Measuring Space: Perimeter and Area

In my school, the playground is too small to have a 400 m track, so the school constructed a 200 m track instead. Does this mean that we need a smaller stagger for the race tracks in my school (i.e., smaller than the stagger used in the Olympics), for the same 4 × 100 m relay race?

Answer

No, the stagger does not need to be smaller. In fact, it must be larger.

The stagger compensates for the extra distance that runners in outer lanes cover on the curved portions of the track.

On a 400 m track:

A 4 × 100 m relay covers a total of 400 m, which equals one full lap. The runners traverse the two semicircular curves once, which together form one complete circle.

∴ Extra distance for an outer lane = 2π × (difference in radius)

On a 200 m track:

A 4 × 100 m relay still covers 400 m, but this equals two full laps. The runners traverse the curves twice, completing two full circles' worth of curve.

∴ Extra distance for an outer lane = 2 × 2π × (difference in radius) = 4π × (difference in radius)

So the stagger on the 200 m track for the same 4 × 100 m relay is about twice as large as on a 400 m track, not smaller.

What is the connection between this question and the one about the 400 m athletics track?

Answer

(i) The perimeter of a circle is called its circumference.

For a circle of radius r units,

Circumference = 2πr

Since the diameter d = 2r,

we can also write:

Circumference = πd

Hence, the perimeter of the circle is 2πr units or πd units.

(ii) The stagger in the athletics track depends on the difference in the lengths of the circular arcs in different lanes.

To calculate this difference, we need to know the perimeter (circumference) of a circle.

Thus, the question about the athletics track is directly connected to finding the circumference of circles.

So finding the perimeter (circumference) of a circle is essential for working out lane lengths and the staggers in the athletics track problem.

What is the difference in radius between the first and second lanes? Use the Fig. 6.11 to find the stagger needed by the runner in the second lane. Will an equal stagger be needed between the third and second lanes?

Answer

From Fig. 6.11,

Width of each lane = 1.22 m

Therefore, the difference in radius between the first and second lanes = 1.22 m.

On the straight parts, both runners cover the same distance. The extra distance is only along the two curved parts.

The two curved parts together form one complete circle.

Stagger needed by the runner in the second lane

= Difference in circumference of the two circular paths

= 2π(r₂ - r₁)

= 2π x 1.22

= 2 x 3.1416 x 1.22

= 7.67 m (approx.)

Hence, the stagger needed by the runner in the second lane is approximately 7.67 m.

The difference in radius between any two adjacent lanes is the same, i.e., 1.22 m.

Therefore, the stagger needed between the third and second lanes will also be the same.

Hence, an equal stagger will be needed between the third and second lanes.

The perimeter of a circle is 44 cm. What is its radius?

Answer

Given:

Perimeter (Circumference) of circle = 44 cm

π = $\dfrac{22}{7}$

We know the formula,

Circumference of a circle = 2πr

44 cm = 2 × $\dfrac{22}{7}$ × r $\quad$[Substituting the values]

44 cm = $\dfrac{44}{7}$ × r

⇒ r = $\dfrac{44 \times 7}{44}$ cm

⇒ r = 7 cm

The radius of the circle = 7 cm.

Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.

Answer

(i) radius 7 cm

Given:

r = 7 cm, π = $\dfrac{22}{7}$

Circumference of a circle = 2πr

= 2 × $\dfrac{22}{7}$ × 7 cm $\quad$[Substituting the values]

= 2 × 22 cm

= 44 cm

Circumference = 44.0 cm (to 3 significant figures).

(ii) radius 10 cm

Given:

r = 10 cm, π = $\dfrac{22}{7}$

Circumference of a circle = 2πr

= 2 × $\dfrac{22}{7}$ × 10 cm $\quad$[Substituting the values]

= $\dfrac{440}{7}$ cm

= 62.857... cm

Circumference ≈ 62.9 cm (to 3 significant figures).

(iii) radius 12 cm

Given:

r = 12 cm, π = $\dfrac{22}{7}$

Circumference of a circle = 2πr

= 2 × $\dfrac{22}{7}$ × 12 cm $\quad$[Substituting the values]

= $\dfrac{528}{7}$ cm

= 75.428... cm

Circumference ≈ 75.4 cm (to 3 significant figures).

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°, and (ii) the radius is 6.3 m and the angle at the centre is 120°.

Answer

We know the formula:

Length of arc = 2πr × $\dfrac{\theta°}{360°}$

(i) radius = 3.5 cm, angle at centre = 60°

Length of arc = 2 × $\dfrac{22}{7} \times 3.5 \times \dfrac{60°}{360°}$ cm $\quad$[Substituting the values]

$= 2 \times \dfrac{22}{7} \times 3.5 \times \dfrac{1}{6} \text{ cm} \\[1em] = 2 \times \dfrac{22}{1} \times 0.5 \times \dfrac{1}{6} \text{ cm} \\[1em] = \dfrac{22}{6} \text{ cm} \\[1em] = \dfrac{11}{3} \text{ cm} \\[1em]$

Length of arc = $\mathbf{\dfrac{11}{3}}$ cm.

(ii) radius = 6.3 m, angle at centre = 120°

Length of arc = 2 × $\dfrac{22}{7}$ × 6.3 × $\dfrac{120°}{360°}$ m $\quad$[Substituting the values]

$= 2 \times \dfrac{22}{7} \times 6.3 \times \dfrac{1}{3} \text{ m} \\[1em] = 2 \times \dfrac{22}{1} \times 0.9 \times \dfrac{1}{3} \text{ m} \\[1em] = \dfrac{39.6}{3} \text{ m} \\[1em] = 13.2 \text{ m}$

Length of arc = 13.2 m.

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°.

Answer

Given:

Radius (r) = 14 cm

Sector angle (θ) = 75°

π = $\dfrac{22}{7}$

The perimeter of a sector consists of the arc length plus the two radii.

Length of arc = 2πr × $\dfrac{\theta°}{360°}$

$= 2 \times \dfrac{22}{7} \times 14 \times \dfrac{75°}{360°} \text{ cm} \\[1em] = 2 \times \dfrac{22}{1} \times 2 \times \dfrac{75}{360} \text{ cm} \\[1em] = 88 \times \dfrac{5}{24} \text{ cm} \\[1em] = \dfrac{440}{24} \text{ cm} \\[1em] = \dfrac{55}{3} \text{ cm}$

Perimeter of sector = Arc length + 2 × radius

$= \dfrac{55}{3} \text{ cm} + 2 \times 14 \text{ cm} \\[1em] = \dfrac{55}{3} \text{ cm} + 28 \text{ cm} \\[1em] = \dfrac{55}{3} \text{ cm} + \dfrac{84}{3} \text{ cm} \\[1em] = \dfrac{139}{3} \text{ cm} \\[1em]$

Perimeter of sector = $\mathbf{\dfrac{139}{3}}$ cm.

Find the perimeter of the following shape (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate)

Answer

The given shape is made up of a rectangle and two semicircles.

Length of rectangle = 80 m

Diameter of each semicircle = 60 m

Radius of each semicircle = $\dfrac{60}{2}$ = 30 m

Perimeter of the shape = Sum of lengths of two straight sides + Circumference of one complete circle

= 80 + 80 + 2πr

= 160 + 2 x $\dfrac{22}{7}$ x 30

= 160 + $\dfrac{1320}{7}$

= $\dfrac{1120 + 1320}{7}$

= $\dfrac{2440}{7}$ m

= $348\dfrac{4}{7}$ m

Hence, the perimeter of the shape is $\mathbf{348\dfrac{4}{7}}$ m.

Find the perimeter of the following shape (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate)

Answer

The given shape is a semi-circular ring.

Outer diameter = 12 cm

Outer radius = $\dfrac{12}{2}$ = 6 cm

Inner diameter = 8 cm

Inner radius = $\dfrac{8}{2}$ = 4 cm

Width of the ring on each side = $\dfrac{12 - 8}{2}$ = 2 cm

Perimeter of the shape = Outer semicircular arc + Inner semicircular arc + 2 straight sides

= π x 6 + π x 4 + 2 + 2

= 10π + 4

= 10 x $\dfrac{22}{7}$ + 4

= $\dfrac{220}{7}$ + 4

= $\dfrac{248}{7}$ cm

= $35\dfrac{3}{7}$ cm

Hence, the perimeter of the shape is $\mathbf{35\dfrac{3}{7}}$ cm.

Find the perimeter of the following shape (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate)

Answer

The boundary of the shape is made up of four semicircular arcs.

Diameter of each semicircle = 10 cm

Radius of each semicircle = $\dfrac{10}{2}$ = 5 cm

Perimeter of the shape = 4 x Length of one semicircular arc

= 4 x πr

= 4 x $\dfrac{22}{7}$ x 5

= $\dfrac{440}{7}$ cm

= $62\dfrac{6}{7}$ cm

Hence, the perimeter of the shape is $\mathbf{62\dfrac{6}{7}}$ cm.

