Measuring Space: Perimeter and Area Class 9 (Ganita Manjari)

The perimeter of a circle is 44 cm. What is its radius?
Now,
Perimeter of circle = Circumference of circle
44 = 𝟐𝝅𝒓
44 = 2 ×22/(7\ ) × 𝑟
44 = 44/(7\ ) × 𝑟
44 ×7/44= 𝑟
7=𝑟
𝒓=𝟕

(i) Calculate, correct to 3 significant figures, the circumference of a circle with:
(i) radius 7 cm.
Radius = r = 7 cm

(i) Calculate the length of the arc of a circle if:
(i) the radius is 3.5 cm and the angle at the centre is 60°
Given
Radius = r = 3.5 cm
Angle = 𝜃 = 60°

Find the perimeter of a sector (i.e., the curved portion as well as the
two straight portions) of a circle of radius 14 cm and sector angle 75°.
Given
Radius = r = 14 cm
Angle = 𝜃 = 75°

Find the perimeters of the following shapes (taking the arcs to
be quarter or half or three-quarters of a circle, as appropriate)
Let’s label all sides and radius

Find the perimeters of the following shapes (taking the arcs to
be quarter or half or three-quarters of a circle, as appropriate)
Let’s label all sides and radius

Find the perimeters of the following shapes (taking the arcs to
be quarter or half or three-quarters of a circle, as appropriate)
Here,
Perimeter of given shape
= 4 × Length of curved blue line
= 4 × Circumference of half circle

Find the perimeters of the following shapes (taking the arcs to
be quarter or half or three-quarters of a circle, as appropriate)
Here, we have an equilateral triangle of side 12 cm
On each side, there is a semicircle of Diameter = Side = 12 cm

Find the perimeters of the following shapes (taking the arcs to
be quarter or half or three-quarters of a circle, as appropriate)
Quarter circle with radius = 14 cm
Our figure has
Quarter circles
Semi circles

Find the perimeters of the following shapes (taking the arcs to
be quarter or half or three-quarters of a circle, as appropriate)
Here, 28 cm is divided into 4 equal parts
∴ Length of each part = 28/4
= 7 cm

Find the perimeters of the following shapes (taking the arcs to
be quarter or half or three-quarters of a circle, as appropriate)
Here, we have 3 semicircles of different Diameters
Let’s find the Hypotenuse of the Right angled triangle

Find the perimeters of the following shapes (taking the arcs to
be quarter or half or three-quarters of a circle, as appropriate)
Here, we have
1 big semicircle with Diameter = 4 + 4 + 4 = 12 cm
3 small semicircle with Diameter = 4 cm

Find the perimeters of the following shapes (taking the arcs to
be quarter or half or three-quarters of a circle, as appropriate)
Here, we have
1 big semicircle with Diameter = 10 + 10 = 20 cm
2 small semicircle with Diameter = 10 cm

If the diameter of a car tyre is 56 cm, then:
(i) How far does the car need to travel for the tyre to complete one revolution?
(ii) How many revolutions does the tyre make if the car travels 10 km?

Find the total perimeter of all the petals in each of the given flowers.
We can separate this into 4 semicircles of Diameter 14 cm
Now,
Perimeter of shape
= 4 × Circumference of semicircle with Diameter 14 cm
= 4 × Circumference of semicircle with Radius 7 cm
= 𝟒 × 𝟏/𝟐 × 𝟐 × 𝝅 × 𝟕
= 4 × 𝜋 × 7
= 𝟐𝟖 × 𝝅
= 28 ×22/7
= 4 × 22
= 88 cm

Find the total perimeter of all the petals in each of the given flowers.
Let’s look at 1 arc
We observe that in 1 arc,
we get form an equilateral triangle of side 42 cm

The ratio of the perimeters of two circles is 5:4. What is the ratio of
their radii?
Let the radius of two circles be 𝒓_𝟏 and 𝒓_𝟐

Find the area of triangle ADE in Fig. 6.31.
We can use formula
Area of Triangle = 𝟏/𝟐 × Base × Height

The parallel sides of a trapezium are 40 cm and 20 cm . If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.
Let’s draw the figure

Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.
Let sides be a, b, c
∴ a = 8 cm, b = 11 cm, c = ?

