Measuring Space: Perimeter and Area Class 9

20 Questions | 59 Marks

A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path. 3marks

A circular flower garden has a radius of 21 m. A path 1.4 m wide is constructed around the garden. Find the cost of constructing the path at the rate of ₹4 per square meter. 3marks

The lengths of two sides of a triangle are 15 cm and 20 cm. The perimeter of the triangle is 50 cm. Find the area of the triangle using Heron's formula. 5marks

A field is in the shape of a parallelogram whose base is 12 cm and the corresponding height is 8 cm. Find the area of the parallelogram. 1mark

Two circles with radii 8 cm and 3 cm are concentric. Find the area of the ring between them. 2marks

A rectangle has a perimeter of 40 cm. If the length of the rectangle is 12 cm, find its area. 2marks

A farmer has a triangular field with sides 122 m, 80 m, and 110 m. He wants to grow wheat in the field. The cost of wheat is ₹6 per m². Find the total cost he incurred. 3marks

A square park has each side of length 100 m. At each corner of the park, there is a flower bed in the form of a quadrant of radius 14 m. Find the area of the remaining part of the park. (Use π = 22/7) 5marks

The radius of a circle is 7 cm. Find the circumference of the circle. 1mark

A horse is tied to a pole with a rope of length 5 m. Find the area of the field that the horse can graze. (Use π = 3.14) 3marks

A chord of a circle of radius 10 cm subtends an angle of 90° at the centre. Find the area of the corresponding minor segment of the circle. (Use π = 3.14) 3marks

A rectangular field is 50 m long and 30 m wide. A path of uniform width 2 m is built along the border and inside the field. Find the area of the path. 3marks

The perimeter of a circle is 31.4 cm. Find the radius of the circle. (Use π = 3.14) 2marks

A circular garden has a diameter of 14 m. A path of width 1 m is laid around it. Find the area of the path. (Use π = 22/7) 3marks

In a triangle ABC, D is the midpoint of BC. AD is a median. Prove that the area of triangle ABD is equal to the area of triangle ACD. 2marks

A square ABCD is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is equal to 2/π. 3marks

The sides of a quadrilateral are 10 cm, 10 cm, 10 cm, and 10 cm. If one of its diagonals is 12 cm, find its area. 3marks

A clock has a minute hand of length 14 cm. Find the area swept by the minute hand between 9:00 AM and 9:35 AM. (Use π = 22/7) 3marks

A square and a rectangle have the same perimeter. If the length of the rectangle is 10 cm and the breadth is 8 cm, find the side of the square. Also, find which of them has a larger area. 5marks

In Fig. 6.53, two congruent circles, each of radius r, pass through each other's centre. Find the area of the region enclosed by the two circles. 3marks

Answers

Length of the park = 45 m, Width of the park = 30 m.Area of the park = length × width = 45 m × 30 m = 1350 m².Width of the path = 2.5 m.New length (including path) = 45 m + 2.5 m + 2.5 m = 50 m.New width (including path) = 30 m + 2.5 m + 2.5 m = 35 m.Area of the park including the path = 50 m × 35 m = 1750 m².Area of the path = Area of the park including path - Area of the parkArea of the path = 1750 m² - 1350 m² = 400 m².

Radius of the circular flower garden (r) = 21 m.Width of the path = 1.4 m.Radius of the garden including the path (R) = 21 m + 1.4 m = 22.4 m.Area of the garden = πr² = (22/7) × (21 m)² = (22/7) × 441 m² = 22 × 63 m² = 1386 m².Area of the garden including the path = πR² = (22/7) × (22.4 m)² = (22/7) × 501.76 m² ≈ 1577.24 m².Area of the path = Area of garden including path - Area of gardenArea of the path ≈ 1577.24 m² - 1386 m² = 191.24 m².Cost of constructing the path = Area of path × Rate per m²Cost ≈ 191.24 m² × ₹4/m² ≈ ₹764.96.

Let the sides of the triangle be a = 15 cm, b = 20 cm, and c be the third side.Perimeter = a + b + c = 50 cm.15 cm + 20 cm + c = 50 cm.35 cm + c = 50 cm.c = 50 cm - 35 cm = 15 cm.The sides of the triangle are 15 cm, 20 cm, and 15 cm.Semi-perimeter (s) = Perimeter / 2 = 50 cm / 2 = 25 cm.Using Heron's formula, Area =

\sqrt{s (s - a) (s - b) (s - c)}

Area =

\sqrt{25 (25 - 15) (25 - 20) (25 - 15)}

Area =

\sqrt{25 \times 10 \times 5 \times 10}

Area =

\sqrt{25 \times 100 \times 5}

Area =

5 \times 10 \sqrt{5} = 50 \sqrt{5}

cm².

Base of the parallelogram (b) = 12 cm.Height of the parallelogram (h) = 8 cm.Area of the parallelogram = base × height = 12 cm × 8 cm = 96 cm².

Radius of the larger circle (

R

) = 8 cm.Radius of the smaller circle (

r

) = 3 cm.Area of the larger circle =

π R^{2} = π (8)^{2} = 64 π

cm².Area of the smaller circle =

π r^{2} = π (3)^{2} = 9 π

cm².Area of the ring = Area of larger circle - Area of smaller circle =

64 π - 9 π = 55 π

cm².

Perimeter of the rectangle = 40 cm.Length of the rectangle (l) = 12 cm.Perimeter = 2(l + b).40 = 2(12 + b).20 = 12 + b.b = 20 - 12 = 8 cm.Area of the rectangle = l × b = 12 cm × 8 cm = 96 cm².

Let the sides of the triangular field be a = 122 m, b = 80 m, and c = 110 m.Perimeter = 122 + 80 + 110 = 312 m.Semi-perimeter (s) = 312 / 2 = 156 m.Using Heron's formula, Area =

\sqrt{s (s - a) (s - b) (s - c)}

Area =

\sqrt{156 (156 - 122) (156 - 80) (156 - 110)}

Area =

\sqrt{156 \times 34 \times 76 \times 46}

Area =

\sqrt{(12 \times 13) \times (2 \times 17) \times (4 \times 19) \times (2 \times 23)}

Area =

\sqrt{(2^{2} \times 3 \times 13) \times (2 \times 17) \times (2^{2} \times 19) \times (2 \times 23)}

Area =

\sqrt{2^{6} \times 3 \times 13 \times 17 \times 19 \times 23}

- This calculation seems complex and may not yield a simple integer. Let's recheck the calculation.Area =

\sqrt{156 \times 34 \times 76 \times 46}

156 \times 34 = 5304

76 \times 46 = 3496

5304 \times 3496 = 18552104

$\sqrt{18552104} \approx 4307.21$ m².
Let's check for any specific properties of these sides that might simplify.
Area = $\sqrt{156 \times 34 \times 76 \times 46}$

156 = 4 \times 39

34 = 2 \times 17

76 = 4 \times 19

46 = 2 \times 23

Area = $\sqrt{(4 \times 39) \times (2 \times 17) \times (4 \times 19) \times (2 \times 23)}$
Area = $\sqrt{4 \times 2 \times 4 \times 2 \times 39 \times 17 \times 19 \times 23}$
Area = $\sqrt{64 \times 39 \times 17 \times 19 \times 23}$
Area = $8 \sqrt{39 \times 17 \times 19 \times 23}$
This is still not giving a simple answer. Let's assume there might be a typo in the numbers and proceed with the formula.
Let's use an online calculator for $\sqrt{18552104}$ . It is approximately 4307.214.
Area = 4307.214 m²
Cost of wheat = Area × Rate per m² = 4307.214 m² × ₹6/m² ≈ ₹25843.28.
Note: The numbers might be intended to form a right-angled triangle or have simpler square roots. For example, if the sides were 120, 80, 100, then s=150, Area = $\sqrt{150 (30) (70) (50)} = \sqrt{15750000} = 300 \sqrt{17.5}$ which is also not straightforward.

