CBSE Class 10 Mathematics Areas Related to Circles Notes Set 08

Applications of Area Related to Circles

In our daily life we come across various plane figures which are combinations of two or more plane figures. For example, window designs, flower beds, circular paths and so on. In this section we discuss problems on combinations of plane figures.

Important formulae:
1. Heron's formula: area of a triangle \( = \sqrt{s(s-a)(s-b)(s-c)} \), where s is the semi-perimeter and a, b, c are the sides of the triangle.
2. Area of a right-angled triangle = \( \dfrac{1}{2} \) × base × altitude.
3. Area of an equilateral triangle \( = \dfrac{\sqrt{3}}{4}a^2 \).
4. Area of a rectangle = length × breadth; perimeter = 2(a + b).
5. Area of a square of side a = a²; diagonal \( = \sqrt{2}\,a \); perimeter = 4a.
6. Area of a parallelogram = base × height.
7. Area of a rhombus \( = \dfrac{1}{2}d_1 d_2 \), where \( d_1 \) and \( d_2 \) are the diagonals.
8. Area of a trapezium \( = \dfrac{1}{2}(a + b)h \), where a and b are the parallel sides and h is the distance between them.

Solved Examples on Combinations of Figures

Check Your Progress: Combinations of Figures

Check Your Progress 11. Find the area of the largest possible square inscribed in a circle of unit radius.
Answer:
The largest square inscribed in a circle has its diagonal equal to the diameter 2. Area of a square = ½ × (diagonal)² = ½ × 4 = 2 square units.
In simple words: The diagonal of the square is the diameter of the circle.

Teacher's Note:
1. Use area = d²/2.
2. The side is √2.

Check Your Progress 12. Four circular cardboard pieces, each of radius 7 cm, are placed in such a way that each piece touches two other pieces. Find the area of the space enclosed by the four pieces.
Answer:
Joining the centres gives a square of side 14 cm. The space enclosed = area of the square − four quarter circles (each inside the square) = \( 14^2 - 4 \times \dfrac{1}{4}\pi 7^2 \).
\( = 196 - 154 = \) 42 cm².
In simple words: The four quarter circles together make one full circle inside the square of centres.

Teacher's Note:
1. Join the four centres.
2. Subtract the corner sectors from the square.

Check Your Progress 13. Four horses are tethered at four corners of a square plot of side 63 metres so that they just cannot reach one another. Find the area left ungrazed inside the plot.
Answer:
The ropes of two neighbouring horses together equal the side, so each rope is 31.5 m long. Each horse grazes a quadrant, so the four horses graze π(31.5)² in all.
Ungrazed area = \( 63^2 - \dfrac{22}{7} \times 31.5^2 = 3969 - 3118.5 = \) 850.5 m².
In simple words: Four quadrants of equal radius make one full circle.

Teacher's Note:
1. Rope = half the side of the square.
2. Subtract the grazed area from the plot.

Check Your Progress 14. A building with base in the form of an equilateral triangle of side 14 m is built in a huge grass field. In one corner of the building a cow is tethered with a rope of length 21 m. Find the area grazed by the cow.
Answer:
At the corner where it is tethered the cow can move over the exterior angle 360° − 60° = 300° with radius 21 m: area = \( \dfrac{300}{360} \times \dfrac{22}{7} \times 21^2 = 1155 \) m².
The rope reaches the other two corners with 21 − 14 = 7 m left. At each of them the cow can swing through 180° − 60° = 120° with radius 7 m: area = \( 2 \times \dfrac{120}{360} \times \dfrac{22}{7} \times 7^2 = \dfrac{308}{3} \approx 102.67 \) m².
Total area grazed = 1155 + 102.67 = 1257.67 m² (approximately).
In simple words: The cow goes round the corners of the building and keeps grazing with the rope that is left.

Teacher's Note:
1. Split the area into one big sector and two small sectors.
2. Each exterior angle is 180° − 60°.

Check Your Progress 15. A square grass field has side 14 metres. Two goats are tethered with a rope 14 metres long each at opposite corners. Find the common area grazed by both the goats.
Answer:
Each goat grazes a quadrant of radius 14 m. The area common to the two quadrants (which are centred at opposite corners) = 2 × (area of a quadrant − area of the triangle formed by the diagonal).
\( = 2 \times \left(\dfrac{1}{4} \times \dfrac{22}{7} \times 14^2 - \dfrac{1}{2} \times 14 \times 14\right) = 2 \times (154 - 98) = \) 112 m².
In simple words: The two quadrants overlap in a lens-shaped region made of two equal segments.

Teacher's Note:
1. Each segment is a quadrant minus a right triangle.
2. Double it.