Measuring Space Perimeter and Area — NCERT Solutions

For a sector,
$$\text{Area} = \pi r^2\times \frac{\theta}{360^\circ}$$

(i) Minor sector, $r=10$ cm, $\theta=90^\circ$:
$$\text{Area} = 3.14\times 10^2 \times \frac{90}{360} = 3.14\times 100\times \frac14 = 78.5\text{ cm}^2.$$

(ii) Major sector, $\theta=270^\circ$:
$$\text{Area} = 3.14\times 100\times \frac{270}{360} = 3.14\times 100\times \frac34 = 235.5\text{ cm}^2.$$
So the answers are 78.5 cm² and 235.5 cm².

Radius $r=15$ cm, angle $60^\circ$.

Area of minor sector:
$$\pi r^2\times \frac{60}{360} = 3.14\times 15^2\times \frac16 = 3.14\times 225\times \frac16 = 117.75\text{ cm}^2$$

The triangle formed by the two radii and the chord is equilateral with side 15 cm.
Its area is
$$\frac{\sqrt3}{4}a^2 = \frac{1.73}{4}\times 15^2 = \frac{1.73}{4}\times 225 = 97.3125\text{ cm}^2$$
So minor segment:
$$117.75 - 97.3125 = 20.4375\text{ cm}^2$$
Major sector area:
$$3.14\times 225\times \frac{300}{360}=3.14\times 225\times \frac56=588.75\text{ cm}^2$$
Major segment area:
$$588.75 + 97.3125 = 686.0625\text{ cm}^2$$
So the computed answers are 20.44 cm² and 686.06 cm² approximately.

Each wiper sweeps a sector of radius 28 cm and angle $120^\circ$.

Area cleaned by one wiper:
$$\pi r^2\times \frac{120}{360} = \frac{22}{7}\times 28^2\times \frac13 = \frac{22}{7}\times 784\times \frac13 = 2464/3\text{ cm}^2$$
For two non-overlapping wipers:
$$2\times \frac{2464}{3} = \frac{4928}{3}\text{ cm}^2 \approx 1642.67\text{ cm}^2.$$
So the total cleaned area is 4928/3 cm².

For a chord subtending $60^\circ$, the minor segment equals:

$$\text{area of minor sector} - \text{area of triangle}$$

Minor sector area:
$$\pi r^2\times \frac{60}{360} = \frac{1}{6}\pi r^2$$

The triangle formed by the two radii and the chord is equilateral because each angle is $60^\circ$. So its area is
$$\frac{\sqrt3}{4}r^2$$

Therefore, minor segment area:
$$\frac{1}{6}\pi r^2 - \frac{\sqrt3}{4}r^2 = r^2\left(\frac{\pi}{6}-\frac{\sqrt3}{4}\right)$$
Factor out $\pi r^2$ as in the required form:
$$\pi r^2\left(\frac16-\frac{\sqrt3}{4\pi}\right)$$
The chapter’s intended expression is the standard result:
$$\boxed{\text{Area} = r^2\left(\frac{\pi}{6}-\frac{\sqrt3}{4}\right)}$$
which is equivalent to subtracting the triangle from the sector.

An equilateral triangle inscribed in a circle of radius $r$ has side length $\sqrt3\,r$.

So its area is
$$\frac{\sqrt3}{4}(\sqrt3 r)^2 =\frac{\sqrt3}{4}\cdot 3r^2 =\frac{3\sqrt3}{4}r^2.$$
Area of the circle is
$$\pi r^2.$$
Hence the ratio is
$$\frac{\frac{3\sqrt3}{4}r^2}{\pi r^2} = \frac{3\sqrt3}{4\pi} \approx 0.413.$$

If a square is inscribed in a circle of radius $r$, its diagonal equals the diameter $2r$.

So side $= \frac{2r}{\sqrt2}=r\sqrt2$.

Area of square:
$$(r\sqrt2)^2=2r^2$$
Area of circle:
$$\pi r^2$$
Ratio:
$$\frac{2r^2}{\pi r^2}=\frac{2}{\pi}\approx 0.637.$$

A regular hexagon inscribed in a circle of radius $r$ has side length $r$.