The sides of a triangular plot are in the ratio 3: 5: 7; its perimeter is 300 m. Find its area.
Since sides are in ratio 3: 5: 7
We can assume sides are
3x, 5x, 7x

One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm2, find the length of the shorter diagonal.
Given that
One diagonal of a rhombus is twice as long as the other diagonal
Let First diagonal = d cm
Thus,
Second Diagonal = 2d cm

ABCD is a parallelogram. P and Q are any two points on side AB . What can you say about the ratio area (△PCD ): area (△QCD )?
Let’s draw the diagram
Here,
∆ PCD & ∆ QCD have same base and lie between same parallel lines

O is any point on the diagonal PR of a parallelogram PQRS. Prove
that the areas of triangles PSO and PQO are equal.
Let’s draw the figure

If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)
Given: Let ABCD is a quadrilateral
P, Q, R and S are mid-points of the sides
AB, BC, CD and DA respectively

In △ABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on 𝐴𝐷. Show that area (△𝐴𝐵𝑃)=area(△𝐴𝐶𝑃).
We know that
Median divides a triangle into two equal area

Given a square ABCD , let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region (△PAB and △PCD ) and the green region (△PBC and △PDA )?
Let side of square be a

In △ABC,D is the midpoint of AB . P is any point on BC, and Q is a point on AB such that CQ‖PD.PQ is joined (Fig. 6.34). Prove that Area(△BPQ)=1/2 Area(△ABC).
Since D is mid-point of AB
CD is the median

Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60^∘.
Given that,
Radius = r = 7 cm
& Angle of the sector = θ = 60°

Find the area of a quadrant of a circle whose circumference is 44 cm.
Quadrant is ¼th of a circle
So, we can write
Area of quadrant = 1/4× area of circle
= 𝟏/𝟒 𝝅𝒓^𝟐

The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.
We need to find Area swept by the minute hand in 10 minutes.
That will be the area of sector

(i) A chord of a circle of radius 10 cm subtends 90^∘ at the centre. Find the area of the corresponding:
(i) minor sector (that subtends 90^∘ at the centre). (Use 𝜋≈3.14.)
Let the minor sector be OAPB
Given that
Radius = r = 10 cm
𝛉=𝟗𝟎°

A chord of a circle of radius 15 cm subtends an angle of 60^∘ at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use 𝜋≈3.14 and √3≈1.73.)

A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120^∘. Find the total area cleaned at each sweep of the blades.
Given that
Radius = r = 29 cm
Sweeping angle = 𝜃 = 120°

A chord of a circle of radius 𝑟 subtends an angle of 60^∘ at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to 𝜋𝑟^2 (1/6−√3/4).
Let the minor segment be APB

An equilateral triangle is inscribed in a circle of radius 𝑟. Show that the ratio of the area of the triangle to the area of the circle is equal to (3√3)/4𝜋≈0.413.
Let’s look at it step-by-step
Step 1 of 8
The Circle's Area
We start with a circle of radius r
. The area of the entire circle is simply 𝜋r^2.
Step 2 of 8
Inscribe the Triangle
We draw an equilateral triangle inside the circle so that its corners touch the edge perfectly.
Step 3 of 8
Split from the Center
By drawing lines from the center (radii) to the three corners, we split the equilateral triangle into 3 identical isosceles triangles.Step 4 of 8
Drop a Height
Let's focus on one of these slices. It has an angle of 120^∘ at the center. Dropping a straight line down splits it into two 30^∘-60^∘-90^∘ right triangles.
Step 5 of 8
Find Base and Height
In a 30-60-90 triangle with hypotenuse r, the short side (height) is r/2, and the long side is (r√3)/2. So the full base of our slice is r√3.
Step 6 of 8
Area of One Piece
Area =1/2× base × height
=1/2×(r√3)×(r/2)=√3/4 r^2.
Step 7 of 8
Total Triangle Area
Since there are 3 identical pieces, the total area of the equilateral triangle is 3 ×√3/4 r^2=(3√3)/4 r^2.
Step 8 of 8
Calculate the Ratio
To find the final ratio, we divide the Triangle Area by the Circle Area. The r^2 terms cancel out perfectly!
"Ratio "=(3√3/4)/π=(3√3)/4π≈0.413