Find the perimeter of the following shape (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate)

Answer

The boundary of the shape is made up of three semicircular arcs.

Diameter of each semicircle = 12 cm

Radius of each semicircle = $\dfrac{12}{2}$ = 6 cm

Perimeter of the shape = 3 x Length of one semicircular arc

= 3 x πr

= 3 x $\dfrac{22}{7}$ x 6

= $\dfrac{396}{7}$ cm

= $56\dfrac{4}{7}$ cm

Hence, the perimeter of the shape is $\mathbf{56\dfrac{4}{7}}$ cm.

Find the perimeter of the following shape (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate)

Answer

The perimeter consists of:

Four semicircular arcs:

Radius = 7 cm

Total length = 4 x πr

= 4 x $\dfrac{22}{7}$ x 7

= 4 x $\dfrac{22}{1}$ x 1 cm

= 88 cm

Four quarter-circular arcs:

Radius = 14 cm

Four quarter-circles make one complete circle.

Total length = 2 x πr

= 2 x $\dfrac{22}{7}$ x 14

= 2 x $\dfrac{22}{1}$ x 2 cm

= 88 cm

Perimeter of the shape = Total length of 4 semi-circular arcs + Total length of 4 quarter-circular arcs

= 88 cm + 88 cm

= 176 cm

Hence, the perimeter of the shape is 176 cm.

Find the perimeter of the following shape (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate)

Answer

From the figure,

Diameter of the large semicircle = 28 cm

Radius of the large semicircle = $\dfrac{28}{2}$ = 14 cm

The diameter of the large semicircle is divided into 4 equal parts.

Therefore, diameter of each small semicircle = $\dfrac{28}{4}$ = 7 cm

Radius of each small semicircle = $\dfrac{7}{2}$ cm

Perimeter of the shape = Length of large semicircular arc + Lengths of 4 small semicircular arcs

= π x 14 + 4 x π x $\dfrac{7}{2}$

= 14π + 14π

= 28π

= 28 x $\dfrac{22}{7}$

= 88 cm

Hence, the perimeter of the shape is 88 cm.

Find the perimeter of the following shape (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate):

Answer

From the figure,

The three arcs are semicircles with diameters 8 cm, 6 cm and the slant side of the right triangle.

By Baudhāyana–Pythagoras theorem,

Slant side² = 8² + 6²

= 64 + 36

= 100

⇒ Slant side = $\sqrt{100}$

= 10 cm

Therefore, the radii of the three semicircles are:

r₁ = $\dfrac{8}{2}$ cm = 4 cm

r₂ = $\dfrac{6}{2}$ cm = 3 cm

r₃ = $\dfrac{10}{2}$ cm = 5 cm

Length of a semicircular arc of radius 4 cm

= πr₁

= $\dfrac{22}{7}$ x 4 cm

= $\dfrac{88}{7}$ cm

Length of a semicircular arc of radius 3 cm

= πr₂

= $\dfrac{22}{7}$ x 3 cm

= $\dfrac{66}{7}$ cm

Length of a semicircular arc of radius 5 cm

= πr₃

= $\dfrac{22}{7}$ x 5 cm

= $\dfrac{110}{7}$ cm

Perimeter = Length of all 3 semicircular arc:

= $\dfrac{88}{7}$ cm + $\dfrac{66}{7}$ cm + $\dfrac{110}{7}$ cm

= $\dfrac{88 + 66 + 110}{7}$ cm

= $\dfrac{264}{7}$ cm

= $37\dfrac{5}{7}$ cm

Hence, the perimeter of the shape is $37\dfrac{5}{7}$ cm.

Find the perimeter of the following shape (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate):

Answer

From the figure,

Diameter of the large semicircle = 12 cm

Diameter of each small semicircle = 4 cm

Therefore, the radii are:

R = $\dfrac{12}{2}$ cm = 6 cm

r = $\dfrac{4}{2}$ cm = 2 cm

Length of the large semicircular arc

= πR

= $\dfrac{22}{7}$ x 6 cm

= $\dfrac{132}{7}$ cm

Length of one small semicircular arc

= πr

= $\dfrac{22}{7}$ x 2 cm

= $\dfrac{44}{7}$ cm

Length of 3 small semicircular arcs

= 3 x $\dfrac{44}{7}$ cm

= $\dfrac{132}{7}$ cm

Perimeter of the shape = Length of large semicircular arc + Lengths of 3 small semicircular arcs

= $\dfrac{132}{7}$ cm + $\dfrac{132}{7}$ cm

= $\dfrac{264}{7}$ cm

= $37\dfrac{5}{7}$ cm

Hence, the perimeter of the shape is $37\dfrac{5}{7}$ cm.

Find the perimeter of the following shape (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate):

Answer

From the figure,

Diameter of the large semicircle = 10 + 10 = 20 cm

Diameter of each small semicircle = 10 cm

Therefore, the radii are:

R = $\dfrac{20}{2}$ cm = 10 cm

r = $\dfrac{10}{2}$ cm = 5 cm

Length of the large semicircular arc

= πR

= $\dfrac{22}{7}$ x 10 cm

= $\dfrac{220}{7}$ cm

Length of one small semicircular arc

= πr

= $\dfrac{22}{7}$ x 5 cm

= $\dfrac{110}{7}$ cm

Length of 2 small semicircular arcs

= 2 x $\dfrac{110}{7}$ cm

= $\dfrac{220}{7}$ cm

Perimeter of the shape = Length of large semicircular arc + Lengths of 2 small semicircular arcs

= $\dfrac{220}{7}$ cm + $\dfrac{220}{7}$ cm

= $\dfrac{440}{7}$ cm

= $62\dfrac{6}{7}$ cm

Hence, the perimeter of the shape is $62\dfrac{6}{7}$ 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?

Answer

Given:

Diameter (d) = 56 cm

π = $\dfrac{22}{7}$

(i) Distance covered in one revolution = Circumference of the tyre

Circumference = πd

$= \dfrac{22}{7} \times 56 \text{ cm} \\[1em] = \dfrac{22}{1} \times 8 \text{ cm} \\[1em] = 176 \text{ cm}$

The car needs to travel 176 cm for the tyre to complete one revolution.

(ii) Total distance = 10 km

Converting distance into cm:

1 km = 100000 cm

∴ 10 km = 10 × 100000 cm = 1000000 cm

Number of revolutions = $\dfrac{\text{Total Distance}}{\text{Circumference}}$

$= \dfrac{1000000}{176} \\[1em] = \dfrac{125000}{22} \\[1em] = \dfrac{62500}{11} \\[1em]$

The tyre makes $\mathbf{\dfrac{62500}{11}}$ revolutions if the car travels 10 km.

Find the total perimeter of all the petals in each of the given flowers.

Answer

(i) In the first flower,

Side of the square = 14 cm

∴ Radius of each arc = $\dfrac{14}{2}$ = 7 cm

Each petal is made up of 2 quarter circular arcs.

So, 4 petals are made up of 8 quarter circular arcs.

Total perimeter of all the petals

= 8 × length of one quarter circular arc

= 8 × $2πr \times \dfrac{90^\circ}{360^\circ}$

= 8 × $\dfrac{πr}{2}$

= 4πr

= $4 \times \dfrac{22}{7} \times 7$

= 88 cm

Hence, the total perimeter of all the petals in the first flower is 88 cm.

(ii) In the second flower,

Side of the regular hexagon = 42 cm

∴ Radius of each arc = 42 cm

Each arc subtends an angle of 60° at the centre.

Each petal is made up of 2 such arcs.

So, 6 petals are made up of 12 such arcs.

Total perimeter of all the petals

= 12 × length of one arc of angle 60°

= 12 × $2πr \times \dfrac{60^\circ}{360^\circ}$

= 12 × $\dfrac{πr}{3}$

= 4πr

= $4 \times \dfrac{22}{7} \times 42$

= 528 cm

Hence, the total perimeter of all the petals in the second flower is 528 cm.

The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?

Answer

Let the radii of the two circles be r₁ and r₂.

Perimeters of the two circles: C₁ = 2πr₁ and C₂ = 2πr₂

Given:

$\dfrac{C_1}{C_2} = \dfrac{5}{4}$

$\dfrac{2\pi r_1}{2\pi r_2} = \dfrac{5}{4} \\[1em] \dfrac{r_1}{r_2} = \dfrac{5}{4}$

The ratio of their radii is 5 : 4.

What happens if the parallelogram is 'thin' (Fig. 6.18) and the foot of the perpendicular from C to AD does not lie on side AD? The construction then does not seem to work. How do we fix this 'gap'?

Answer

When the parallelogram is thin, the perpendicular from C may fall outside the side AD. So, the usual method of cutting a triangle from one side and moving it to the other side does not directly give a rectangle.

To fix this gap, extend the side DA beyond A.

Choose a point D' on DA close to D, and choose a point A' on DA produced such that

A'A = D'D

Then A'BCD' is also a parallelogram.

Also, the triangle cut off on one side is congruent to the triangle added on the other side.