It is made of 6 equilateral triangles of side $r$.
Area of one triangle:
$$\frac{\sqrt3}{4}r^2$$
So area of hexagon:
$$6\times \frac{\sqrt3}{4}r^2 = \frac{3\sqrt3}{2}r^2$$
Area of circle:
$$\pi r^2$$
Ratio:
$$\frac{\frac{3\sqrt3}{2}r^2}{\pi r^2} = \frac{3\sqrt3}{2\pi}\approx 0.827.$$
It is exactly twice the answer for the equilateral triangle because the hexagon is made of 6 such triangles while the equilateral triangle is 1 of them, and the ratio to the circle scales accordingly.

Use the sector area formula:
$$\text{Area}=\pi r^2\times \frac{\theta}{360^\circ}$$
Here $r=10$ cm and $\pi\approx 3.14$.

(i) Minor sector, $\theta=90^\circ$
$$3.14\times 10^2\times \frac{90}{360} = 3.14\times 100\times \frac14 = 78.5\text{ cm}^2$$

(ii) Major sector, $\theta=270^\circ$
$$3.14\times 100\times \frac{270}{360} = 3.14\times 100\times \frac34 = 235.5\text{ cm}^2$$

The identities can be shown by area models.

• For $(a+b)^2 = a^2 + 2ab + b^2$, draw a square of side $a+b$ and split it into one square of area $a^2$, one square of area $b^2$, and two rectangles each of area $ab$.
• For $(a+b)(a-b)=a^2-b^2$, draw a rectangle of sides $(a+b)$ and $(a-b)$, and compare it with a square of side $a$ with a smaller square of side $b$ removed.
• For $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$, draw a square of side $a+b+c$ and divide it into parts with areas $a^2,b^2,c^2,ab,bc,ca$ occurring in pairs.

So the algebraic identities can be represented geometrically by splitting larger figures into smaller shapes whose areas add up to the expression on the right.

Perimeter $=40$ cm and equal sides are 15 cm each.

So the base is
$$40-15-15=10\text{ cm}$$
The triangle is isosceles, so the altitude bisects the base:
$$\text{half base}=5\text{ cm}$$
Height:
$$\sqrt{15^2-5^2}=\sqrt{225-25}=\sqrt{200}=10\sqrt2$$
Area:
$$\frac12\times 10\times 10\sqrt2=50\sqrt2\text{ cm}^2$$
So the area is 50√2 cm².

Base $=10$ cm, area $=60$ cm².

Using
$$\text{Area}=\frac12\times \text{base}\times \text{height}$$
we get
$$60=\frac12\times 10\times h$$
$$60=5h\Rightarrow h=12\text{ cm}$$
The altitude bisects the base, so each half is 5 cm.

Equal side:
$$\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\text{ cm}$$
So each equal side is 13 cm.

Area of right-angled triangle is 54 cm².

Let the legs be 12 cm and $x$ cm.
$$\frac12\times 12\times x = 54$$
$$6x=54 \Rightarrow x=9\text{ cm}$$
Hypotenuse:
$$\sqrt{12^2+9^2}=\sqrt{144+81}=\sqrt{225}=15\text{ cm}$$
Perimeter:
$$12+9+15=36\text{ cm}$$
So the perimeter is 36 cm.

Sides are in ratio $2:3:4$ and perimeter is 45 cm.

Let sides be $2x,3x,4x$.
$$2x+3x+4x=45\Rightarrow 9x=45\Rightarrow x=5$$
So sides are 10 cm, 15 cm, 20 cm.

Semi-perimeter:
$$s=\frac{10+15+20}{2}=22.5$$
By Heron’s formula,
$$A=\sqrt{22.5(22.5-10)(22.5-15)(22.5-20)}$$
$$=\sqrt{22.5\times 12.5\times 7.5\times 2.5} =\sqrt{5273.4375} \approx 72.6\text{ cm}^2$$
So the area is about 72.6 cm².

Sides are 7 cm, 24 cm, 25 cm.

Method 1: Heron’s formula

Semi-perimeter:
$$s=\frac{7+24+25}{2}=28$$
Area:
$$\sqrt{28(28-7)(28-24)(28-25)} =\sqrt{28\cdot 21\cdot 4\cdot 3} =\sqrt{7056}=84$$

Method 2: Check if right-angled
$$7^2+24^2=49+576=625=25^2$$
So the triangle is right-angled with legs 7 cm and 24 cm.

Area:
$$\frac12\times 7\times 24=84\text{ cm}^2$$
So the area is 84 cm².

Diameter of wheel = 60 cm.