A square 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≈0.637.
Let’s look at it step-by-step
Step 1 of 7
The Circle's Area
Once again, we have a circle of radius r. Area = 𝜋r^2.
Step 2 of 7
Inscribe the Square
We draw a square inside the circle. Its corners touch the circle's edge.
Step 3 of 7
Split with Diagonals
By drawing the two diagonals of the square (which are also diameters of the circle), we split the square into 4 identical triangles.Step 4 of 7
The 〖90〗^∘ Angle
Because a square's diagonals intersect at exactly 90^∘, each of the 4 pieces is a perfect right-angled triangle. Its two short sides are simply the radii of the circle.
Step 5 of 7
Area of One Piece
Look at one triangle. Its base is r and its height is r.
" Area "=1/2×" base × height "=1/2 r^2
Step 6 of 7
Total Square Area
Since there are 4 identical pieces, the total area of the square is 4×1/2 r^2=2r^2.
Step 7 of 7
Calculate the Ratio
Divide the Square Area by the Circle Area. The r^2 terms cancel out.
Ratio =(2r^2)/(πr^2 )=2/π≈0.637

A hexagon is inscribed in a circle of radius 𝑟. Show that the ratio of the area of the hexagon to the area of the circle is equal to (3√3)/2𝜋≈0.827. Can you see why the answer is exactly twice the answer to Question 8?
Let’s look at it step-by-step
Step 1 of 9
The Circle's Area
Circle of radius r
. Area = 𝜋r^2.Step 2 of 9
Inscribe the Hexagon
We draw a regular hexagon inside the circle.
Step 3 of 9
Split from the Center
By drawing radii to all 6 vertices, we split the hexagon into 6 identical triangles. Because 360^∘/6=60^∘, these are all perfect equilateral triangles with side length r!
Step 4 of 9
Drop a Height
Let's focus on one equilateral triangle at the bottom. We drop a perpendicular line down the middle to find its height. This splits its base r perfectly in half
Step 5 of 9
Find the Height
Using the Pythagorean theorem (or 30-60-90 rules), if the hypotenuse is r and the base is r/2, the height is
(r√3)/2Step 6 of 9
Area of One Piece
Area " =1/2× base × height =1/2×r×((r√3)/2)=√3/4 r^2.)
Step 7 of 9
Total Hexagon Area
Since there are 6 identical pieces, the total area is 6×√3/4 r^2=(6√3)/4 r^2=(3√3)/2 r^2.
Step 8 of 9
Calculate the Ratio
Divide the Hexagon Area by the Circle Area. The r^2 terms cancel out.
Ratio =(3√3/2)/π=(3√3)/2π≈0.827
Step 9 of 9
Why exactly TWICE Question 8?
Look closely: The hexagon is made of 6 identical equilateral triangles. The triangle from Question 8 (highlighted) perfectly covers exactly 3 of them! Since 6 is exactly double 3 , the hexagon's area is exactly double the triangle's area.