Therefore,

Area of parallelogram A'BCD' = Area of parallelogram ABCD

Now, this process can be repeated until the foot of the perpendicular lies on the base. Hence, the parallelogram can still be converted into a rectangle with the same base and height.

So,

Area of parallelogram = base x height

Hence, even for a thin parallelogram, the formula for area remains base x height.

The area of a rectangle can be found when we know the lengths of its sides. Is the same true for a parallelogram? That is, can we find the area of a parallelogram when we know the lengths of its sides? Why or why not?

Answer

No, knowing only the lengths of the sides of a parallelogram is not enough to find its area.

The area of a parallelogram depends on both:

• the base, and

• the corresponding height.

Area of parallelogram = base x height

Two parallelograms may have the same side lengths but different heights. In that case, their areas will be different.

Therefore, unlike a rectangle, the area of a parallelogram cannot be determined from the side lengths alone.

Hence, to find the area of a parallelogram, we need additional information such as the height, an angle, or the length of a diagonal.

Since ΔABD and ΔACD have equal area, you may wonder — Can we divide ΔABD using straight cuts into two or more pieces that we can then rearrange to exactly cover ΔACD? What do you think? Is it possible?

Answer

Yes, it is possible.

Since AD is a median of ΔABC, D is the midpoint of BC.

∴ BD = DC

Now, ΔABD and ΔACD have equal bases BD and DC and the same height from A to BC.

So, area (ΔABD) = area (ΔACD)

Hence, ΔABD can be divided into smaller pieces using straight cuts and rearranged to exactly cover ΔACD.

Hence, it is possible to cut and rearrange ΔABD to cover ΔACD exactly.

Suppose we are given two polygons P and Q with equal area. Will it always be possible to divide one of them using straight cuts into two or more pieces and then rearrange the pieces to exactly cover the other polygon? Try this out for familiar shapes, e.g.,

1. A square and non-square rectangle with equal area,
2. Two triangles with different shapes but equal area,
3. A triangle and a square with equal area. Formulate a conjecture of your own about this.

Answer

Yes, it is always possible.

This is the Bolyai–Gerwien theorem (also known as the Wallace–Bolyai–Gerwien theorem) which states:

Any two polygons with equal area can be cut into a finite number of pieces with straight cuts and rearranged to form one another.

Examples:

1. A square and non-square rectangle with equal area: Yes, this can be done. Cuts can transform a square into any rectangle of the same area (or vice versa).

2. Two triangles with different shapes but equal area: Yes, this is possible.

3. A triangle and a square with equal area: Yes, this is possible. The process is called "squaring the triangle" (constructing a square equal in area to a given triangle).

Conjecture: Any two polygons (regardless of shape) with the same area can be dissected into a finite number of pieces using straight cuts and rearranged to form one another.

Think of various rectangles with perimeter 40 units (the sides do not have to be integers).

1. How many such rectangles are there?
2. Among them, is there one whose area is the largest? What are its dimensions?
3. Among all these rectangles, is there one whose area is the smallest? What are its dimensions? Do either of these answers come as a surprise to you?

Answer

(1) Let the length and breadth of the rectangle be l units and b units respectively.

Given perimeter = 40 units.

So,

2(l + b) = 40

l + b = 20

Hence, any pair of positive numbers whose sum is 20 will form a rectangle.

There are infinitely many rectangles with perimeter 40 units.

(2) Area of rectangle

l x b

Suppose:

l = 19, b = 1

Area = 19

l = 18, b = 2

Area = 36

l = 15, b = 5

Area = 75

l = 10, b = 10

Area = 100

We observe that the area increases as the rectangle becomes more like a square.

Hence, the rectangle with the largest area is the square:

Length = 10 units

Breadth = 10 units

Maximum area

= 10 x 10

= 100 square units

The rectangle with the largest area is a square of side 10 units, with area = 100 sq. units.

(3)

As l approaches 0 (or 20), the rectangle becomes very thin, and the area approaches 0. But the area is never exactly 0 (since both sides must be positive).

So there is no rectangle with the smallest area — the area can be made as small as we like, but never reaches a minimum.

There is no rectangle with the smallest area; the area can be made arbitrarily small but never reaches a minimum value.

Yes, it is interesting that:

the maximum area occurs for a square, but there is no minimum area.

What procedure would you use to square a given triangle? Here, the task is to construct a square whose area is equal to the area of some given triangle. Think carefully. How would you proceed?

Answer

To square a given triangle (i.e., construct a square with the same area as the given triangle), we proceed in two steps:

Step 1: Convert the triangle into a rectangle of the same area.

For a triangle with base b and height h:

Area of triangle = $\dfrac{1}{2}$ × b × h

Construct a rectangle with the same base b and height $\dfrac{h}{2}$ (or equivalently, base $\dfrac{b}{2}$ and height h). This rectangle has area:

$\text{Area of rectangle} = b \times \dfrac{h}{2} = \dfrac{1}{2}bh$

which equals the area of the triangle.

Step 2: Convert the rectangle into a square of the same area.

Now transform this rectangle into an equivalent square.

If the side of the square is s, then

s² = $\dfrac{1}{2}bh$

Therefore,

s = $\sqrt{\dfrac{1}{2}bh}$

Construct a square with side s.

Example:

Suppose,

b = 8 cm and h = 4 cm

Area of triangle

= $\dfrac{1}{2}$ x 8 x 4

= 16 cm²

Now for the square:

s² = 16

s = $\sqrt{16}$ = 4 cm

The resulting square has area equal to the area of the rectangle, which in turn equals the area of the original triangle.

Hence, the triangle is squared.

Find the area of triangle ADE in Fig. 6.31.

Answer

From the figure,

AD = BC = 8 cm

The perpendicular distance of E from AD = DC = 10 cm

Area of ΔADE = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x AD x DC

= $\dfrac{1}{2}$ x 8 x 10 cm²

= 4 x 10 cm²

= 40 cm²

Hence, area of triangle ADE = 40 cm².

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.

Answer

Given:

Parallel sides: AB = 40 cm and DC = 20 cm

Non-parallel sides: AD = BC = 26 cm (it is an isosceles trapezium)

Since the trapezium is isosceles, if we drop perpendiculars from C and D to AB, the foot of each perpendicular is at a distance of $\dfrac{40 - 20}{2}$ = 10 cm from the corresponding endpoint of AB.

Let h be the height (perpendicular distance between the parallel sides).

Using the Baudhāyana–Pythagoras theorem on the right triangle formed:

(non-parallel side)² = (horizontal projection)² + h²

26² = 10² + h²

676 = 100 + h²

⇒ h² = 676 - 100

⇒ h² = 576

⇒ h = $\sqrt{576}$ cm

⇒ h = 24 cm

Area of trapezium = $\dfrac{1}{2}$ × (sum of parallel sides) × height

$= \dfrac{1}{2} \times (40 + 20) \text{ cm} \times 24 \text{ cm} \\[1em] = \dfrac{1}{2} \times 60 \text{ cm} \times 24 \text{ cm} \\[1em] = \dfrac{1}{1} \times 30 \text{ cm} \times 24 \text{ cm} \\[1em] = 720 \text{ cm}^2$

Area of the trapezium = 720 cm².

Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

Answer

Given:

Two sides: a = 8 cm, b = 11 cm

Perimeter = 32 cm

We know that,

Perimeter of triangle = a + b + c

32 cm = 8 cm + 11 cm + c

⇒ c = 32 - (8 + 11)

⇒ c = 32 - 19

⇒ c = 13 cm

Semi-perimeter (s) = $\dfrac{\text{Perimeter}}{2}$ = $\dfrac{32}{2}$ cm = 16 cm

Using Heron's formula:

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{16 \times (16 - 8) \times (16 - 11) \times (16 - 13)} \text{ cm}^2 \\[1em] = \sqrt{16 \times 8 \times 5 \times 3} \text{ cm}^2 \\[1em] = \sqrt{16 \times 120} \text{ cm}^2 \\[1em] = \sqrt{16} \times \sqrt{120} \text{ cm}^2 \\[1em] = 4 \times \sqrt{4 \times 30} \text{ cm}^2 \\[1em] = 4 \times 2\sqrt{30} \text{ cm}^2 \\[1em] = 8\sqrt{30} \text{ cm}^2 \\[1em]$

Area of the triangle = $\mathbf{8\sqrt{30}}$ cm².

The sides of a triangular plot are in the ratio 3 : 5 : 7; its perimeter is 300 m. Find its area.

Answer

Given:

Sides in the ratio 3 : 5 : 7

Perimeter = 300 m

Let the sides be 3x, 5x and 7x.

