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Lesson Example Discussion Quiz: Class Homework

Step-4

Title: Area of Circle

Grade: 6-a Lesson: S2-L1

Explanation: Hello Students, time to practice and review the steps for the problem.

Lesson Steps

Step	Type	Explanation	Answer
1	Problem	A chord of 8 mm is 6 mm from the center of the circle. Calculate the area of the circle.
2	Step	The given values are	Diameter 1 = 5 and diameter 2= 6
3	Formula	Applying Pythagoras theorem	\$r^2 = ("Distance from center to chord")^2+(("Chord length")/2)^2\$
4	Step	r is equal to the radius of circle(r)
5	Step	Substitute the values	\$r^2 = 6^2 + (8/2) ^2\$
6	Step	After simplification	\$r^2 = 36 + 16\$
7	Step	After simplification	\$r^2 = 52\$
8	Step	After simplification	\$ r = \sqrt 52\$ stem:[r = 7.2 mm]
9	Step	After simplification	r = 6.5 mm
10	Step	Area of circle	\$ A = πr^2\$
11	Step	Substitute the values	\$A = 3.141 times (7.2)^2\$
12	Step	After simplification	A = 162.82 sq.mm
13	Step	The area of a circle is 162.82 sq.mm
14	Answer	Option	B
15	SumUp	Can you summarize what you’ve understood in the above steps?	Audio

Step

Type

Explanation

Answer

Problem

A chord of 8 mm is 6 mm from the center of the circle. Calculate the area of the circle.

Step

The given values are

Diameter 1 = 5 and diameter 2= 6

Formula

Applying Pythagoras theorem

\$r^2 = ("Distance from center to chord")^2+(("Chord length")/2)^2\$

Step

r is equal to the radius of circle(r)

Step

Substitute the values

\$r^2 = 6^2 + (8/2) ^2\$

Step

After simplification

\$r^2 = 36 + 16\$

Step

After simplification

\$r^2 = 52\$

Step

After simplification

\$ r = \sqrt 52\$

stem:[r = 7.2 mm]

Step

After simplification

r = 6.5 mm

Step

Area of circle

\$ A = πr^2\$

Step

Substitute the values

\$A = 3.141 times (7.2)^2\$

Step

After simplification

A = 162.82 sq.mm

Step

The area of a circle is 162.82 sq.mm

Answer

Option

SumUp

Can you summarize what you’ve understood in the above steps?

Audio