Identities in algebra can sometimes be shown as area relationships. For example:
The figure shown corresponds to the identity

An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.
Let sides be a, b, c
∴ a = 15 cm, b = ?, c = 15 cm

An isosceles triangle has base 10 cm , and its area is 60〖" " cm〗^2. What are the lengths of the equal sides?
Since two sides are equal
Let the equal sides be a
So,
a = a, b = 10, c = a

The area of a right-angled triangle is 54 sq.cm. One of its legs has length 12 cm. Find its perimeter.
Let the other two sides of right triangle be a & c

The sides of a triangle are in the ratio 2:3:4, and its perimeter is 45 cm. Find its area.
Since sides are in ratio 2: 3: 4
We can assume sides are
2x, 3x, 4x

The sides of a triangle have lengths 7" " cm,24" " cm,25" " cm. Find the area of the triangle in two different ways.
Given our sides 7 cm, 24 cm, 25 cm
This follows Pythagoras theorem
〖𝟐𝟓〗^𝟐=𝟕^𝟐+〖𝟐𝟒〗^𝟐

If the wheel of a bicycle has a diameter of 60 cm , find how far a cyclist will have travelled after the wheel has rotated 100 times.
In 1 rotation, wheel travels distance = Circumference

Find the area of a quadrant of a circle whose circumference is 66 cm.
Quadrant is ¼th of a circle
So, we can write
Area of quadrant = 1/4× area of circle
= 𝟏/𝟒 𝝅𝒓^𝟐

The wheel of a car has an outer radius of 28 cm. Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of 1 km.
Circumference of wheel
In 1 rotation, wheel travels distance = Circumference

Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?
Let’s look at it step-by-step
Step 1 of 6
What does 'Congruent' mean?
In geometry, two shapes are congruent if they have the exact same size and shape. You can slide, reflect, or rotate one to fit perfectly on top of the other.
Look at the animation: Shape B is congruent to Shape A because it can be moved to perfectly overlap it!tep 2 of 6
The Algebraic Setup
Imagine Rectangle 1 has length
and width
, and Rectangle 2 has length
and width
.
We know they have the exact same area and perimeter. Let's see what this means mathematically.
Step 3 of 6
Condition 1: Same Perimeter
Perimeter is
Length + Width
.
Since the perimeters are the same, the sum of their length and width must be exactly the same:Step 4 of 6
Condition 2: Same Area
Area is Length × Width.
Since the areas are the same, the product of their length and width must be exactly the same:
Step 5 of 6
The Number Pair Rule
In mathematics, if you have two numbers and you know their exact sum and their exact product, there is only ONE possible pair of numbers that can exist.
For example, if two numbers add to 7 and multiply to 12 , those numbers MUST be 3 and 4 .Step 6 of 6
The Verdict
Therefore, both rectangles must share the exact same dimensions.
Even if Rectangle 2 is rotated (Length becomes Width), the physical measurements are identical, meaning they are perfectly congruent.

You know that the area of a parallelogram is base × height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides × height, i.e., 1/2(𝑎+𝑏)ℎ.
We can divide the given trapezium into
Parallelogram
Triangle

By dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height (the same formula as the one given above).
Let’s draw the figure

Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?
Let’s draw figure
So, by joining identical trapeziums ABCD and MNOP we get Parallelogram APMD

(i) Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra.
A kite has two diagonals intersecting at right angles
Let AC = d1
DB = d2

(i)
Three problems about fitting congruent shapes together:
(i) Rectangle ABCD has sides 𝑎,𝑏, and rectangle PQRS has sides 2𝑎,2𝑏. Show that PQRS has 4 times the area of ABCD . Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see
Now,
Area of ABCD =𝑎 × 𝑏=𝒂𝒃
Area of PQRS =2𝑎 × 2𝑏 =𝟒𝒂𝒃

– Part 1
What fraction of the triangle is shaded?

– Part 1
What fraction of the rectangle is covered by the circles?