Perimeter = 3x + 5x + 7x = 15x

300 m = 15x

⇒ x = $\dfrac{300}{15}$ m

⇒ x = 20 m

∴ Sides of triangle:

a = 3x = (3 x 20 m) = 60 m

b = 5x = (5 x 20 m) = 100 m

c = 7x = (7 x 20 m) = 140 m

Semi-perimeter (s) = $\dfrac{\text{Perimeter}}{2}$ = $\dfrac{300}{2}$ m = 150 m

Using Heron's formula:

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{150 \times (150 - 60) \times (150 - 100) \times (150 - 140)} \text{ m}^2 \\[1em] = \sqrt{150 \times 90 \times 50 \times 10} \text{ m}^2 \\[1em] = \sqrt{6750000} \text{ m}^2 \\[1em] = \sqrt{2^4 \times 3^3 \times 5^6} \text{ m}^2 \\[1em] = 2^2 \times 3 \times 5^3 \times \sqrt{3} \text{ m}^2 \\[1em] = 4 \times 3 \times 125 \times \sqrt{3} \text{ m}^2 \\[1em] = 1500\sqrt{3} \text{ m}^2 \\[1em]$

Area of the triangular plot = $\mathbf{1500\sqrt{3}}$ m².

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.

Answer

Given:

Area of rhombus = 128 cm²

One diagonal is twice the other.

Let the shorter diagonal be d₁ = d cm, and the longer diagonal be d₂ = 2d cm.

We know the formula:

Area of rhombus = $\dfrac{1}{2}$ × d₁ × d₂

$128 \text{ cm}^2 = \dfrac{1}{2} \times d \times 2d \\[1em] 128 \text{ cm}^2 = \dfrac{1}{2} \times 2d^2 \\[1em] 128 \text{ cm}^2 = d^2 \\[1em] \Rightarrow d = \sqrt{128} \text{ cm} \\[1em] \Rightarrow d = \sqrt{64 \times 2} \text{ cm} \\[1em] \Rightarrow d = 8\sqrt{2} \text{ cm} \\[1em]$

The length of the shorter diagonal = $\mathbf{8\sqrt{2}}$ 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)?

Answer

Given: ABCD is a parallelogram, with P and Q any two points on side AB.

In ΔPCD:

Base = CD

Apex = P (which lies on AB)

Height = perpendicular distance from P to line CD

In ΔQCD:

Base = CD

Apex = Q (which lies on AB)

Height = perpendicular distance from Q to line CD

Since AB ∥ CD (opposite sides of a parallelogram are parallel), the perpendicular distance from any point on AB to line CD is the same. This perpendicular distance equals the height of the parallelogram.

∴ Height of ΔPCD = Height of ΔQCD

Both triangles have the same base CD and the same height.

∴ Area(ΔPCD) = Area(ΔQCD)

⇒ $\dfrac{\text{Area}(\Delta PCD)}{\text{Area}(\Delta QCD)} = \dfrac{1}{1}$

Area(∆PCD) : Area(∆QCD) = 1 : 1.

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

Answer

Given: PQRS is a parallelogram. O is any point on the diagonal PR.

To prove: Area(ΔPSO) = Area(ΔPQO).

Proof:

In parallelogram PQRS, the diagonal PR divides it into two triangles ΔPQR and ΔPSR.

A diagonal of a parallelogram divides it into two triangles of equal area.

∴ Area(ΔPQR) = Area(ΔPSR)

Now ΔPQR and ΔPSR share the same base PR. Since their areas are equal and they have the same base, their corresponding heights (perpendicular distances from Q and S to line PR) must be equal.

Let h be this common perpendicular distance from Q and S to line PR.

Now consider triangles ΔPQO and ΔPSO:

Both have the same base PO (a portion of diagonal PR).

Height of ΔPQO = perpendicular distance from Q to line PR = h

Height of ΔPSO = perpendicular distance from S to line PR = h

Area(ΔPQO) = $\dfrac{1}{2}$ × PO × h

Area(ΔPSO) = $\dfrac{1}{2}$ × PO × h

∴ Area(ΔPQO) = Area(ΔPSO)

Hence, the areas of triangles PSO and PQO are equal.

Hence proved.

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.)

Answer

Given:

Let ABCD be a 4-gon (quadrilateral).

Let P, Q, R and S be the mid-points of sides AB, BC, CD and DA respectively.

To Prove: Area (PQRS) = $\dfrac{1}{2}$ x Area (ABCD)

Construction: Join BD and BS.

Proof:

We know that the median of a triangle divides it into two triangles of equal areas.

BS is the median of the △BAD.

Area (△BAS) = $\dfrac{1}{2}$ x Area (△BAD) $\quad$....(1)

Similarly,

PS is the median of the △BAS.

∴ Area (△APS) = $\dfrac{1}{2}$ x Area (△BAS) $\quad$....(2)

Similarly, in △ACD, S is the mid-point of AD and R is the mid-point of CD.

From (1) and (2),

Area (△APS) = $\dfrac{1}{2}$ x $\dfrac{1}{2}$ x Area (△BAD)

= $\dfrac{1}{4}$ x Area (△BAD) $\quad$....(3)

Similarly,

Area (△CQR) = $\dfrac{1}{4}$ x Area (△BCD) $\quad$....(4)

Adding (3) and (4),

Area (△APS) + Area (△CQR) = $\dfrac{1}{4}$ x Area (△BAD) + $\dfrac{1}{4}$ x Area (△BCD)

= $\dfrac{1}{4}$ x Area (ABCD) $\quad$....(5)

Similarly,

Area (△DRS) + Area (△BQP) = $\dfrac{1}{4}$ x Area (ABCD) $\quad$....(6)

Adding (5) and (6),

Area (△APS) + Area (△CQR) + Area (△DRS) + Area (△BQP) = $\dfrac{1}{4}$ x Area (ABCD) + $\dfrac{1}{4}$ x Area (ABCD)

= $\dfrac{1}{2}$ x Area (ABCD) $\quad$....(7)

From the figure,

Area (△APS) + Area (△CQR) + Area (△DRS) + Area (△BQP) + Area (PQRS) = Area (ABCD)

From (7),

$\dfrac{1}{2}$ x Area (ABCD) + Area (PQRS) = Area (ABCD)

Area (PQRS) = Area (ABCD) - $\dfrac{1}{2}$ x Area (ABCD)

= Area (ABCD) x $\left(1 - \dfrac{1}{2}\right)$

= $\dfrac{1}{2}$ x Area (ABCD)

∴ Area of the parallelogram PQRS = $\bold {\dfrac{1}{2}}$ x Area of the 4-gon ABCD.

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)

Answer

Given,

D is the midpoint of BC.

∴ BD = DC

In ΔABC, AD is a median.

∴ area (ΔABD) = area (ΔACD) .....(1)

In ΔPBC, PD is a median.

∴ area (ΔPBD) = area (ΔPCD) .....(2)

Subtracting (2) from (1), we get:

area (ΔABD) - area (ΔPBD) = area (ΔACD) - area (ΔPCD)

⇒ area (ΔABP) = area (ΔACP)

Hence proved.

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)?

Answer

Let the side of square ABCD be a units.

Let the perpendicular distances of P from AB and CD be h₁ and h₂ respectively.

Since AB ∥ CD,

h₁ + h₂ = a

Area of red region = area (ΔPAB) + area (ΔPCD)

= $\dfrac{1}{2}$ x AB x h₁ + $\dfrac{1}{2}$ x CD x h₂

= $\dfrac{1}{2}$ x a x h₁ + $\dfrac{1}{2}$ x a x h₂

= $\dfrac{1}{2}$ x a x (h₁ + h₂)

= $\dfrac{1}{2}$ x a x a

= $\dfrac{a^2}{2}$

Similarly, let the perpendicular distances of P from BC and AD be k₁ and k₂ respectively.

Since BC ∥ AD,

k₁ + k₂ = a

Area of green region = area (ΔPBC) + area (ΔPDA)

= $\dfrac{1}{2}$ x BC x k₁ + $\dfrac{1}{2}$ x AD x k₂

= $\dfrac{1}{2}$ x a x (k₁ + k₂)

= $\dfrac{1}{2}$ x a x a

= $\dfrac{a^2}{2}$

Thus,

Area of red region : Area of green region = $\dfrac{a^2}{2}$ : $\dfrac{a^2}{2}$

= 1 : 1

Hence, the required ratio is 1 : 1.

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) = $\dfrac{1}{2}$Area (∆ABC).

Answer

Given:

ABC is a triangle.

D is the midpoint of AB.

∴ BD = $\dfrac{1}{2}$ AB

Also, CQ ∥ PD

To Prove: Area (∆BPQ) = $\dfrac{1}{2}$ x Area (∆ABC)

Construction: Join PQ and CD.

Proof:

Since D is the midpoint of AB then CD is the median of the triangle ABC.

And

We know that the median of a triangle divides it into two triangles of equal areas.

So,

Area (ΔBCD) = $\dfrac{1}{2}$ x Area (ΔABC)

From the figure,

Area (ΔBCD) = Area (ΔBPD) + Area (ΔDPC)

Now,

Area (ΔBPD) + Area (ΔDPC) = $\dfrac{1}{2}$ x Area (ΔABC) $\quad$....(1)

We know that the area of triangles on the same base and between the same parallel lines are equal.

ΔDPQ and ΔDPC are on the same base DP and between the same parallel lines DP and CQ.

