Ex 11.1, 9 - Chapter 11 Class 10 Areas related to Circles

Last updated at April 16, 2024 by Teachoo

Transcript

Ex 11.1, 9
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. Find :
the total length of the silver wire required
Brooch is made with silver wire in form of a circle.
Diameter of the brooch = 35 mm
Radius of brooch = 𝟑𝟓/𝟐 mm
Since wire is used in making 5 diameter & circle.
So,
Total wire is used = Length of wire in circle + Wire used in 5 diameters
Length of wire in circle
Length of wire in circle = Circumference of Circle
= 2𝜋𝑟
= 2×22/7×35/2
= 22 × 5
= 110 mm
Wire used in 5 diameters
Silver wire used in 5 diameters = 5 × Diameter of circle
= 5 × 35
= 175 mm
Now,
Total wire is used = Length of wire in circle + Wire used in 5 diameters
= 110 + 175
= 285 mm
Ex 11.1, 9
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. Find :
(ii) the area of each sector of the brooch.
Since the wire divides the circle into 10 equal sectors
Area of each sector of the brooch
= 𝟏/𝟏𝟎× Area of all sectors of the brooch
Now,
Area of all sectors of the brooch = Area of circle
= 𝝅𝒓𝟐
= 22/7×(35/2)^2
= 22/7×35/2×35/2
= (𝟏𝟏 × 𝟓 × 𝟑𝟓)/𝟐
Hence,
Area of each sector of the brooch
= 𝟏/𝟏𝟎× Area of all sectors of the brooch
= 1/10×(11 × 5 × 35)/2
= 𝟑𝟖𝟓/𝟒 mm2
∴ Area of each sector = 385/4 mm2

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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