Use the above to make a conjecture about the area occupied by circles fitted into a rectangle in the manner shown. Test your conjecture for particular cases: 10 circles; 20 circles; 50 circles. Then prove your conjecture!
Let’s consider n circles inside a rectangle

The figure shows nine identical rectangles fitted together to make a large rectangle whose area is 72 cm^2. Find the perimeter of each small rectangle.
Let the Longer side of smaller rectangle be a
And Shorter side of smaller rectangle be b
Labelling our figure

Show that the areas of the shaded blue triangle and the shaded red triangle are equal.
Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.
Since base of big triangle is trisected
Length of base in grey, white and red triangle is the same
Let’s first show that Areas are equal
tep 1 of 5
The Problem
We have a large triangle. Lines are drawn from the top vertex to the points of trisection on the opposite side.
Show that the Blue and Red triangles have equal area.
Step 2 of 5
Base and Height
Because they are trisection points, the base is cut into 3 exactly equal pieces. Let's call each base
.
Notice that both the Blue and Red triangles reach up to the exact same top vertex. This means they share the exact same height (h).Step 3 of 5
Equal Area Proof
The area of any triangle is 1/2× base × height.
Blue Area =1/2×h
Red Area =1/2×h
Mathematically, their areas are identically equal!
Now, we need to answer
Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.
Because they have the exact same base and height, geometry guarantees we can cut one up to build the other. Here is a simple way to do it:
Cut the blue triangle horizontally exactly halfway up its height.
Cut the top small blue triangle straight down the middle.
Because the red triangle leans further to the right, you can take those two top pieces from the blue triangle and slide them down the slanted edge of the bottom piece to rebuild the rightward-leaning red triangle!

The figure shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions A and B. Show that A and B have equal area.
Let’s look at it step by step
Step 1 of 5
The Problem
The figure shows a quarter circle in a square. Its centre is at the bottom-left vertex. There are two semicircles on the adjacent sides (left and bottom) as diameters.
We need to prove that the area of Region A equals Region B. Let the square's side length be 2r.
Step 2 of 5
Area of the Quarter Circle
The large Quarter Circle has a radius of 2 r .
Area " =1/4×π(2r)^2 Area " =1/4×4πr^2=πr^2)
Step 3 of 5
Area of the Semicircles
Now look at the two smaller
Semicircles. Each has a diameter of 2r, meaning their radius is just r.
Combined Area =2×(1/2 πr^2 )=πr^2
Step 4 of 5
The Overlap Principle
Wait! The Quarter Circle Area ( πr^2 ) is exactly equal to the Sum of the Semicircles ( πr^2 ).
The semicircles cover almost the exact same space, EXCEPT they doublecount Region A (overlap), and completely miss Region B!
tep 5 of 5
Balancing the Equation
If two sets of shapes have the exact same total area, and occupy the same boundary:
The area they Double-Count MUST be perfectly equal to the Area they Miss!
Therefore, Area
.

In Fig. 6.50, four semicircles have been drawn within the given square whose side is 2 units. The centres of these semicircles are the midpoints of the sides. They create a 4-petalled flower (shown in blue). Find the perimeter and the area of this flower.
Let’s do it step by step
Step 1 of 6
The Setup
We have a square with a side length of 2 units. Inside, four semicircles are drawn using each side as a diameter.
This creates a beautiful solid blue 4petalled flower in the center.
Step 2 of 6
Finding the Perimeter
Let's find the perimeter of the flower. Look at the top edge: it forms exactly one semicircle.
The length of a semicircular arc is πr. Since Side =2,r=1, so this arc is π(1)= π.
There are 4 identical petals, so the total boundary is 4×π=4π
Step 3 of 6
Area: One Semicircle
Now let's find the area. We look at just one complete semicircle (the top one).
Area of a semicircle is 1/2 πr^2. Area =1/2 π(1)^2=π/2.
If we sum all 4 semicircles: 4×(π/2)=2π.
Step 4 of 6
Covering the Square
Notice that these 4 semicircles completely cover the entire pink square.
The Area of the Square =2×2=4.
Wait! Our pieces add up to 2π(≈6.28), which is bigger than 4 !
Step 5 of 6
The Double Count
Why is it bigger? Because the Semicircles overlap to form the solid blue petals!
Because their colors multiply, you can visibly see exactly which parts are counted twice (the darker petals). The math rule says:
Sum of Pieces = Square Area + Overlap Area.
Step 6 of 6
The Final Answer
Using simple algebra:
Area of Flower = Sum of Semicircles Area of Square
Area
.