So, area (DPQ) = area (DPC) $\quad$....(2)

Substituting (2) in (1),

Area (ΔBPD) + Area (ΔDPQ) = $\dfrac{1}{2}$ x Area (ΔABC)

From the figure,

Area (ΔBPQ) = Area (ΔBPD) + Area (ΔDPQ)

Therefore,

Area (ΔBPQ) = $\dfrac{1}{2}$ x Area (ΔABC)

Hence proved.

Why were human beings so fond of using circular shapes? Was this only for practical reasons, or could there have been other reasons too? What kinds of uses have human beings found for the circular shape?

Answer

Human beings have always been fond of the circular shape for both practical and non-practical reasons.

Practical reasons:

1. Ease of construction: A circle can be drawn easily with a rope and a peg.

2. Strength and stability: Circular shapes distribute force evenly, making round columns, arches, domes and wheels strong.

3. Maximum area: A circle encloses the greatest area for a given perimeter, making it useful for tanks, silos and shelters.

4. Smooth motion: Wheels and rollers reduce friction and help move heavy loads easily.

5. No corners: Circular pots, plates and bowls are easier to clean because dirt does not collect in corners.

Other reasons:

1. Aesthetic beauty: Circles are perfectly symmetrical and visually pleasing.

2. Symbolic meaning: They represent wholeness, eternity, the sun, the moon, the universe and the cycle of life.

3. Connection with nature: The sun, moon, fruits, flowers, eyes, ripples and tree rings are circular.

Uses of the circular shape:

• Wheels and rollers
• Pots, bowls and plates
• Wells, tanks and silos
• Coins, clocks and gauges
• Gears, fans and turbines
• Domes, pipes and drums
• Balls, lenses and mirrors

Thus, circles are valued for their usefulness, beauty, symbolism and connection with nature.

Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.

Answer

Given:

Radius (r) = 7 cm

Angle of sector (θ) = 60°

We know the formula,

Area of sector = $\dfrac{\theta}{360°}$ x πr²

$= \dfrac{60°}{360°} \times \dfrac{22}{7} \times (7 \text{ cm})^2 \\[1em] = \dfrac{1}{6} \times \dfrac{22}{7} \times 49 \text{ cm}^2 \\[1em] = \dfrac{1}{6} \times \dfrac{22}{1} \times 7 \text{ cm}^2 \\[1em] = \dfrac{22 \times 7}{6} \text{ cm}^2 \\[1em] = \dfrac{154}{6} \text{ cm}^2 \\[1em] = \dfrac{77}{3} \text{ cm}^2$

Area of the sector = $\mathbf{\dfrac{77}{3}}$ cm²

Find the area of a quadrant of a circle whose circumference is 44 cm.

Answer

Given:

Circumference of circle = 44 cm

We know the formula,

Circumference of circle = 2πr

$\phantom{=}44 \text{ cm} = 2 \times \dfrac{22}{7} \times r \\[1em] \phantom{=}44 \text{ cm} = \dfrac{44}{7} \times r \\[1em] \Rightarrow r = \dfrac{44 \times 7}{44} \text{ cm} \\[1em] \Rightarrow r = 7 \text{ cm}$

A quadrant of a circle subtends an angle of 90° at the centre.

Area of quadrant = $\dfrac{90°}{360°}$ x πr²

$= \dfrac{1}{4} \times \dfrac{22}{7} \times (7 \text{ cm})^2 \\[1em] = \dfrac{1}{4} \times \dfrac{22}{7} \times 49 \text{ cm}^2 \\[1em] = \dfrac{1}{4} \times \dfrac{22}{1} \times 7 \text{ cm}^2 \\[1em] = \dfrac{154}{4} \text{ cm}^2 \\[1em] = 38.5 \text{ cm}^2$

Area of the quadrant = 38.5 cm²

The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

Answer

Given:

Length of the minute hand (r) = 7 cm

Time = 10 minutes

In 60 minutes, the minute hand traces an angle of 360° (one complete revolution).

∴ In 1 minute, the angle traced = $\dfrac{360°}{60}$ = 6°

In 10 minutes, the angle traced (θ) = 10 x 6° = 60°

Area swept by the minute hand = Area of the sector with angle 60°

Area of sector = $\dfrac{\theta}{360°}$ x πr²

$= \dfrac{60°}{360°} \times \dfrac{22}{7} \times (7 \text{ cm})^2 \\[1em] = \dfrac{1}{6} \times \dfrac{22}{7} \times 49 \text{ cm}^2 \\[1em] = \dfrac{1}{6} \times \dfrac{22}{1} \times 7 \text{ cm}^2 \\[1em] = \dfrac{154}{6} \text{ cm}^2 \\[1em] = \dfrac{77}{3} \text{ cm}^2$

Area swept by the minute hand in 10 minutes = $\mathbf{\dfrac{77}{3}}$ cm².

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), and (ii) major sector (that subtends 270° at the centre). (Use π ≈ 3.14.)

Answer

Given:

Radius (r) = 10 cm

π = 3.14

(i) Area of the minor sector (angle 90°)

Area of minor sector = $\dfrac{\theta}{360°}$ x πr²

$= \dfrac{90°}{360°} \times 3.14 \times (10 \text{ cm})^2 \\[1em] = \dfrac{1}{4} \times 3.14 \times 100 \text{ cm}^2 \\[1em] = \dfrac{314}{4} \text{ cm}^2 \\[1em] = 78.5 \text{ cm}^2$

Area of the minor sector = 78.5 cm²

(ii) Area of the major sector (angle 270°)

Area of major sector = $\dfrac{\theta}{360°}$ x πr²

$= \dfrac{270°}{360°} \times 3.14 \times (10 \text{ cm})^2 \\[1em] = \dfrac{3}{4} \times 3.14 \times 100 \text{ cm}^2 \\[1em] = \dfrac{3 \times 314}{4} \text{ cm}^2 \\[1em] = \dfrac{942}{4} \text{ cm}^2 \\[1em] = 235.5 \text{ cm}^2$

Area of the major sector = 235.5 cm²

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 $\sqrt{3}$ ≈ 1.73.)

Answer

Given:

Radius (r) = 15 cm

Angle (θ) = 60°

π = 3.14, $\sqrt{3}$ ≈ 1.73

Let the chord be AB, subtending 60° at centre O.

Area of minor segment = Area of minor sector OAB − Area of △OAB

Area of minor sector OAB = $\dfrac{\theta}{360°}$ x πr²

$= \dfrac{60°}{360°} \times 3.14 \times (15 \text{ cm})^2 \\[1em] = \dfrac{1}{6} \times 3.14 \times 225 \text{ cm}^2 \\[1em] = \dfrac{706.5}{6} \text{ cm}^2 \\[1em] = 117.75 \text{ cm}^2$

Since OA = OB = r and ∠AOB = 60°, the triangle OAB is equilateral with side 15 cm.

Area of △OAB (equilateral) = $\dfrac{\sqrt{3}}{4}$ x (side)²

$= \dfrac{\sqrt{3}}{4} \times (15 \text{ cm})^2 \\[1em] = \dfrac{1.73}{4} \times 225 \text{ cm}^2 \\[1em] = \dfrac{389.25}{4} \text{ cm}^2 \\[1em] = 97.3125 \text{ cm}^2$

Area of minor segment = 117.75 cm² − 97.3125 cm² = 20.4375 cm²

Area of minor segment = 20.4375 cm²

Area of major segment = Area of circle − Area of minor segment

Area of circle = πr² = 3.14 x 225 cm² = 706.5 cm²

Area of major segment = 706.5 cm² − 20.4375 cm² = 686.0625 cm²

Area of major segment ≈ 686.06 cm²

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.

Answer

Given:

Length of each wiper blade (r) = 28 cm

Angle swept (θ) = 120°

Number of wipers = 2 (non-overlapping)

Area cleaned by one wiper = Area of sector with r = 28 cm and θ = 120°

Area of sector = $\dfrac{\theta}{360°}$ x πr²

$= \dfrac{120°}{360°} \times \dfrac{22}{7} \times (28 \text{ cm})^2 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 784 \text{ cm}^2 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{1} \times 112 \text{ cm}^2 \\[1em] = \dfrac{22 \times 112}{3} \text{ cm}^2 \\[1em] = \dfrac{2464}{3} \text{ cm}^2$

Since the two wipers do not overlap,

Total area cleaned = 2 x Area cleaned by one wiper

$= 2 \times \dfrac{2464}{3} \text{ cm}^2 \\[1em] = \dfrac{4928}{3} \text{ cm}^2$

Total area cleaned at each sweep = $\mathbf{\dfrac{4928}{3}}$ cm².

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 $r^2 \left(\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4}\right)$.

Answer

Given:

Radius = r

Angle subtended at centre (θ) = 60°

Let the chord be AB, subtending 60° at centre O.