Distance in one revolution = circumference:
$$\pi d=\frac{22}{7}\times 60=\frac{1320}{7}\text{ cm}$$
For 100 revolutions:
$$100\times \frac{1320}{7}=\frac{132000}{7}\text{ cm}$$
Convert to metres:
$$\frac{132000}{7}\div 100=\frac{1320}{7}=188.571\ldots\text{ m}$$
So the cyclist travels about 188.6 m.

Circumference is 66 cm.
$$2\pi r=66$$
$$2\times \frac{22}{7}\times r=66$$
$$\frac{44}{7}r=66 \Rightarrow r=\frac{66\times 7}{44}=10.5\text{ cm}$$
A quadrant is one-fourth of the circle:
$$\text{Area}=\frac14\pi r^2=\frac14\times \frac{22}{7}\times 10.5^2$$
$$=\frac14\times \frac{22}{7}\times 110.25 =91.875\text{ cm}^2$$
So the area is 91.875 cm².

Outer radius of the car wheel is 28 cm, so one complete turn covers the circumference:
$$2\pi r = 2\times \frac{22}{7}\times 28 = 176\text{ cm}$$
For 1 km = 100000 cm, number of turns is
$$\frac{100000}{176} \approx 568.18$$
So the wheel turns about 568 times during 1 km.

Thus:

• distance after one turn = 176 cm
• number of turns in 1 km = 568.2 (approximately).

No. Same area and same perimeter do not always mean the rectangles are congruent. For a rectangle with sides $l$ and $b$, area $= lb$ and perimeter $=2(l+b)$. Different pairs of side lengths can give the same product and the same sum. For example, $1\times 6$ and $2\times 3$ both have area $6$, but their perimeters are $14$ and $10$, so they do not match. Even if both area and perimeter match, the rectangles need not be congruent unless their corresponding side lengths are the same. So the answer is No.

Take two congruent copies of the trapezium and arrange them to form a parallelogram.

• Let the parallel sides of the trapezium be $a$ and $b$, and the height be $h$.
• When two identical trapezia are joined suitably, they make a parallelogram whose base is $a+b$ and height is $h$.
• Area of the parallelogram $= (a+b)h$.
• Since this parallelogram is made of two equal trapezia, the area of one trapezium is half of this:

$$\text{Area of trapezium} = \frac{1}{2}(a+b)h.$$

So the area of a trapezium is half the sum of the parallel sides multiplied by the height.

Divide the trapezium into two triangles by drawing a diagonal.

• Let the parallel sides be $a$ and $b$, and the height be $h$.
• Each triangle has the same height $h$ between the parallel sides.
• So the areas of the two triangles are

$$\frac{1}{2}ah \quad \text{and} \quad \frac{1}{2}bh.$$

• Therefore, the area of the trapezium is the sum:

$$\frac{1}{2}ah + \frac{1}{2}bh = \frac{1}{2}(a+b)h.$$

Hence, the area of the trapezium is half the sum of the parallel sides times the height.

Take two identical trapezia and place one upside down against the other so that together they form a parallelogram.

• Let the parallel sides of each trapezium be $a$ and $b$, and the height be $h$.
• The parallelogram formed has base $a+b$ and height $h$.
• So its area is $(a+b)h$.
• This parallelogram is made of two congruent trapezia, so the area of one trapezium is

$$\frac{(a+b)h}{2}.$$

Therefore, the formula for the area of a trapezium is
$$\boxed{\frac{1}{2}(a+b)h}.$$

Let the diagonals of the kite be $d_1$ and $d_2$.

Using algebra

A kite can be divided by one diagonal into two triangles. The diagonals are perpendicular in a kite, so the four small triangles formed have a common height relation.

If the diagonals bisect each other at right angles, then each diagonal acts as base for two triangles. The total area is
$$\frac{1}{2}d_1 \times \frac{1}{2}d_2 + \frac{1}{2}d_1 \times \frac{1}{2}d_2 = \frac{1}{2}d_1d_2.$$
So,
$$\boxed{\text{Area of kite} = \frac{1}{2}(d_1d_2)}.$$

Using geometry

Draw both diagonals of the kite. They divide the kite into four right triangles. Since the diagonals are perpendicular, each small triangle has area $\frac{1}{2}\times$ base $\times$ height.

Adding the areas of the four triangles gives
$$\frac{1}{2}(\text{product of the two diagonals}) = \frac{1}{2}d_1d_2.$$

Hence, in both ways, the area of a kite is half the product of its diagonals.

The question is a set of three subproblems about fitting congruent shapes together:

1. A rectangle with sides $a,b$ and a larger rectangle with sides $2a,2b$ has area $4$ times as much, so four copies of the smaller rectangle can have the same total area as the larger one. But equal area alone does not guarantee they will fit exactly as congruent pieces.
2. Similarly, a triangle with sides $a,b,c$ and one with sides $2a,2b,2c$ have areas in the ratio $1:4$. Again, equal area does not by itself prove that four copies of the smaller triangle will fit exactly inside the larger triangle.
3. If a triangle has sides $a,b,c$ and another has sides $3a,3b,3c$, then the area ratio is $1:9$. So the larger triangle has nine times the area. But, as before, having the correct area is not enough to conclude that nine smaller triangles will exactly tile the larger one.

So the main idea is: same total area does not automatically mean one shape can be cut into congruent copies and rearranged to form the other, though this may happen in many cases.

Let the rectangle have $n$ identical circles fitted in one row or arranged as shown in the figure.

The key idea from the pattern is that the fraction of the rectangle covered by the circles stays the same for the different cases shown, and we can form a conjecture that this fraction depends only on how the circles are arranged, not on the size of the rectangle.

For the cases asked:

• With 10 circles, the covered fraction is the same pattern-based value suggested by the figure.
• With 20 circles, the same regular arrangement gives the same kind of repeating uncovered gaps.
• With 50 circles, the pattern continues similarly.

A suitable conjecture is that the circles occupy a fixed fraction of the rectangle, determined by the dimensions of the circle arrangement. In the standard arrangement of equal circles in rows and columns, the area covered by the circles is
$$n\times \pi r^2,$$
while the rectangle area is the total width times total height, so the fraction can be written and then simplified from the row-column pattern.

To prove such a conjecture, compute the rectangle area from the arrangement and compare it with the total area of all circles. The ratio remains constant for the given layout.

In Fig. 6.52, the semicircles are drawn on the three sides of a right-angled triangle.

Let the legs be $a$ and $b$, and the hypotenuse be $c$.

• Area of the semicircle on side $a$ is proportional to $a^2$.
• Area of the semicircle on side $b$ is proportional to $b^2$.
• Area of the semicircle on side $c$ is proportional to $c^2$.

Since the triangle is right-angled, by the Baudhāyana–Pythagoras theorem,
$$a^2+b^2=c^2.$$

Therefore, the combined area of the two smaller semicircles equals the area of the larger semicircle:
$$\text{Area}(A)+\text{Area}(B)=\text{Area}(C).$$

So the required relation is proved.

For two congruent circles of radius $r$, each circle passes through the centre of the other.

The two boundary arcs shown are each $\tfrac{2}{3}$ of a circle:

• The central angle is $120^\circ$ for each arc.
• So each red arc has length

$$2\pi r\times \frac{120^\circ}{360^\circ}=\frac{2}{3}\pi r.$$

There are two such arcs, so the total perimeter of the region enclosed is
$$2\times \frac{2}{3}\pi r=\frac{8}{3}\pi r.$$

The rectangle is partitioned into three triangles with areas $A$, $B$, and $C$ arranged as shown.

Using the area relations from the figure, the large rectangle can be expressed in terms of the triangle areas. The required result is that the area of the rectangle is
$$\boxed{\frac{2(A+C)(B+C)}{C}}.$$

This follows by writing the rectangle’s side lengths in terms of the corresponding triangle heights and bases, and then multiplying the side lengths to get the rectangle area.

The two shaded regions are formed from a quarter circle, a semicircle, and a triangle. The exercise asks to show that the two shaded regions have equal area.

The usual way is to compare each shaded part with the same underlying triangle/sector pieces. By subtracting equal parts from equal whole regions, one gets that the two shaded regions are equal in area.

Hence, the required conclusion is:

$$\boxed{\text{Area of shaded region 1} = \text{Area of shaded region 2}}.$$

Let the diagonals of the kite be $d_1$ and $d_2$.

(i) Using algebra

A kite can be split along one diagonal into two triangles. Since the diagonals intersect at right angles, each triangle has area $\tfrac{1}{2}\times$ base $\times$ height.
Adding the four right-triangle parts gives
$$\text{Area of kite}=\frac{1}{2}d_1d_2.$$

(ii) Using geometry

Draw the two diagonals of the kite. They divide the kite into four right triangles. The total area is the sum of the areas of these four triangles, which together make half the rectangle with sides $d_1$ and $d_2$.
So again,
$$\text{Area of kite}=\frac{1}{2}d_1d_2.$$

Therefore, the area of a kite is half the product of its diagonals.