In Fig. 6.51 we see two concentric circles with a common centre O. A chord BC of the larger circle is drawn, touching the smaller circle at A. The length of BC is 𝑙. Show that the area of the green region enclosed between the two circles is 1/4 𝜋𝑙^2.
Let’s answer this step by step
Problem 23: Concentric Circles
Find the area of the green region enclosed between two concentric circles.
1 Here we have two concentric circles (circles with the common center O ). The green ring is the area we want to find.
Problem 23: Concentric Circles
Find the area of the green region enclosed between two concentric circles.
2 Let the radius of the large circle be
and the small circle be
. The area of the green ring is Area
Problem 23: Concentric Circles
Find the area of the green region enclosed between two concentric circles.
3 A chord BC of length
is drawn on the large circle, touching the small circle at point A. This means
is a tangent to the small circle.
Problem 23: Concentric Circles
Find the area of the green region enclosed between two concentric circles.
4 Draw a line from center O to point A. A radius drawn to a tangent is always perpendicular
. So,
and
.Problem 23: Concentric Circles
Find the area of the green region enclosed between two concentric circles.
5 Draw a line from
to
. This is the radius of the large circle, so
. We now have a right-angled triangle OAB .
Problem 23: Concentric Circles
Find the area of the green region enclosed between two concentric circles.
6 A perpendicular line from the center to a chord bisects (cuts in half) the chord. This means
is exactly half of
. So,
Problem 23: Concentric Circles
Find the area of the green region enclosed between two concentric circles.
7 Let's use the Pythagorean theorem on triangle OAB: Base ^2+ Height ^2= Hypotenuse ^2. This gives us: (I/2)^2+r^2=R^2
Problem 23: Concentric Circles
Find the area of the green region enclosed between two concentric circles.
8 Let's rearrange this equation by subtracting r^2 from both sides: R^2-r^2=(l/2)^2=l^2/4.
Problem 23: Concentric Circles
Find the area of the green region enclosed between two concentric circles.
9 Remember our area formula from Step 2? Area =Π(R^2-r^2 ). Substituting what we just found, we get: Area =n(l^2/4)=1/4πl^2. Proved!
Proof Complete! Great job!

In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A) + Area (B) = Area (C).
Let’s answer this step by step
Problem 24: Semicircles on a Triangle
Show that Area
.
1 Start with the right-angled triangle. Let's call its area
. Its sides are
(left),
(bottom), and hypotenuse
.
Problem 24: Semicircles on a Triangle
Show that Area
Area
rea
.
2 Draw a semicircle on side
. Let's call the area of this entire semicircle Small Semi 1.Problem 24: Semicircles on a Triangle
Show that Area (A)+ Area (B)= Area (C).
3 Draw a semicircle on side b. Let's call the area of this one Small Semi 2.
Problem 24: Semicircles on a Triangle
Show that Area (A)+ Area (B)= Area (C).
4 If we color all of this green, our Total Area is made of three pieces added together:
Total Area =( Small Semi
+(" Small Semi 2 ")+ Triangle CProblem 24: Semicircles on a Triangle
Show that Area (A)+ Area (B)= Area (C).
5 Pythagoras' theorem (a^2+b^2 ┤=c^2 ) gives us a neat circle rule: The area of the two small semicircles added together exactly equals the area of a big semicircle drawn on the hypotenuse c.
Problem 24: Semicircles on a Triangle
Show that Area (A)+ Area (B)= Area (C).
6 Let's draw that big pink semicircle on c. Because it's a right triangle, this large semicircle perfectly covers the entire triangle and part of the small semicircles!
Problem 24: Semicircles on a Triangle
Show that Area (A)+ Area (B)= Area (C).
7 Look at the green shapes that are NOT covered by the pink semicircle. Those are our Lunes, labeled A and B.
Problem 24: Semicircles on a Triangle
Show that Area (A)+ Area (B)= Area (C).
8 This means the area of A+B is just whatever is left over from our Total Area after subtracting the big pink semicircle.
Area (A)+Area(B)=( Total Area) - (Big Semi)Problem 24: Semicircles on a Triangle
Show that Area (A)+ Area (B)= Area (C).
7 Look at the green shapes that are NOT covered by the pink semicircle. Those are our Lunes, labeled A and B.
Problem 24: Semicircles on a Triangle
Show that Area (A)+ Area (B)= Area (C).
8 This means the area of A+B is just whatever is left over from our Total Area after subtracting the big pink semicircle.
Area (A)+Area(B)=( Total Area) - (Big Semi

