel make,

so as to keep a speed of 66 km/h?

Solution. Question. 5 A cow is tied with a rope of length 14m at the corner of a rectangular field of dimensions 20m ´ 16m.

Find the area of the field in which the cow can graze.

Solution.

The area covered by cow for grazing is in the form of a sector with radius 14m (= length of the rope). Question. 6 Find the area of the flower bed (with semi-circular ends) shown in figure. Solution.

The flower bed is a combination of a rectangle and two semicircles as shown in the figure given below: Question. 7 In figure, AB is a diameter of the circle, AC = 6cm and BC = 8cm. Find the area of the shaded region, (use π = 3.14) Solution.

Given figure is a combination of a circle and a ΔACB.

Since, angle in a semi circle is equal to 90°.

∴          ∠C = 90°

Thus, applying Pythagoras theorem in right angled ΔACB, Solution. Given shaded region is a combination of a rectangle and a semi circle as shown below: NCERT Solutions for CBSE Class 10 Maths

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