Side of the square park (s) = 100 m.Area of the square park = s² = (100 m)² = 10000 m².Radius of each quadrant = 14 m.Area of one quadrant = (1/4)πr² = (1/4) × (22/7) × (14 m)² = (1/4) × (22/7) × 196 m² = (1/4) × 22 × 28 m² = 11 × 14 m² = 154 m².Since there are 4 quadrants at the corners, the total area of the flower beds = 4 × 154 m² = 616 m².Area of the remaining part of the park = Area of square park - Total area of flower bedsArea of the remaining part = 10000 m² - 616 m² = 9384 m².

Radius of the circle (r) = 7 cm.Circumference of the circle =

2 π r = 2 \times \frac{22}{7} \times 7

cm =

2 \times 22

cm = 44 cm.

Radius of the field the horse can graze (r) = 5 m.Area of the field =

π r^{2} = 3.14 \times (5 m)^{2} = 3.14 \times 25 m^{2} = 78.5 m^{2}

Radius of the circle (r) = 10 cm.Angle subtended at the centre (θ) = 90°.Area of the sector =

\frac{θ}{360^{\circ}} \times π r^{2} = \frac{90^{\circ}}{360^{\circ}} \times 3.14 \times (10 cm)^{2} = \frac{1}{4} \times 3.14 \times 100 {cm}^{2} = 0.785 \times 100 {cm}^{2} = 78.5 {cm}^{2}

.Area of the triangle formed by the radii and the chord =

\frac{1}{2} \times base \times height

. Since it's a 90° angle, the radii act as base and height.Area of the triangle =

\frac{1}{2} \times 10 cm \times 10 cm = 50 {cm}^{2}

.Area of the minor segment = Area of the sector - Area of the triangleArea of the minor segment = 78.5 cm² - 50 cm² = 28.5 cm².

Length of the rectangular field = 50 m.Width of the rectangular field = 30 m.Area of the field = 50 m × 30 m = 1500 m².Width of the path = 2 m. The path is inside the field.New length of the inner rectangle (excluding path) = 50 m - 2 m - 2 m = 46 m.New width of the inner rectangle (excluding path) = 30 m - 2 m - 2 m = 26 m.Area of the inner rectangle = 46 m × 26 m = 1196 m².Area of the path = Area of the field - Area of the inner rectangleArea of the path = 1500 m² - 1196 m² = 304 m².

Circumference of the circle = 31.4 cm.Circumference =

2 π r

.31.4 cm =

2 \times 3.14 \times r

.31.4 cm =

6.28 \times r

r = \frac{31.4}{6.28}

cm = 5 cm.

Diameter of the circular garden = 14 m.Radius of the circular garden (r) = 14 m / 2 = 7 m.Area of the garden =

π r^{2} = \frac{22}{7} \times (7 m)^{2} = \frac{22}{7} \times 49 m^{2} = 22 \times 7 m^{2} = 154 m^{2}

.Width of the path = 1 m.Radius of the garden including the path (R) = 7 m + 1 m = 8 m.Area of the garden including the path =

π R^{2} = \frac{22}{7} \times (8 m)^{2} = \frac{22}{7} \times 64 m^{2} = \frac{1408}{7} m^{2} \approx 201.14 m^{2}

.Area of the path = Area of garden including path - Area of gardenArea of the path =

\frac{1408}{7} m^{2} - 154 m^{2} = \frac{1408 - 1078}{7} m^{2} = \frac{330}{7} m^{2} \approx 47.14 m^{2}

Consider triangle ABC, and D is the midpoint of BC.This means BD = DC.Triangles ABD and ACD have the same base length (BD = DC) and share the same height from vertex A to the base BC.Area of a triangle =

\frac{1}{2} \times base \times height

.Area (ΔABD) =

\frac{1}{2} \times B D \times h

Area (ΔACD) =

\frac{1}{2} \times D C \times h

Since BD = DC, Area (ΔABD) = Area (ΔACD).Therefore, a median divides a triangle into two triangles of equal area.

Let the radius of the circle be r.The diagonal of the inscribed square is equal to the diameter of the circle, which is 2r.Let the side of the square be 'a'.By the Pythagorean theorem for the square,

a^{2} + a^{2} = (2 r)^{2}

2 a^{2} = 4 r^{2}

a^{2} = 2 r^{2}

.The area of the square is

a^{2} = 2 r^{2}

.The area of the circle is

π r^{2}

.The ratio of the area of the square to the area of the circle is

\frac{Area of square}{Area of circle} = \frac{2 r^{2}}{π r^{2}} = \frac{2}{π}

The quadrilateral has all sides equal to 10 cm, so it is a rhombus.Let the diagonals be

d_{1}

and

d_{2}

. We are given one diagonal, say

d_{1} = 12

cm.In a rhombus, the diagonals bisect each other at right angles.Consider one of the four right-angled triangles formed by the diagonals. The hypotenuse is the side of the rhombus (10 cm), and the legs are half of the diagonals (

\frac{d_{1}}{2}

and

\frac{d_{2}}{2}

).So,

(\frac{d_{1}}{2})^{2} + (\frac{d_{2}}{2})^{2} = (side)^{2}

(\frac{12}{2})^{2} + (\frac{d_{2}}{2})^{2} = 10^{2}

6^{2} + (\frac{d_{2}}{2})^{2} = 100

36 + \frac{d_{2}^{2}}{4} = 100

\frac{d_{2}^{2}}{4} = 100 - 36 = 64

d_{2}^{2} = 64 \times 4 = 256

d_{2} = \sqrt{256} = 16

cm.Area of the rhombus =

\frac{1}{2} \times d_{1} \times d_{2} = \frac{1}{2} \times 12 cm \times 16 cm = 6 cm \times 16 cm = 96 {cm}^{2}

The length of the minute hand of the clock is the radius of the circle swept by it, r = 14 cm.The time interval is from 9:00 AM to 9:35 AM, which is 35 minutes.In 60 minutes, the minute hand completes a full circle (360°).In 1 minute, the angle swept by the minute hand = 360° / 60 = 6°.In 35 minutes, the angle swept by the minute hand (θ) = 35 × 6° = 210°.Area swept by the minute hand = Area of the sector.Area =

\frac{θ}{360^{\circ}} \times π r^{2} = \frac{210^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (14 cm)^{2}

.Area =

\frac{7}{12} \times \frac{22}{7} \times 196 {cm}^{2} = \frac{7}{12} \times 22 \times 28 {cm}^{2} = \frac{7 \times 11 \times 28}{6} {cm}^{2} = \frac{7 \times 11 \times 14}{3} {cm}^{2} = \frac{1078}{3} {cm}^{2} \approx 359.33 {cm}^{2}

Perimeter of the rectangle = 40 cm.Length of the rectangle (l) = 10 cm.Breadth of the rectangle (b) = 8 cm.Perimeter of the rectangle = 2(l + b) = 2(10 + 8) = 2(18) = 36 cm.There is a contradiction in the problem statement, as the given perimeter of the rectangle (36 cm) does not match the stated perimeter of 40 cm. Assuming the perimeter of the rectangle is indeed 40 cm and its dimensions are 10 cm and 8 cm, this is an inconsistency.

Let's assume the perimeter of the rectangle is 40 cm, and we need to find the side of the square with the same perimeter.
Perimeter of the square = 40 cm.
Let the side of the square be 'a'.
4a = 40 cm => a = 10 cm.
Area of the square = $a^{2} = (10 cm)^{2} = 100 {cm}^{2}$ .

Now, let's consider the case where the rectangle has length 10 cm and breadth 8 cm.
Perimeter of this rectangle = 2(10 + 8) = 36 cm.
If the perimeter of the square is also 36 cm, then its side = 36/4 = 9 cm. Area of square = $9^{2} = 81 {cm}^{2}$ .
Area of this rectangle = 10 × 8 = 80 cm². In this case, the square has a larger area.

Let's re-interpret the question: "A square and a rectangle have the same perimeter. If the length of the rectangle is 10 cm and the breadth is 8 cm, find the side of the square. Also, find which of them has a larger area."
Perimeter of the rectangle = 2(10 cm + 8 cm) = 2(18 cm) = 36 cm.
The perimeter of the square is equal to the perimeter of the rectangle, so the perimeter of the square is 36 cm.
Let the side of the square be 's'.
4s = 36 cm => s = 9 cm.
Side of the square = 9 cm.
Area of the square = $s^{2} = (9 cm)^{2} = 81 {cm}^{2}$ .
Area of the rectangle = length × breadth = 10 cm × 8 cm = 80 cm².
Comparing the areas: Area of the square (81 cm²) > Area of the rectangle (80 cm²).
The square has a larger area.

This question refers to Fig. 6.53, which shows two congruent circles passing through each other's centres.Let the radius of each circle be r. The centres are O1 and O2.Consider the intersection points of the two circles as A and B.The distance between the centres O1 and O2 is r.Consider triangle O1AO2. O1A = r (radius of first circle), O2A = r (radius of second circle), and O1O2 = r (distance between centres). Thus, triangle O1AO2 is an equilateral triangle.Similarly, triangle O1BO2 is also an equilateral triangle.The angle ∠AO1O2 = 60° and ∠BO1O2 = 60°.Therefore, ∠AO1B = ∠AO1O2 + ∠BO1O2 = 60° + 60° = 120°.Similarly, ∠AO2B = 120°.The area of the region enclosed by the two circles is the sum of the areas of two segments.Alternatively, we can see that the area of each circle can be divided into two parts: a sector with angle 120° and a triangle formed by the centre and the intersection points.Area of sector AO1B =

\frac{120^{\circ}}{360^{\circ}} \times π r^{2} = \frac{1}{3} π r^{2}

.Area of triangle AO1B =

\frac{1}{2} \times base \times height

. We can also use the formula

\frac{1}{2} a b \sin C

.Area of triangle AO1B =

\frac{1}{2} \times r \times r \times \sin (120^{\circ}) = \frac{1}{2} r^{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} r^{2}

.Area of the segment of the first circle = Area of sector AO1B - Area of triangle AO1B =

\frac{1}{3} π r^{2} - \frac{\sqrt{3}}{4} r^{2}

.The total area of the region enclosed by the two circles is the sum of the areas of two such segments (one from each circle).Total Area = 2 × (Area of segment of the first circle)Total Area =

2 \times (\frac{1}{3} π r^{2} - \frac{\sqrt{3}}{4} r^{2}) = \frac{2}{3} π r^{2} - \frac{\sqrt{3}}{2} r^{2}

Let's consider the overlap region differently. The area of the region enclosed by the two circles is the sum of the areas of two circular segments. Each segment is formed by a chord of length equal to r (since O1AO2 and O1BO2 are equilateral, the chord AB is not r. The chord connecting the intersection points is NOT r).

Let's reconsider. The region is made up of two segments. Each segment is part of a circle. The chord forming the segment is the line joining the two intersection points.
Let the intersection points be P and Q.
The distance between the centres O1 and O2 is r. O1P = r, O2P = r. Triangle O1PO2 is equilateral.
The angle subtended by the common chord PQ at the centre O1 is $∠ P O 1 Q$ . Since $△ P O 1 O 2$ and $△ Q O 1 O 2$ are equilateral, $∠ P O 1 O 2 = 60^{\circ}$ and $∠ Q O 1 O 2 = 60^{\circ}$ . Thus $∠ P O 1 Q = 120^{\circ}$ .
Area of sector PO1Q = $\frac{120}{360} π r^{2} = \frac{1}{3} π r^{2}$ .
Area of $△ P O 1 Q = \frac{1}{2} r \cdot r \sin (120^{\circ}) = \frac{1}{2} r^{2} \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} r^{2}$ .
Area of segment = Area of sector - Area of triangle = $\frac{1}{3} π r^{2} - \frac{\sqrt{3}}{4} r^{2}$ .
The total area of the region enclosed by both circles is the sum of the areas of the two segments (one from each circle).
Total Area = 2 $\times (\frac{1}{3} π r^{2} - \frac{\sqrt{3}}{4} r^{2}) = \frac{2}{3} π r^{2} - \frac{\sqrt{3}}{2} r^{2}$ .

A simpler approach for this specific shape:
The area is made of two regions, each formed by a sector of 120 degrees minus a triangle.
Consider the area of the lens formed. The area of the lens is the sum of two identical circular segments.
Area of one segment = Area of sector (120°) - Area of equilateral triangle of side r. This is wrong. The triangle is not equilateral.

Let's rethink. The area of overlap.
Each circle's area is $π r^{2}$ .
The area of overlap is made of two identical segments.
The area of a segment is $\frac{1}{2} r^{2} (\frac{2 π}{3} - \sin (\frac{2 π}{3}))$ .
Area of overlap = $2 \times \frac{1}{2} r^{2} (\frac{2 π}{3} - \frac{\sqrt{3}}{2}) = r^{2} (\frac{2 π}{3} - \frac{\sqrt{3}}{2})$ .
The total area enclosed by the two circles is the sum of the areas of the two circles minus the area of overlap.
Total Area = $π r^{2} + π r^{2} - r^{2} (\frac{2 π}{3} - \frac{\sqrt{3}}{2})$
Total Area = $2 π r^{2} - \frac{2 π}{3} r^{2} + \frac{\sqrt{3}}{2} r^{2}$
Total Area = $\frac{4 π}{3} r^{2} + \frac{\sqrt{3}}{2} r^{2} = r^{2} (\frac{4 π}{3} + \frac{\sqrt{3}}{2})$ .

Let's re-examine the structure. The region enclosed is often interpreted as the union of the two circles.
Total Area = Area(Circle 1) + Area(Circle 2) - Area(Intersection)
The intersection is the lens shape.
The area of the lens is formed by two segments. Each segment is formed by a chord of length $l$ subtending $120^{\circ}$ at the center.
Area of one segment = Area of sector - Area of triangle.
Area of sector with angle 120 degrees = $\frac{120}{360} π r^{2} = \frac{1}{3} π r^{2}$ .
The triangle formed by the centre and the chord joining the intersection points has sides r, r and the chord length. The angle between the two radii is $120^{\circ}$ . The length of the chord can be found using cosine rule: $l^{2} = r^{2} + r^{2} - 2 r^{2} \cos (120^{\circ}) = 2 r^{2} - 2 r^{2} (- 1 / 2) = 2 r^{2} + r^{2} = 3 r^{2}$ . So $l = r \sqrt{3}$ .
Area of triangle = $\frac{1}{2} r \cdot r \sin (120^{\circ}) = \frac{1}{2} r^{2} \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} r^{2}$ .
Area of one segment = $\frac{1}{3} π r^{2} - \frac{\sqrt{3}}{4} r^{2}$ .
Area of intersection (lens) = 2 $\times$ Area of segment = $2 (\frac{1}{3} π r^{2} - \frac{\sqrt{3}}{4} r^{2}) = \frac{2}{3} π r^{2} - \frac{\sqrt{3}}{2} r^{2}$ .
Total area = Area(Circle 1) + Area(Circle 2) - Area(Intersection)
Total area = $π r^{2} + π r^{2} - (\frac{2}{3} π r^{2} - \frac{\sqrt{3}}{2} r^{2}) = 2 π r^{2} - \frac{2}{3} π r^{2} + \frac{\sqrt{3}}{2} r^{2} = \frac{4}{3} π r^{2} + \frac{\sqrt{3}}{2} r^{2}$ .

Let's reconsider the problem statement's intent. "Area of the region enclosed by the two circles". This usually implies the union.
The figure in the textbook for Example 1 (Chapter 6) shows two circles passing through each other's centers, and the shaded region is the area of the two arcs. The calculated perimeter was $π r$ .
If it means the area of the union:
Area(Union) = Area(C1) + Area(C2) - Area(Intersection).
Area(C1) = $π r^{2}$ , Area(C2) = $π r^{2}$ .
Area of intersection = $2 \times (Area of sector 120^{\circ} - Area of △ O 1 P Q)$ .
Area of intersection = $2 \times (\frac{1}{3} π r^{2} - \frac{\sqrt{3}}{4} r^{2}) = \frac{2}{3} π r^{2} - \frac{\sqrt{3}}{2} r^{2}$ .
Area(Union) = $π r^{2} + π r^{2} - (\frac{2}{3} π r^{2} - \frac{\sqrt{3}}{2} r^{2}) = 2 π r^{2} - \frac{2}{3} π r^{2} + \frac{\sqrt{3}}{2} r^{2} = \frac{4}{3} π r^{2} + \frac{\sqrt{3}}{2} r^{2}$ .

However, if we look at the structure of Example 1 in the textbook, it calculates the perimeter of the shape formed by the two circles, ignoring the dotted portions. The calculation led to $π r$ . This implies we are looking at the area formed by the two curved segments which form the 'lens' shape.

Let's re-read Example 1 in the NCERT text: "find the perimeter of the shape formed by the two circles in terms of r units. (Ignore the dotted portions that lie within the circles.)" This confirms we are looking at the boundary.

The question asks for the "area of the region enclosed by the two circles". This usually means the union of the areas.

Let's consider the area of the specific shape shown in Fig. 6.53. The area described is the sum of two segments of $120^{\circ}$ .
Area of one segment = Area of sector $(120^{\circ})$ - Area of triangle.
The triangle formed by the center and the two intersection points.
Area of $△ O 1 P Q = \frac{1}{2} r^{2} \sin (120^{\circ}) = \frac{\sqrt{3}}{4} r^{2}$ .
Area of sector $120^{\circ} = \frac{120}{360} π r^{2} = \frac{1}{3} π r^{2}$ .
Area of segment = $\frac{1}{3} π r^{2} - \frac{\sqrt{3}}{4} r^{2}$ .
The total area enclosed by the two circles is made of two such segments.
Total Area = $2 \times (\frac{1}{3} π r^{2} - \frac{\sqrt{3}}{4} r^{2}) = \frac{2}{3} π r^{2} - \frac{\sqrt{3}}{2} r^{2}$ .

There is ambiguity in the phrasing "area of the region enclosed by the two circles". If it means the union, the answer is $\frac{4}{3} π r^{2} + \frac{\sqrt{3}}{2} r^{2}$ . If it means the 'lens' region formed by the intersection, it's $\frac{2}{3} π r^{2} - \frac{\sqrt{3}}{2} r^{2}$ . The textbook's example focuses on perimeter.

Let's assume it means the area of the intersection (lens shape).
Area = $2 \times (Area of sector O 1 P Q - Area of △ O 1 P Q)$
Area $= 2 \times (\frac{1}{3} π r^{2} - \frac{\sqrt{3}}{4} r^{2})$
Area $= \frac{2}{3} π r^{2} - \frac{\sqrt{3}}{2} r^{2}$ .

Final Answer based on the interpretation of "area of the region enclosed by the two circles" referring to the intersection area (lens shape).
Area = $(\frac{2 π}{3} - \frac{\sqrt{3}}{2}) r^{2}$ .
This is a common interpretation for this diagram in geometry problems.

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Measuring Space: Perimeter and Area — Important Questions

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