46
Practical exercise and application
Textbook: pp. 124–125
GoalApply the formulas for circumference, area of a disc, sector and segment in mixed problems; split a complex figure into simple parts.
New words
area of a composite figure · murakkab shakl yuzisplitting into parts · bo‘laklarga ajratishC = 2πR · C = 2πRS = πR² · S = πR²
Explanation
To find the area of a composite figure we split it into simple parts such as discs, sectors, triangles and squares, or subtract an extra part from a larger figure. The answer often contains π and it is best to leave it exact, for example 25π − 50. A circle inscribed in a square of side a has radius a/2, and the circumscribed circle has radius a√2/2. When the angles of several sectors with the same radius add up to a known total, we can find their combined area at once instead of one by one. Before solving, mark the needed radii, angles and sides on the figure.
Worked examples
A square of side 10 with an inscribed disc: S = π·5² = 25π. The corner region is 100 − 25π.
Three sectors of radius 2 at the vertices of a triangle with angles adding up to 180°: S = π·4·180/360 = 2π.
Class activity
“Cut and compare”: draw a square and a disc inside it on paper; with an adult's help with scissors cut off the corner regions and check that their area is 100 − 25π for side 10.
Practice
1
What is the circumference of the circle circumscribed about a square of side 8?
8√2π
2
A 90° sector is cut from a disc of radius 6. What is the area of the rest?
27π
3
A square with side 10√2 is inscribed in a disc of radius 10. What is the area between them?
100π − 200
4
Why is the diagonal of an inscribed square equal to the diameter?
The vertices lie on the circle and the diagonal passes through the centre.