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In the following figure, O is the centre of a circular arc and AOB is a straight line. Find the perimeter and area of the shaded region correct to one decimal place (Take π = 3.14). - Mathematics

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Question

In the following figure, O is the centre of a circular arc, and AOB is a straight line. Find the perimeter and area of the shaded region correct to one decimal place (Take π = 3.14).

Sum
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Solution

Given: O is the centre, A–O–B is a straight line. So, AB is a diameter. The arc and the triangle meet at C with AC = 12 cm and BC = 16 cm.

Step-wise calculation:

1. Since AB is a diameter, triangle ACB is right-angled at C Thales.

`AB = sqrt(AC^2 + BC^2)`

= `sqrt(12^2 + 16^2)`

= `sqrt(144 + 256)`

= `sqrt(400)`

= 20 cm

2. Radius `r = "AB"/2` 

= `20/2` 

= 10 cm

3. Perimeter of shaded region = Length of semicircular arc AB + AC + CB.

Semicircular arc length

= πr = 3.14 × 10 

= 31.4 cm

AC + CB

= 12 + 16 

= 28 cm

Perimeter

= 31.4 + 28 

= 59.4 cm

4. Area of shaded region = Area of semicircle – Area of triangle ACB.

Area of semicircle

= `1/2 πr^2` 

= 0.5 × 3.14 × 102 

= 157.0 cm2

Area of triangle

= `1/2 xx AC xx BC` 

= 0.5 × 12 × 16 

= 96 cm2

Shaded area

= 157.0 – 96 

= 61.0 cm2

Perimeter = 59.4 cm to 1 decimal place.

Area = 61.0 cm2 to 1 decimal place.

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Chapter 16: Mensuration - Exercise 16C [Page 335]

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Nootan Mathematics [English] Class 9 ICSE
Chapter 16 Mensuration
Exercise 16C | Q 32. | Page 335
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