Area of minor segment = Area of minor sector OAB − Area of △OAB

Area of minor sector OAB = $\dfrac{\theta}{360°}$ x πr²

$= \dfrac{60°}{360°} \times \pi r^2 \\[1em] = \dfrac{1}{6} \pi r^2 \\[1em] = \dfrac{\pi r^2}{6}$

Since OA = OB = r and ∠AOB = 60°, the remaining angles ∠OAB and ∠OBA are also each 60°. So △OAB is equilateral with side r.

Area of △OAB (equilateral with side r) = $\dfrac{\sqrt{3}}{4}$ x r²

Area of minor segment = Area of minor sector − Area of △OAB

$= \dfrac{\pi r^2}{6} - \dfrac{\sqrt{3}}{4} r^2 \\[1em] = r^2 \left(\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4}\right)$

∴ Area of the minor segment = $\mathbf{r^2 \left(\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4}\right)}$.

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 $\dfrac{3\sqrt{3}}{4\pi}$ ≈ 0.413.

Answer

Let the equilateral triangle be ABC, inscribed in a circle with centre O and radius r. Since A, B, C all lie on the circle, the segments to the centre are all radii:

OA = OB = OC = r

Now draw the line from A to the midpoint D of the opposite side BC. Two facts force A, O, D onto one straight line:

• AB = AC, so A lies on the perpendicular bisector of BC.
• OB = OC, so O lies on the perpendicular bisector of BC too.

The perpendicular bisector of a chord is a single line passing through the centre, so A, O, D are collinear, and AD ⊥ BC with D the midpoint of BC. [∵ the perpendicular from the centre to a chord bisects it]

Step 1 — Height of the triangle

Let the side be a, so BD = $\dfrac{a}{2}$.

In right triangle ABD (right angle at D), Pythagoras gives:

AD² = AB² - BD²

= a² - $\dfrac{a^2}{4}$

= $\dfrac{3a^2}{4}$

⇒ AD = $\dfrac{\sqrt{3}}{2}$a.

This AD is the height of the equilateral triangle.

Step 2 — Use the radii to find a in terms of r

Since O lies on AD with OA = r:

OD = AD - OA

= $\dfrac{\sqrt{3}}{2}$a - r.

Apply Pythagoras in right triangle OBD (right angle at D), with OB = r and BD = $\dfrac{a}{2}$:

r² = OD² + BD²

= $\left(\dfrac{\sqrt{3}}{2}a - r\right)^2 + \left(\dfrac{a}{2}\right)^2$.

Expanding the square:

$r^2 = \dfrac{3}{4}a^2 - \sqrt{3}ar + r^2 + \dfrac{1}{4}a^2 \\[1em] = a^2 - \sqrt{3}ar + r^2.$

The r² cancels from both sides, leaving:

$0 = a^2 - \sqrt{3},ar = a\left(a - \sqrt{3},r\right).$

Since a ≠ 0, we get

a = $\sqrt{3}$r

Step 3 — The two areas

Area of triangle = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × a × $\dfrac{\sqrt{3}}{2}$a

= $\dfrac{\sqrt{3}}{4}$a²

= $\dfrac{\sqrt{3}}{4}\big(\sqrt{3},r\big)^2 = \dfrac{\sqrt{3}}{4} × 3r^2 = \dfrac{3\sqrt{3}}{4}r^2.$

Area of circle = πr²

Step 4 — The ratio

$\dfrac{\text{Area of triangle}}{\text{Area of circle}} = \dfrac{\dfrac{3\sqrt{3}}{4}r^2}{\pi r^2} = \dfrac{3\sqrt{3}}{4\pi}.$

The r² cancels, so the ratio doesn't depend on the size of the circle.

Numerically, $3\sqrt{3} \approx 5.196$ and $4\pi \approx 12.566$, giving

$\dfrac{3\sqrt{3}}{4\pi} \approx 0.413.$

∴ Ratio of area of equilateral triangle to area of circle = $\mathbf{\dfrac{3\sqrt{3}}{4\pi}}$ ≈ 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 $\dfrac{2}{\pi}$ ≈ 0.637.

Answer

Given:

A square is inscribed in a circle of radius r.

When a square is inscribed in a circle, the diagonal of the square equals the diameter of the circle.

∴ Diagonal of square = 2r

Let the side of the square be a.

Using the Baudhāyana–Pythagoras theorem:

(diagonal)² = (side)² + (side)²

$(2r)^2 = a^2 + a^2 \\[1em] 4r^2 = 2a^2 \\[1em] \Rightarrow a^2 = 2r^2$

Area of the square = a² = 2r²

Area of the circle = πr²

Ratio = $\dfrac{\text{Area of square}}{\text{Area of circle}}$

$= \dfrac{2r^2}{\pi r^2} \\[1em] = \dfrac{2}{\pi}$

Approximate value: $\dfrac{2}{\pi} \approx \dfrac{2}{3.1416} \approx 0.637$

∴ Ratio of area of square to area of circle = $\mathbf{\dfrac{2}{\pi}}$ ≈ 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 $\dfrac{3\sqrt{3}}{2\pi}$ ≈ 0.827. Can you see why the answer is exactly twice the answer to Question 8?

Answer

Given:

A regular hexagon is inscribed in a circle of radius r with centre O.

A regular hexagon can be divided into 6 congruent equilateral triangles by joining each vertex to the centre O.

Each triangle has two sides equal to the radius r, and the central angle at O is $\dfrac{360°}{6}$ = 60°.

Since two sides are equal (each = r) and the included angle is 60°, the other two angles must also be 60°, making each triangle equilateral with side r.

Area of one equilateral triangle (side r) = $\dfrac{\sqrt{3}}{4}$ x r²

Area of regular hexagon = 6 x Area of one equilateral triangle

$= 6 \times \dfrac{\sqrt{3}}{4} r^2 \\[1em] = \dfrac{6\sqrt{3}}{4} r^2 \\[1em] = \dfrac{3\sqrt{3}}{2} r^2$

Area of the circle = πr²

Ratio = $\dfrac{\text{Area of hexagon}}{\text{Area of circle}}$

$= \dfrac{\dfrac{3\sqrt{3}}{2} r^2}{\pi r^2} \\[1em] = \dfrac{3\sqrt{3}}{2\pi}$

Approximate value: $\dfrac{3\sqrt{3}}{2\pi} \approx \dfrac{3 \times 1.732}{2 \times 3.1416} \approx \dfrac{5.196}{6.2832} \approx 0.827$

∴ Ratio of area of hexagon to area of circle = $\mathbf{\dfrac{3\sqrt{3}}{2\pi}}$ ≈ 0.827.

Reason why this is exactly twice the answer to Question 8:

The ratio in Question 8 is $\dfrac{3\sqrt{3}}{4\pi}$ and the ratio in Question 10 is $\dfrac{3\sqrt{3}}{2\pi} = 2 \times \dfrac{3\sqrt{3}}{4\pi}$.

This is because a regular hexagon inscribed in a circle of radius r has area $\dfrac{3\sqrt{3}}{2} r^2$, while an equilateral triangle inscribed in the same circle has area $\dfrac{3\sqrt{3}}{4} r^2$ — exactly half. Geometrically, if we take alternate vertices of the regular hexagon, they form an equilateral triangle whose area is exactly half that of the hexagon. (Equivalently, the hexagon is made up of 6 equilateral triangles of side r, while the inscribed equilateral triangle can be shown to contain exactly 3 such equilateral pieces of area.)

Identities in algebra can sometimes be shown as area relationships. For example:

The figure shown corresponds to the identity (a + b)² = a² + 2ab + b². Do you see how?

Draw figures corresponding to the identities (a + b)(a – b) = a² – b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

Answer

(i) For (a + b)(a – b) = a² – b²

Draw a square of side a units. From one corner, remove a square of side b units.

Area of big square = a²

Area of small square = b²

Area of remaining shaded region = a² – b² .....(1)

Now, this remaining region can be rearranged to form a rectangle whose length is (a + b) units and breadth is (a – b) units.

Area of the rectangle = length × breadth

= (a + b)(a – b)

Since both figures have the same area,

(a + b)(a – b) = a² – b²

Hence, the identity is shown by the area relationship.

(ii) For (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Draw a square of side (a + b + c) units and divide each side into three parts of lengths a, b and c units.

Area of the whole square = (a + b + c)²

The square is divided into:

Area of square of side a = a²

Area of square of side b = b²

Area of square of side c = c²

Area of two rectangles of sides a and b = 2ab

Area of two rectangles of sides b and c = 2bc

Area of two rectangles of sides c and a = 2ca

So,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Hence, the identity is shown by the area relationship.

An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.

Answer

Given:

Perimeter = 40 cm

Equal sides = 15 cm each

Let the base be b cm.

Perimeter = Sum of all sides

40 cm = 15 cm + 15 cm + b

40 cm = 30 cm + b

⇒ b = 40 cm - 30 cm

⇒ b = 10 cm

So, sides of the triangle are a = 15 cm, b = 15 cm, c = 10 cm.

Using Heron's formula:

s = $\dfrac{a + b + c}{2}$ = $\dfrac{15 + 15 + 10}{2}$ cm = $\dfrac{40}{2}$ cm = 20 cm

Area of the triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{20(20 - 15)(20 - 15)(20 - 10)} \text{ cm}^2 \\[1em] = \sqrt{20 \times 5 \times 5 \times 10} \text{ cm}^2 \\[1em] = \sqrt{(2 \times 2 \times 5) \times 5 \times 5 \times (2 \times 5)} \text{ cm}^2 \\[1em] = \sqrt{2^2 \times 5^4 \times 2} \text{ cm}^2 \\[1em] = 2 \times 5^2 \times \sqrt{2} \text{ cm}^2 \\[1em] = 50\sqrt{2} \text{ cm}^2$

Area of the triangle = $\mathbf{50\sqrt{2}}$ cm².

An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?

Answer

Given:

Base of the isosceles triangle = 10 cm

Area of the triangle = 60 cm²

Let the equal sides be a cm each, and let h be the height from the apex to the base.

Area of the triangle = $\dfrac{1}{2}$ x base x height

$60 \text{ cm}^2 = \dfrac{1}{2} \times 10 \times h \\[1em] 60 \text{ cm}^2 = 5 \times h \\[1em] \Rightarrow h = \dfrac{60}{5} \text{ cm} \\[1em] \Rightarrow h = 12 \text{ cm}$

In an isosceles triangle, the altitude from the apex to the base bisects the base. So, the foot of the perpendicular divides the base into two equal halves of 5 cm each.

Using the Baudhāyana–Pythagoras theorem in the right-angled triangle formed:

a² = h² + (half of base)²

a² = (12)² + (5)²

a² = 144 + 25

a² = 169 cm²

a = $\sqrt{169}$ cm

a = 13 cm

The length of each equal side = 13 cm.

The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.

Answer

Given:

Area of the right-angled triangle = 54 cm²

One leg (say base) = 12 cm

Let the other leg (height) be h cm.

Area of right-angled triangle = $\dfrac{1}{2}$ x base x height

$54 \text{ cm}^2 = \dfrac{1}{2} \times 12 \times h \\[1em] 54 \text{ cm}^2 = 6 \times h \\[1em] \Rightarrow h = \dfrac{54}{6} \text{ cm} \\[1em] \Rightarrow h = 9 \text{ cm}$

So, the two legs are 12 cm and 9 cm.

Using the Baudhāyana–Pythagoras theorem to find hypotenuse:

(Hypotenuse)² = (12)² + (9)²

$= 144 + 81 \\[1em] = 225 \text{ cm}^2$

Hypotenuse = $\sqrt{225}$ cm = 15 cm

Perimeter = Sum of all sides

= 12 cm + 9 cm + 15 cm

= 36 cm

Perimeter of the right-angled triangle = 36 cm.

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

Answer

Given:

Sides are in ratio 2 : 3 : 4

Perimeter = 45 cm

Let the sides be 2x, 3x and 4x.

Perimeter = Sum of all sides

45 cm = 2x + 3x + 4x

45 cm = 9x

⇒ x = $\dfrac{45}{9}$

⇒ x = 5 cm

So, the sides are:

a = 2x = 2 x 5 = 10 cm

b = 3x = 3 x 5 = 15 cm

c = 4x = 4 x 5 = 20 cm

Using Heron's formula:

s = $\dfrac{a + b + c}{2}$ = $\dfrac{45}{2}$ cm = 22.5 cm

Area of the triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{22.5 \times (22.5 - 10) \times (22.5 - 15) \times (22.5 - 20)} \text{ cm}^2 \\[1em] = \sqrt{22.5 \times 12.5 \times 7.5 \times 2.5} \text{ cm}^2 \\[1em] = \sqrt{\dfrac{45}{2} \times \dfrac{25}{2} \times \dfrac{15}{2} \times \dfrac{5}{2}} \text{ cm}^2 \\[1em] = \sqrt{\dfrac{45 \times 25 \times 15 \times 5}{16}} \text{ cm}^2 \\[1em] = \sqrt{\dfrac{84375}{16}} \text{ cm}^2 \\[1em] = \dfrac{\sqrt{84375}}{4} \text{ cm}^2 \\[1em] = \dfrac{\sqrt{5625 \times 15}}{4} \text{ cm}^2 \\[1em] = \dfrac{75\sqrt{15}}{4} \text{ cm}^2$

Area of the triangle = $\mathbf{\dfrac{75\sqrt{15}}{4}}$ cm²

The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.

Answer

Given:

Sides of triangle: a = 7 cm, b = 24 cm, c = 25 cm

Method 1: Using Heron's formula

s = $\dfrac{a + b + c}{2}$ = $\dfrac{7 + 24 + 25}{2}$ cm = $\dfrac{56}{2}$ cm = 28 cm

Area of the triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{28(28 - 7)(28 - 24)(28 - 25)} \text{ cm}^2 \\[1em] = \sqrt{28 \times 21 \times 4 \times 3} \text{ cm}^2 \\[1em] = \sqrt{(4 \times 7) \times (3 \times 7) \times 4 \times 3} \text{ cm}^2 \\[1em] = \sqrt{4^2 \times 7^2 \times 3^2} \text{ cm}^2 \\[1em] = 4 \times 7 \times 3 \text{ cm}^2 \\[1em] = 84 \text{ cm}^2$

Method 2: Using the right-angled triangle property

Let us check if the triangle is right-angled:

7² + 24² = 25²

49 + 576 = 625

625 = 625

Since 7² + 24² = 25², by the converse of the Baudhāyana–Pythagoras theorem, the triangle is right-angled with the right angle between the sides of length 7 cm and 24 cm.

So, base = 7 cm, height = 24 cm.

Area of the triangle = $\dfrac{1}{2}$ x base x height

$= \dfrac{1}{2} \times 7 \times 24 \text{ cm}^2 \\[1em] = \dfrac{1}{1} \times 7 \times 12 \text{ cm}^2 \\[1em] = 84 \text{ cm}^2$

Area of the triangle = 84 cm² (by both methods).

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.

Answer

Given:

Diameter of wheel (d) = 60 cm

Number of revolutions = 100

In one revolution, the wheel travels a distance equal to its circumference.

Circumference of wheel = πd

$= \dfrac{22}{7} \times 60 \text{ cm} \\[1em] = \dfrac{1320}{7} \text{ cm}$

Total distance travelled = Circumference x Number of revolutions

$= \dfrac{1320}{7} \times 100 \text{ cm} \\[1em] = \dfrac{132000}{7} \text{ cm} \\[1em]$

Converting into metres:

1 m = 100 cm

∴ Distance = $\dfrac{132000}{7 \times 100}$ m = $\dfrac{1320}{7}$ m

The cyclist will have travelled $\mathbf{\dfrac{1320}{7}}$ m.

Find the area of a quadrant of a circle whose circumference is 66 cm.

Answer

Given:

Circumference of circle = 66 cm

We know the formula,

Circumference = 2πr

$66 \text{ cm} = 2 \times \dfrac{22}{7} \times r \\[1em] 66 \text{ cm} = \dfrac{44}{7} \times r \\[1em] \Rightarrow r = \dfrac{66 \times 7}{44} \text{ cm} \\[1em] \Rightarrow r = \dfrac{3 \times 7}{2} \text{ cm} \\[1em] \Rightarrow r = \dfrac{21}{2} \text{ cm} \\[1em] \Rightarrow r = 10.5 \text{ cm}$

A quadrant subtends an angle of 90° at the centre.

Area of quadrant = $\dfrac{90°}{360°}$ x πr²

$= \dfrac{1}{4} \times \dfrac{22}{7} \times (10.5 \text{ cm})^2 \\[1em] = \dfrac{1}{4} \times \dfrac{22}{7} \times 110.25 \text{ cm}^2 \\[1em] = \dfrac{1}{4} \times \dfrac{22 \times 110.25}{7} \text{ cm}^2 \\[1em] = \dfrac{1}{4} \times \dfrac{2425.5}{7} \text{ cm}^2 \\[1em] = \dfrac{1}{4} \times 346.5 \text{ cm}^2 \\[1em] = 86.625 \text{ cm}^2$

Area of the quadrant = 86.625 cm²

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.

Answer

Given:

Outer radius (r) = 28 cm

π = $\dfrac{22}{7}$

The distance covered by the car in one complete turn of the wheel is equal to the circumference of the wheel.

Circumference of the wheel = 2πr

= 2 x $\dfrac{22}{7}$ x 28 cm $\quad$[Substituting the values]

= 2 x $\dfrac{22}{1}$ x 4 cm

= 2 x 22 x 4 cm

= 176 cm

∴ Distance covered by the car in one complete turn of the wheel = 176 cm

Now, let's find the number of turns of the wheel during a journey of 1 km.

Converting 1 km into cm:

1 km = 100000 cm

∴ Total distance = 100000 cm

Number of turns of the wheel = $\dfrac{\text{Total Distance}}{\text{Distance covered in one turn}}$

$= \dfrac{100000}{176} \quad\text{[Substituting the values]} \\[1em] = \dfrac{6250}{11} \\[1em] = 568\dfrac{2}{11}$

∴ The wheel turns $\mathbf{568\dfrac{2}{11}}$ times during a journey of 1 km.

Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

Answer

Let the two rectangles have sides (l₁, b₁) and (l₂, b₂).

Given:

Both rectangles have the same area and the same perimeter.

From the same perimeter:

2(l₁ + b₁) = 2(l₂ + b₂)

⇒ l₁ + b₁ = l₂ + b₂ $\quad$...(1)

From the same area:

l₁ x b₁ = l₂ x b₂ $\quad$...(2)

From (1) and (2), the pairs (l₁, b₁) and (l₂, b₂) have the same sum and the same product. Hence, both pairs are roots of the same quadratic equation:

x² - (l₁ + b₁)x + (l₁ x b₁) = 0

This equation has only two roots. So, either:

(l₁, b₁) = (l₂, b₂), or

(l₁, b₁) = (b₂, l₂)

In both cases, the two rectangles have exactly the same dimensions (length and breadth), only the labels for length and breadth are possibly swapped.

∴ Yes, two rectangles that have the same area and the same perimeter must be congruent to each other.

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., $\dfrac{1}{2}$(a + b)h.

Answer

Let the parallel sides of the trapezium be a and b, and let its height be h.

Take another congruent copy of the same trapezium and join it with the given trapezium as shown below.

The two congruent trapezia together form a parallelogram.

Base of the parallelogram = a + b

Height of the parallelogram = h

Area of parallelogram = base × height

= (a + b)h

Since the parallelogram is made up of two congruent trapezia,

Area of one trapezium = $\dfrac{1}{2}$ × Area of parallelogram

= $\dfrac{1}{2}$ × (a + b)h

= $\dfrac{1}{2}$(a + b)h

Hence, area of trapezium = $\dfrac{1}{2}$(sum of parallel sides) × height.

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).

Answer

Consider a trapezium ABCD in which AB ∥ CD.

Let AB = a, CD = b, and the height (perpendicular distance between AB and CD) = h.

Join the diagonal AC. This divides the trapezium ABCD into two triangles: △ABC and △ACD.

For △ABC:

base = AB = a

The height of △ABC = perpendicular distance from C to AB = h (since AB ∥ CD).

We have the formula,

Area of △ABC = $\dfrac{1}{2}$ x base x height

$= \dfrac{1}{2} \times a \times h \\[1em] = \dfrac{1}{2} ah$

For △ACD:

base = CD = b

The height of △ACD = perpendicular distance from D to AB = h (since AB ∥ CD).

Area of △ACD = $\dfrac{1}{2}$ x base x height

$= \dfrac{1}{2} \times b \times h \\[1em] = \dfrac{1}{2} bh$

Area of trapezium ABCD = Area of △ABC + Area of △ACD

$= \dfrac{1}{2} ah + \dfrac{1}{2} bh \\[1em] = \dfrac{1}{2} h(a + b) \\[1em] = \dfrac{1}{2} (a + b) \times h$

∴ Area of trapezium = $\bold {\dfrac{1}{2}}$ x (sum of parallel sides) x height.

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?

Answer

Consider a trapezium ABCD with AB ∥ CD.

Let AB = a, CD = b, and the height (perpendicular distance between AB and CD) = h.

Take a second identical copy of trapezium ABCD, say A'B'C'D'.

Rotate the second trapezium A'B'C'D' through 180° and place it next to ABCD such that the non-parallel side BC of the first trapezium coincides with the corresponding non-parallel side of the second trapezium.

The resulting figure is a parallelogram, because:

• The top side of the parallelogram has length = a + b (formed by joining AB and the corresponding side of the rotated copy).

• The bottom side of the parallelogram also has length = a + b.

• Both pairs of opposite sides are equal and parallel.

The height of this parallelogram is the same as the height of the trapezium = h.

We know the formula,

Area of parallelogram = base x height

= (a + b) x h

Since the parallelogram is made up of two identical trapeziums:

Area of parallelogram = 2 x Area of trapezium

⇒ Area of trapezium = $\dfrac{1}{2}$ x Area of parallelogram

= $\dfrac{1}{2}$ x (a + b) x h

∴ Area of trapezium = $\mathbf{\dfrac{1}{2}}$ x (sum of parallel sides) x height.

Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.

Answer

(i) Using algebra:

Consider a kite ABCD in which AB = AD and CB = CD.

Let the diagonals AC and BD intersect at O. Let:

AC = d₁ and BD = d₂

In a kite, the diagonal AC is the perpendicular bisector of the diagonal BD.

So, BO = OD = $\dfrac{d_2}{2}$ and AC ⊥ BD.

The diagonal AC divides the kite into two triangles: △ABC and △ADC.

For △ABC:

base = AC = d₁

height = BO = $\dfrac{d_2}{2}$

Area of △ABC = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2} \times d_1 \times \dfrac{d_2}{2}$

= $\dfrac{1}{4}$ x d₁ x d₂

For △ADC:

base = AC = d₁

height = OD = $\dfrac{d_2}{2}$

Area of △ADC = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2} \times d_1 \times \dfrac{d_2}{2}$

= $\dfrac{1}{4}$ x d₁ x d₂

Area of kite ABCD = Area of △ABC + Area of △ADC

= $\dfrac{1}{4}$ x d₁ x d₂ + $\dfrac{1}{4}$ x d₁ x d₂

= $\dfrac{1}{2}$ x d₁ x d₂

∴ Area of kite = $\mathbf{\dfrac{1}{2}}$ x (product of its diagonals).

(ii) Using geometry:

Consider a kite ABCD with diagonals AC = d₁ and BD = d₂.

Enclose the kite ABCD inside a rectangle PQRS such that:

• The sides PQ and SR of the rectangle are parallel to BD and pass through A and C respectively.

• The sides PS and QR of the rectangle are parallel to AC and pass through B and D respectively.

Then:

PQ = SR = BD = d₂

PS = QR = AC = d₁

Area of rectangle PQRS = PQ x QR

= d₂ x d₁

= d₁ x d₂

Now, the diagonals AC and BD divide the rectangle PQRS into 4 smaller rectangles. The kite occupies exactly half of each of these 4 smaller rectangles (as the four triangles outside the kite are congruent in pairs to the four triangles inside the kite).

⇒ Area of kite = $\dfrac{1}{2}$ x Area of rectangle PQRS

= $\dfrac{1}{2}$ x d₁ x d₂

∴ Area of kite = $\mathbf{\dfrac{1}{2}}$ x (product of its diagonals).

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!

(ii) ∆ABC has sides a, b, c, and ∆PQR has sides 2a, 2b, 2c. Show that ∆PQR has 4 times the area of ∆ABC. Does this mean that 4 copies of ∆ABC will fit into ∆PQR? Check and see!

(iii) ∆ABC has sides a, b, c, and ∆PQR has sides 3a, 3b, 3c. Show that ∆PQR has 9 times the area of ∆ABC. Does this mean that 9 copies of ∆ABC will fit into ∆PQR? Check and see!

Answer

(i) Rectangle ABCD has sides a, b, and rectangle PQRS has sides 2a, 2b.

Given:

Sides of rectangle ABCD = a, b

Sides of rectangle PQRS = 2a, 2b

Area of rectangle ABCD = a x b = ab

Area of rectangle PQRS = 2a x 2b = 4ab

$\dfrac{\text{Area of PQRS}}{\text{Area of ABCD}} = \dfrac{4ab}{ab} = 4$

⇒ Area of PQRS = 4 x Area of ABCD.

Yes, 4 copies of rectangle ABCD will fit exactly into rectangle PQRS.

We can divide rectangle PQRS by drawing one line parallel to its length through its midpoint and another line parallel to its breadth through its midpoint. This divides PQRS into 4 smaller rectangles, each having sides a and b — i.e., each congruent to rectangle ABCD.

∴ 4 copies of rectangle ABCD fit exactly into rectangle PQRS.

(ii) ∆ABC has sides a, b, c, and ∆PQR has sides 2a, 2b, 2c.

Given:

Sides of △ABC = a, b, c

Sides of △PQR = 2a, 2b, 2c

Let s be the semi-perimeter of △ABC.

s = $\dfrac{1}{2}$(a + b + c)

By Heron's formula:

Area of △ABC = $\sqrt{s(s - a)(s - b)(s - c)}$

For △PQR, the semi-perimeter is:

s' = $\dfrac{1}{2}$(2a + 2b + 2c) = (a + b + c) = 2s

Area of △PQR = $\sqrt{s'(s' - 2a)(s' - 2b)(s' - 2c)}$

$= \sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)} \\[1em] = \sqrt{2s \times 2(s - a) \times 2(s - b) \times 2(s - c)} \\[1em] = \sqrt{16 \times s(s - a)(s - b)(s - c)} \\[1em] = 4\sqrt{s(s - a)(s - b)(s - c)} \\[1em] = 4 \times \text{Area of } \triangle ABC$