Fig. 6.53 shows two circles passing through each other's centres. Find the area of the region enclosed by the two circles in terms of the common radius r.
Let’s answer this step by step
Problem 25: Intersecting Circles
Find the area of the region enclosed by two circles passing through each other's centres.
1 We have two identical circles. Let's label their centers
and B. Because they pass through each other's centers, the distance AB is exactly the radius,
.
Problem 25: Intersecting Circles
Find the area of the region enclosed by two circles passing through each other's centres.
2 Let's label the top intersection C and the bottom intersection
D. Draw lines connecting A and B to C.Problem 25: Intersecting Circles
Find the area of the region enclosed by two circles passing through each other's centres.
3 Look at triangle
. The lines
and
are radii of the circles, so they are both length
. AB is also
. Since all sides are equal, it's an equilateral triangle! The angle at A is
.
Problem 25: Intersecting Circles
Find the area of the region enclased by two circles passing through each other's centres.
4 The bottom triangle ABD is also equilateral (
). This means the total angle of the 'pizza slice' (Sector CAD) from center A is
.Problem 25: Intersecting Circles
Find the area of the region enclosed by two circles passing through each other's centres.
5 The area of Sector CAD is
(which is
) of a full circle.
Sector Area
.
Problem 25: Intersecting Circles
Find the area of the region enclosed by two circles passing through each other's centres.
6 The dark red shape we want is made by overlapping two of these sectors: Sector CAD (from circle A) and Sector CBD (from circle B).Problem 25: Intersecting Circles
Find the area of the region enclosed by two circles passing through each other's centres.
7 If we just add the two sectors together, we have a problem: we count the diamond shape in the middle (ACBD) TWICE.
Problem 25: Intersecting Circles
Find the area of the region enclosed by two circles passing through each other's centres.
8 To get the correct area, we add the two sectors and subtract the extra diamond.
Total Area
Sector 1
(Sector 2) - (Diamond ACBD).Problem 25: Intersecting Circles
Find the area of the region enclosed by two circles passing through each other's centres.
9 The diamond ACBD is made of 2 equilateral triangles. The area of one equilateral triangle is (√ 3/4)r^2, so two of them equals (√3/2)r^2.
Problem 25: Intersecting Circles
Find the area of the region enclosed by two circles passing through each other's centres.
10 Now plug it all in:
Total Area =(1/3nr^2 )+(1/3├ nr^2 )-(√ 3/2)r^2
Total Area =r^2 (2n/3-√ 3/2)
Proof Complete! Great job!

In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are 𝐴,𝐵,𝐶, as marked. Show that the area of the rectangle is (2(𝐴+𝐶)(𝐵+𝐶))/𝐶.
Let’s answer this step by step

In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle.