Measuring Space: Perimeter and Area

Measuring Space: Perimeter and Area – NCERT Solutions Class 9 Maths Ganita Manjari includes all the questions with solutions given in NCERT Class 9 Ganita Manjari textbook.

NCERT Solutions Class 9

Measuring Space: Perimeter and Area – NCERT Solutions

Q.1:

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

Solution:

We use the circumference formula $C=2 \pi r$.
Given $C=44 {~cm}$ and taking $\pi \approx \frac{22}{7}$:

$44=2 \times \frac{22}{7} \times r$
$44=\frac{44}{7} \times r$
$r=\frac{44 \times 7}{44}=7 {~cm}$

The radius of the circle is ${7 \ c m}$.

Q.2:

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

Solution:

1. Radius 7 cm: $C=2 \times \pi \times 7 \approx 43.982$ $\Longrightarrow 44.0 {~cm}$
2. Radius 10 cm: $C=2 \times \pi \times 10 \approx 62.831$ $\Longrightarrow {6 2 . 8} {~cm}$
3. Radius 12 cm: $C=2 \times \pi \times 12 \approx 75.398$ $\Longrightarrow {7 5 . 4} {~cm}$

Q.3:

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

Solution:

1. Radius $r=3.5 \ {cm}$, Angle $\theta=60^{\circ}$
   Step 1: Plug values into the formula: $l=\frac{60}{360} \times 2 \times \frac{22}{7} \times 3.5$ Step 2: Simplify the fraction and the radius $\left(3.5=\frac{7}{2}\right)$: $l=\frac{1}{6} \times 2 \times \frac{22}{7} \times \frac{7}{2}$ Step 3: Cancel the common terms (2 and 7): $l=\frac{1}{6} \times 22=\frac{11}{3} \approx 3.67 \ {cm}$
2. Radius $r=6.3 \ {m}$, Angle $\theta=120^{\circ}$
   Step 1: Plug values into the formula: $l=\frac{120}{360} \times 2 \times \frac{22}{7} \times 6.3$ Step 2: Simplify the fraction $\left(\frac{120}{360}=\frac{1}{3}\right)$ and the decimal: $l=\frac{1}{3} \times 2 \times \frac{22}{7} \times 6.3$ Step 3: Divide 6.3 by 7 to get 0.9: $l=\frac{1}{3} \times 2 \times 22 \times 0.9$ Step 4: Multiply the remaining numbers: $l=\frac{1}{3} \times 44 \times 0.9=44 \times 0.3={1 3 . 2} {m}$

Q.4:

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^o.

Solution:

Step 1: Calculate the Arc Length (l)
The formula for the curved part is $l=\frac{\theta}{360} \times 2 \pi r$.
Using $\theta=75^{\circ}, r=14 {~cm}$, and $\pi \approx \frac{22}{7}$:

$l=\frac{75}{360} \times 2 \times \frac{22}{7} \times 14$

• Simplify the fraction: $\frac{75}{360}=\frac{5}{24}$.
• Simplify the radius and $\pi: \frac{14}{7}=2$.
• Multiply: $l=\frac{5}{24} \times 2 \times 22 \times 2=\frac{5}{24} \times 88$.
• Final Arc Length: $l=\frac{55}{3} \approx 18.33 {~cm}$.

Step 2: Calculate the Two Radii
The two straight sides are both radii (r):

$\text { Straight sides }=2 \times r=2 \times 14=28 {~cm}$

Step 3: Find Total Perimeter
Add the curved portion to the straight portions:

$\text { Perimeter }=\text { Arc Length }+2 r$
Perimeter $=18.33+28={4 6 . 3 3} {cm}$

The total perimeter of the sector is approximately ${4 6 . 3 3 ~ c m}$.

Q.5:

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

Solution:

Step 1: Identify the Components
The shape consists of:

• Two straight horizontal segments, each measuring 80 m .
• Two curved ends, which are semicircles. Looking at the vertical dashed line, the diameter of these semicircles is 60 m.

Step 2: Calculate the Straight Sections
The total length of the top and bottom straight edges is:

$\text { Straight Length }=80+80=160 {~m}$

Step 3: Calculate the Curved Sections
Two identical semicircles combine to form one full circle with a diameter $(D)$ of 60 m . Using the formula $C=\pi D$:

$\text { Curved Length }=\pi \times 60$

Using $\pi \approx 3.14$ :

$\text { Curved Length } \approx 3.14 \times 60=188.4 {~m}$

Q.6:

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

Solution:

Step 1: Identify the Diameters and Arcs

• The outer diameter $\left(D_1\right)$ is ${1 2 ~ c m}$, meaning the outer arc is a semicircle.
• The inner diameter $\left(D_2\right)$ is ${8 ~ c m}$, meaning the inner arc is also a semicircle.

Step 2: Calculate the Curved Arcs
Since the perimeter of a full circle is $\pi D$, the length of a semicircle is $\frac{1}{2} \pi D$.

• Outer Arc: $\frac{1}{2} \times \pi \times 12=6$$\pi \approx 6 \times 3.14=18.84 {~cm}$
• Inner Arc: $\frac{1}{2} \times \pi \times 8=4 \pi \approx 4 \times 3.14=12.56 {~cm}$

Step 3: Calculate the Straight Base Segments
The total width of the bottom is 12 cm, and the gap in the middle is 8 cm. The two remaining straight segments at the bottom represent the thickness of the ring.

• Total length of both segments: $12-8=4 {~cm}$
• (Each individual segment is 2 cm ).

Step 4: Find the Total Perimeter
Now, we add all these boundary lengths together:

$\text { Total Perimeter }=$ $\text { Outer Arc }+ \text { Inner Arc }+ \text { Base Segments }$
$\text { Total Perimeter }=18.84+12.56+4={3 5 . 4} {cm}$

The total perimeter of the shape is ${3 5 . 4 ~ c m}$.

Q.7:

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

Solution:

Step 1: Identify the Shape Components

• The central dashed figure is a square with a side length of ${1 0 ~ c m}$.
• Attached to each of the four sides is a semicircle.
• The diameter $(D)$ of each semicircle is equal to the side of the square, which is ${1 0 ~ c m}$.

Step 2: Calculate the Curved Arcs
The perimeter of the shape consists only of the four semicircles (the dashed lines of the square are inside the shape and not part of the perimeter).

• Four semicircles are equivalent to two full circles.
• The circumference $(C)$ of one full circle is $\pi D$.
• For two full circles: Total Curved Length $=2 \times \pi \times D$.

Step 3: Numerical Calculation
Using $D=10 {~cm}$ and $\pi \approx 3.14$ :

$\text { Total Perimeter }=2 \times 3.14 \times 10$
$\text { Total Perimeter }=6.28 \times 10$
$\text { Total Perimeter }={6 2 . 8 ~ c m}$

The total perimeter of the shape is ${6 2 . 8 ~ c m}$.

Q.8:

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

Solution:

Step 1: Analyze the Components

• The central dashed figure is an equilateral triangle with a side length of ${1 2 ~ c m}$.
• Attached to each of the three sides is a semicircle.
• The diameter $(D)$ of each semicircle is equal to the side of the triangle: ${1 2}$ cm.

Step 2: Calculate the Length of One Semicircle
The circumference of a full circle is $\pi D$. Therefore, one semicircle is:

$\text { Arc Length }=\frac{1}{2} \times \pi \times D$

Using $\pi \approx 3.14$:

$\text { Arc Length }=\frac{1}{2} \times 3.14 \times 12=18.84 {~cm}$

Step 3: Calculate the Total Perimeter
The perimeter consists of three such semicircular arcs (the internal dashed lines are not counted):

$\text { Total Perimeter }=3 \times 18.84$
$\text { Total Perimeter }={5 6 . 5 2 ~ c m}$

The total perimeter of the shape is ${5 6 . 5 2 ~ c m}$.

Q.9:

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

Solution:

Step 1: Analyze the Grid Structure
Looking at the internal dashed lines, the central part of the figure is a ${3} \times {3}$ grid of smaller squares.

• The total width of the grid is labeled as ${1 4 ~ c m}$.
• Since there are 3 equal squares across, the side of one small square is $14 \div 3=\frac{14}{3} {~cm}$.

Step 2: Identify the Semicircles
The perimeter is made up of four identical semicircles.

• Each semicircle is attached to the middle square of each outer side.
• The diameter $(D)$ of each semicircle is equal to the side of one small square: $D=\frac{14}{3} {~cm}$.

Step 3: Calculate the Curved Length
Four semicircles are equivalent to two full circles.
The circumference $(C)$ of one circle is $\pi D$.
Total Curved Length $=2 \times \pi \times \frac{14}{3}$
Using $\pi \approx \frac{22}{7}$:

$\text { Total Curved Length }=2 \times \frac{22}{7} \times \frac{14}{3}$
$\text { Total Curved Length }=2 \times 22 \times \frac{2}{3}$
$\text { Total Curved Length }=\frac{88}{3} \approx 29.33 {~cm}$

Step 4: Add the Straight Boundary Segments
The perimeter also includes the straight edges of the outer squares that are not covered by semicircles.

• On each of the 4 sides, there are 2 straight segments (the outer edges of the corner squares).
• Total straight segments $=4$ sides $\times 2=8$ segments.

$\text { Total Straight Length }$$=8 \times \frac{14}{3}=\frac{112}{3} \approx 37.33 {~cm}$

Step 5: Final Perimeter

$\text { Total Perimeter }$$=\frac{88}{3}+\frac{112}{3}=\frac{200}{3}={6 6. 6 7} \ {cm}$

The total perimeter of the shape is ${6 6. 6 7 ~ c m}$.

Q.10:

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

Solution:

Step 1: Analyze the Grid and Diameters
The base of the figure is a dashed line measuring 28 cm.

• Looking at the markings, this base is divided into four equal segments.
• The length of each small segment is $28 \div 4={7 ~ c m}$.

Step 2: Calculate the Large Semicircle
The top part of the shape is a single large semicircle.

• Its diameter $\left(D_{\text {large }}\right)$ is the entire base: ${2 8 ~ c m}$.
• Length $=\frac{1}{2} \times \pi \times 28=14 \pi$.

Step 3: Calculate the Small Semicircles
The bottom part of the shape consists of four small semicircles (two curving up, two curving down).

• The diameter $\left(D_{\text {small }}\right)$ of each small semicircle is ${7} {~ c m}$.
• Combined, four semicircles equal two full circles.
• Length $=2 \times(\pi \times 7)=14 \pi$.

Step 4: Find the Total Perimeter
Add the lengths of the upper and lower boundaries together:

Total Perimeter $=14 \pi($top$)+14 \pi($bottom$)=28 \pi$
Using $\pi \approx \frac{22}{7}$:
Total Perimeter $=28 \times \frac{22}{7}$

Total Perimeter $=4 \times 22={8 8} {cm}$
The total perimeter of the shape is 88 cm.

Q.11:

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

Solution:

Step 1: Find the Missing Side (Hypotenuse)
The central dashed figure is a right-angled triangle with base 8 cm and height 6 cm. Using the Pythagorean theorem:

Hypotenuse² = $6^2+8^2$
Hypotenuse² = $36+64=100$
$\text {Hypotenuse }=\sqrt{100}={1 0}\ {cm}$

Step 2: Identify the Three Semicircles
The perimeter consists of three semicircles with diameters $(D)$ of $6 {~cm}, 8 {~cm}$, and 10 cm.

• Semicircle $1(D=6): \frac{1}{2} \times \pi \times 6=3 \pi$
• Semicircle $2(D=8): \frac{1}{2} \times \pi \times 8=4 \pi$
• Semicircle $3(D=10): \frac{1}{2} \times \pi \times 10=5 \pi$

Step 3: Calculate Total Perimeter
Sum the lengths of all three arcs:

$\text { Total Perimeter }=3 \pi+4 \pi+5 \pi=12 \pi$

Using $\pi \approx 3.14$:

$\text { Total Perimeter }=12 \times 3.14={3 7. 6 8}\ {cm}$

The total perimeter of the shape is ${3 7. 6 8 ~ c m}$.

Q.12:

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

Solution:

Step 1: Analyze the Base and Diameters
The base is divided into three equal segments, each measuring ${4}$ cm.

• The diameter of each small semicircle $\left(D_{\text {small }}\right)$ is ${4} {~ c m}$.
• The diameter of the large semicircle ($D_{\text {large }}$) is the sum of the three segments: $4+4+ 4={1 2}\ {cm}$.

Step 2: Calculate the Large Semicircle Arc
Using the formula for a semicircle $\left(l=\frac{1}{2} \pi D\right)$:

$\text { Large Arc }=\frac{1}{2} \times \pi \times 12=6 \pi$.

Using $\pi \approx 3.14$:

$\text { Large Arc } \approx 6 \times 3.14=18.84 {~cm}$

Step 3: Calculate the Three Small Semicircle Arcs
Each small arc has a diameter of 4 cm:

$\text { One Small Arc }=\frac{1}{2} \times \pi \times 4=2 \pi$

Since there are three such arcs:

$\text { Total Small Arcs }=3 \times 2 \pi=6 \pi$

Total Small Arcs $\approx 6 \times 3.14=18.84 {~cm}$

Step 4: Find the Total Perimeter
Sum the outer arc and the inner arcs:

Total Perimeter $=6 \pi+6 \pi=12 \pi$

Total Perimeter $\approx 12 \times 3.14={3 7 . 6 8} \ {cm}$
The total perimeter of the shape is ${3 7 . 6 8 ~ c m}$.

Q.13:

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

Solution:

Step 1: Analyze the Geometry and Diameters
The base is a dashed line divided into two equal segments of 10 cm each.

• Small Semicircles: There are two smaller semicircles (one curving up, one curving down). Their diameter $\left(D_{\text {small }}\right)$ is 10 cm.
• Large Semicircle: The top arc is one large semicircle. its diameter ($D_{\text {large }}$) is the sum of the two segments: $10+10={2 0 ~ c m}$.

Step 2: Calculate the Large Semicircle Arc
The length of a semicircle is half the circumference: $l=\frac{1}{2} \pi D$.

$\text { Large Arc }=\frac{1}{2} \times \pi \times 20=10 \pi$

Using $\pi \approx 3.14$ :

$\text { Large Arc } \approx 10 \times 3.14={3 1 . 4} {cm}$

Step 3: Calculate the Two Small Semicircle Arcs
Since there are two identical semicircles with a diameter of 10 cm, they combine to form one full circle:

Two Small Arcs $=\pi \times 10$

Two Small Arcs $\approx 3.14 \times 10={3 1 . 4} {cm}$

Step 4: Find the Total Perimeter
Add the outer large arc and the two smaller inner/outer arcs together:

Total Perimeter $=10 \pi+10 \pi=20 \pi$

Total Perimeter $\approx 20 \times 3.14={6 2 . 8} {cm}$
The total perimeter of the shape is ${6 2 . 8 ~ c m}$.

Q.14:

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?

Solution:

Step 1: Calculate the distance for one revolution
The diameter $(D)$ is ${5 6} {~ c m}$. The distance for one revolution is the circumference $(C)$:

$C=\pi D$

Using $\pi \approx \frac{22}{7}$:

$C=\frac{22}{7} \times 56$
$C=22 \times 8={1 7 6} {cm}$

The car travels ${1 7 6 ~ c m}$ (or ${1. 7 6 ~ m}$) in one revolution.

Step 2: Calculate revolutions for 10 km
First, convert the total distance into centimeters to match our circumference unit:

$10 {~km}=10 \times 1,000 \times 100=$ $1,000,000 {~cm}$

Now, divide the total distance by the distance covered in one revolution:

$\text { Revolutions }=\frac{\text { Total Distance }}{\text { Circumference }}$
$\text { Revolutions }=\frac{1,000,000}{176}$
$\text { Revolutions } \approx {5, 6 8 1. 8}$

The tyre makes approximately 5,682 revolutions over 10 km.

Q.15:

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

Solution:

1. Square-based Flower In this figure, the arcs are semicircles because their centers are at the midpoints of the square’s sides.
   Analyze the Arcs: The side of the square is ${1 4 ~ c m}$. Since the center of each arc is the midpoint, the radius ($r$) of each semicircle is 7 cm.
   Identify the Petals: Each of the ${4}$ petals is made of two semicircular arcs. There are ${8}$ semicircles in total forming the boundary of the petals.
   Calculate Length: Eight semicircles equal ${4}$ full circles. $\text { Total Perimeter }=4 \times(\pi \times D)=4 \times \pi \times 14$ Numerical Result: Using $\pi \approx \frac{22}{7}$: $\text { Total Perimeter }=4 \times \frac{22}{7} \times 14=4 \times 22 \times 2={1 7 6} {cm}$
2. Hexagon-based Flower In this figure, the arcs are sectors of a circle where the centers are the vertices of the hexagon.
   Analyze the Arcs: The side of the hexagon is ${4 2 ~ c m}$. The arcs meet at the center, so the radius $(r)$ of each arc is ${4 2 ~ c m}$.
   Identify the Angle: The interior angle of a regular hexagon is ${1 2 0}^{\circ}$. Therefore, each arc is $\frac{120}{360}=\frac{1}{3}$ of a circle.
   Identify the Petals: There are 6 petals, and each petal is formed by two arcs. This gives us 12 arcs in total.
   Calculate Length: Since each arc is $\frac{1}{3}$ of a circle, 12 arcs equal ${4}$ full circles ($12 \times \frac{1}{3}=$ 4). $\text { Total Perimeter }=$ $4 \times(2 \times \pi \times r)=4 \times 2 \times \pi \times 42$ Numerical Result: Using $\pi \approx \frac{22}{7}$: $\text { Total Perimeter }=8 \times \frac{22}{7} \times$$42=8 \times 22 \times 6={1 0 5 6} \ {cm}$

Q.16:

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

Solution:

Step 1: The Core Formula
The circumference $(C)$ is calculated as $C=2 \pi r$. This shows that $C$ is directly proportional to $r$, because $2 \pi$ is a constant value.

Step 2: Mathematical Proof
If we have two circles, their perimeters are $C_1=2 \pi r_1$ and $C_2=2 \pi r_2$. Setting up the ratio:

$\frac{C_1}{C_2}=\frac{2 \pi r_1}{2 \pi r_2}$

Step 3: Final Ratio
Canceling the constant $2 \pi$ leaves us with $\frac{C_1}{C_2}=\frac{r_1}{r_2}$. Given the perimeter ratio is $5: 4$, the radius ratio is also 5:4.

Q.17:

Find the area of triangle ADE in the figure.

Solution:

The area of triangle ADE is found by identifying its base and the perpendicular height relative to that base within the rectangle.

The triangle’s base $A D$ is ${8 ~ c m}$. Its height, the horizontal distance from $A D$ to vertex $E$, is the rectangle’s length, 10 cm.

$\text { Area }=\frac{1}{2} \times \text { base × height }$
$\text { Area }=\frac{1}{2} \times 8 \times 10=40 \ {cm}^2$

Q.18:

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.

Solution:

Step 1: Find the base of the side triangles
When heights are drawn from the top vertices, the bottom side $(40 {~cm})$ is split into a middle section (20 cm) and two equal segments.

$\text { Segment length }=\frac{40-20}{2}=10 {~cm}$

Step 2: Calculate the height ($h$)
Using the Pythagorean theorem on the side triangle with hypotenuse 26 cm and base 10 cm:

$h^2+10^2=26^2$
$h^2+100=676$
$h^2=576 \Rightarrow h=\sqrt{576}=24 {~cm}$

Step 3: Calculate the Area
The area formula is $\frac{1}{2} \times$ (sum of parallel sides) × height:

$\text { Area }=\frac{1}{2} \times(40+20) \times 24$
$\text { Area }=\frac{1}{2} \times 60 \times 24=30 \times 24={7 2 0} {cm}^2$

Q.19:

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

Solution:

Step 1: Find the missing side
The perimeter is the sum of all sides. Let the third side be $c$:

$8+11+c=32$
$19+c=32$
$c=13 {~cm}$

Step 2: Calculate the semi-perimeter ($s$)
The semi-perimeter is half of the total perimeter:

$s=\frac{32}{2}=16 {~cm}$

Step 3: Apply Heron’s Formula
The area $A$ is given by $\sqrt{s(s-a)(s-b)(s-c)}$:

$A=\sqrt{16(16-8)(16-11)(16-13)}$
$A=\sqrt{16 \times 8 \times 5 \times 3}$
$A=\sqrt{1920} \approx {4 3 . 8 2} \ {cm}^2$

Q.20:

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

Solution:

Step 1: Find the side lengths
Let the sides be $3 x, 5 x$, and $7 x$. The perimeter is the sum of these sides:

$3 x+5 x+7 x=300$
$15 x=300 \Rightarrow x=20$

The sides are:

• $a=3 \times 20=60 {~m}$
• $b=5 \times 20=100 {~m}$
• $c=7 \times 20=140 {~m}$

Step 2: Calculate the semi-perimeter ($s$)

$s=\frac{300}{2}=150 {~m}$

Step 3: Apply Heron’s Formula

$\text { Area }=\sqrt{s(s-a)(s-b)(s-c)}$
$\text { Area }=\sqrt{150(150-60)(150-100)(150-140)}$
$\text { Area }=\sqrt{150 \times 90 \times 50 \times 10}$
$\text { Area }=\sqrt{6,750,000}=1500 \sqrt{3} \approx {2 5 9 8. 0 8} \ {m}^2$

Q.21:

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.

Solution:

The area of a rhombus is calculated using the lengths of its two diagonals, which bisect each other at right angles.

Let the length of the shorter diagonal be $d$ and the longer diagonal be $2 d$. The formula for the area is:

$\text { Area }=\frac{1}{2} \times d_1 \times d_2$

Substituting the given values into the equation:

$128=\frac{1}{2} \times d \times 2 d$
$128=d^2$
$d=\sqrt{128}=\sqrt{64 \times 2}=8 \sqrt{2}$

The length of the shorter diagonal is approximately ${1 1 . 3 1 ~ c m}$.

Q.22:

ABCD is a parallelogram. P and Q are any two points on side AB . What can you say about the ratio area ($\triangle {PCD}$): area ($\triangle {QCD}$)?

Solution:

Step 1: Identify the Base
Both $\triangle P C D$ and $\triangle Q C D$ share the same base, which is the side $C D$ of the parallelogram.
Step 2: Determine the Height
The height of a triangle is the perpendicular distance from its base to the opposite vertex. Since vertices ${P}$ and ${Q}$ both lie on the line segment ${A B}$, and line ${A B}$ is parallel to line ${C D}$, the perpendicular distance from ${P}$ to ${C D}$ is the same as the perpendicular distance from ${Q}$ to ${C D}$.

This distance is equal to the height of the parallelogram.
Step 3: Compare Areas
Using the formula Area $=\frac{1}{2} \times$ base × height:

• Area $(\triangle P C D)=\frac{1}{2} \times C D \times$ height
• Area $(\triangle Q C D)=\frac{1}{2} \times C D \times$ height

Since the bases and heights are identical, the areas are equal.
The ratio of area $(\triangle P C D)$ : area $(\triangle Q C D)$ is $1: 1$.

Q.23:

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

Solution:

Both triangles share a base and have equivalent heights relative to that base due to the properties of a parallelogram.

Consider the diagonal PR of parallelogram PQRS. Diagonal PR divides the parallelogram into two triangles of equal area, $\triangle P S R$ and $\triangle P Q R$.

Since ${P R}$ is a diagonal, the perpendicular distance from ${S}$ to ${P R}$ is equal to the perpendicular distance from Q to PR. Both triangles $\triangle P S O$ and $\triangle P Q O$ share the same base PO on the diagonal.

Using the formula Area $=\frac{1}{2} \times$ base $\times$ height, since their bases and heights are equal, their areas must be equal.

Q.24:

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?

Solution:

Step 1: The Varignon Parallelogram
Let the 4-gon be $A B C D$ and its midpoints be $E, F, G, H$. Joining these midpoints always forms a parallelogram because, by the Midpoint Theorem, $E F$ and $H G$ are both parallel to diagonal $A C$ and half its length.

Step 2: Area Relationship
Draw diagonal $A C$. This divides the 4-gon into $\triangle A B C$ and $\triangle A D C$. The triangle formed by the midpoints, $\triangle E B F$, has a base and height each half of $\triangle A B C$. Therefore:

$\operatorname{Area}(\triangle E B F)=\frac{1}{4} \operatorname{Area}(\triangle A B C)$

Step 3: Summing the Corners
Applying this to all four corner triangles $(E B F, F C G, G D H, H A E)$, their total area equals one-half of the entire 4-gon’s area.
Subtracting these corner areas from the total area of the 4-gon leaves the central parallelogram, which must occupy the remaining half of the area.

Q.25:

In $\triangle A B C$, the midpoint of $B C$ is $D$. Median $A D$ is drawn. $P$ is any point on $A D$. Show that area $(\triangle A B P)=$ area $(\triangle A C P)$.

Solution:

Step 1: Large Triangle
In $\triangle A B C, A D$ is the median. A median always divides a triangle into two triangles of equal area, so:

$\operatorname{Area}(\triangle A B D)=\operatorname{Area}(\triangle A C D)$

Step 2: Small Triangle
Now consider $\triangle P B C$. Since $D$ is the midpoint of $B C, P D$ acts as a median for this smaller triangle:

$\operatorname{Area}(\triangle P B D)=\operatorname{Area}(\triangle P C D)$

Step 3: Final Subtraction
To find the remaining areas, we subtract the small triangles from the large ones:

$\operatorname{Area}(\triangle A B D)-\operatorname{Area}(\triangle P B D)=$$\operatorname{Area}(\triangle A C D)-\operatorname{Area}(\triangle P C D)$

This leaves us with the required proof:

$\operatorname{Area}(\triangle A B P)=\operatorname{Area}(\triangle A C P)$

Q.26:

Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD. What is the ratio of the areas of the red region ($\triangle {PAB}$ and $\triangle {PCD}$) and the green region ($\triangle {PBC}$ and $\triangle {PDA}$)?

Solution:

Let the square be $A B C D$ and $P$ be any point inside it. Join $P A, P B, P C, P D$.
Triangles $\triangle P A B$ and $\triangle P C D$ lie on opposite sides of the square, and similarly $\triangle P B C$ and $\triangle P D A$ lie on the other pair of opposite sides.

Now, triangles on the same base and between the same parallels have equal areas.
Using this idea:

• $\triangle P A B+\triangle P C D$ together cover half the square
• $\triangle P B C+\triangle P D A$ together cover the other half

Thus, both regions have equal area. 1 : 1
 

Q.27:

In $\triangle A B C, D$ is the midpoint of $A B . P$ is any point on $B C$, and $Q$ is a point on AB such that ${CQ} \| {PD} . {PQ}$ is joined. Prove that Area $(\triangle B P Q)=\frac{1}{2}$ Area $(\triangle A B C)$.

Solution:

First, join $C D$. Since $D$ is the midpoint of $A B, C D$ is a median, making Area $(\triangle B C D)= \frac{1}{2} \operatorname{Area}(\triangle A B C)$.
Observe that $\triangle P D Q$ and $\triangle P D C$ share the same base $P D$ and lie between parallel lines $P D$ and $Q C$. Therefore, their areas are equal.

Now, $\operatorname{Area}(\triangle B P Q)=$ $\operatorname{Area}(\triangle B P D)+\operatorname{Area}(\triangle P D Q)$. Substituting the equal area, we get $\operatorname{Area}(\triangle B P D)+\operatorname{Area}(\triangle P D C)$, which is $\operatorname{Area}(\triangle B C D)$.
Since $\operatorname{Area}(\triangle B C D)$ is half of the total triangle, we have successfully proved that $\operatorname{Area}(\triangle B P Q)=\frac{1}{2} \operatorname{Area}(\triangle A B C)$.

Q.28:

Find the area of a sector of a circle with radius 7 cm if the angle of the sector is $60^{\circ}$.

Solution:

To find the area, use the formula $\frac{\theta}{360} \times \pi r^2$.
Substitute the radius $r=7 {~cm}$ and the angle $\theta=60^{\circ}$.
Area $=\frac{60}{360} \times \frac{22}{7} \times 7 \times 7$
Area $=\frac{1}{6} \times 22 \times 7=\frac{154}{6}$
Area $=25.67 \ {cm}^2$
This represents the space enclosed by the two radii and the arc.

Q.29:

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

Solution:

A quadrant is a sector with a $90^{\circ}$ angle. Given $C=44 {~cm}$.
Calculate radius: $2 \times \frac{22}{7} \times r=44$, so $r=7 {~cm}$.
Area $=\frac{1}{4} \times \pi r^2=\frac{1}{4} \times \frac{22}{7} \times 7 \times 7$
Area $=\frac{1}{4} \times 154=38.5 \ {cm}^2$
This is exactly one-quarter of the total circle’s area.

Q.30:

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

Solution:

The hand acts as the radius $r=7 \ {cm}$.
In 60 minutes, the hand covers $360^{\circ}$, so in 10 minutes it covers $60^{\circ}$.

$\text { Area }=\frac{60}{360} \times \frac{22}{7} \times 7 \times 7$
$\text { Area }=\frac{1}{6} \times 154=25.67 \ {cm}^2$

The swept area is the region covered by the hand’s movement.

Q.31:

A chord of a circle of radius 10 cm subtends $90^{\circ}$ at the centre. Find the area of the corresponding: (i) minor sector (that subtends $90^{\circ}$ at the centre), and (ii) major sector (that subtends $270^{\circ}$ at the centre). (Use $\pi \approx 3.14$.)

Solution:

For $r=10 {~cm}$ and $\theta=90^{\circ}$, use $\pi=3.14$.

$\text { Minor Area }=\frac{90}{360} \times$ $3.14 \times 10 \times 10=0.25 \times 314$ $=78.5 \ {cm}^2$
$\text { Major Area }=$ $\frac{270}{360} \times 3.14 \times 100=$ $0.75 \times 314=235.5 \ {cm}^2$

The sum of both areas equals the total area of the circle.

Q.32:

A chord of a circle of radius 15 cm subtends an angle of $60^{\circ}$ at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi \approx 3.14$ and $\sqrt{3} \approx 1.73$.)

Solution:

Sector Area $=\frac{60}{360} \times 3.14 \times 225=117.75 \ {cm}^2$.
Triangle Area (equilateral) $=\frac{\sqrt{3}}{4} \times 15^2=0.433 \times 225=97.31 \ {cm}^2$.
Minor Segment Area $=117.75-97.31=20.44 \ {cm}^2$.
Major Segment Area $=(3.14 \times 225)-20.44=686.06 \ {cm}^2$.
Segments are areas bounded by a chord and its corresponding arc.

Q.33:

A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of $120^{\circ}$. Find the total area cleaned at each sweep of the blades.

Solution:

Each blade has $r=28 \ {cm}$ and $\theta=120^{\circ}$. There are two wipers.
Total Area $=2 \times\left(\frac{120}{360} \times \frac{22}{7} \times 28 \times 28\right)$
Total Area $=2 \times \frac{1}{3} \times 22 \times 4 \times 28$
Total Area $=\frac{2 \times 2464}{3}=1642.67 \ {cm}^2$
This total represents the clean surface area per full sweep.

Q.34:

A chord of a circle of radius $r$ subtends an angle of $60^{\circ}$ at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to $\pi r^2\left(\frac{1}{6}-\frac{\sqrt{3}}{4}\right)$.

Solution:

Area of sector $=\frac{60}{360} \pi r^2=\frac{1}{6} \pi r^2$.
Area of equilateral triangle $=\frac{\sqrt{3}}{4} r^2$.
Segment Area $=$ Sector Area – Triangle Area
Segment Area $=\frac{1}{6} \pi r^2-\frac{\sqrt{3}}{4} r^2$
Factoring out $r^2$ gives $\pi r^2\left(\frac{1}{6}-\frac{\sqrt{3}}{4 \pi}\right)$.

Q.35:

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

Solution:

Area of triangle $=\frac{3 \sqrt{3}}{4} r^2$ (using $3 \times$ area of internal triangle).
Area of circle $=\pi r^2$.

$\text { Ratio }=\frac{\text { Triangle Area }}{\text { Circle Area }}=\frac{3 \sqrt{3} r^2}{4 \pi r^2}$
$\text { Ratio }=\frac{3 \sqrt{3}}{4 \pi} \approx 0.413 .$

Q.36:

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 $\frac{2}{\pi} \approx 0.637$.

Solution:

Diagonal of square $=2 r$, so side $s=\frac{2 r}{\sqrt{2}}=r \sqrt{2}$.
Area of square $=s^2=(r \sqrt{2})^2=2 r^2$.
Area of circle $=\pi r^2$.
Ratio $=\frac{2 r^2}{\pi r^2}=\frac{2}{\pi} \approx 0.637$.

Q.37:

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 $\frac{3 \sqrt{3}}{2 \pi} \approx 0.827$. Can you see why the answer is exactly twice the answer to Question 8?

Solution:

A hexagon is made of 6 equilateral triangles: $6 \times \frac{\sqrt{3}}{4} r^2=\frac{3 \sqrt{3}}{2} r^2$.
Area of circle $=\pi r^2$.
Ratio $=\frac{3 \sqrt{3}}{2 \pi} \approx 0.827$.
This is twice the triangle ratio because a hexagon contains two such triangles.

Q.38:

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

The figure shown corresponds to the identity

$(a+b)^2=a^2+2 a b+b^2 .$

Do you see how?

Draw figures corresponding to the identities $(a+b)(a-b)= a^2-b^2$ and $(a+b+c)^2=a^2+b^2+c^2+$$2 a b+2 b c+2 c a$.

Solution:

Identity: $(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$
To represent this, draw a large square with side length $a+b+c$. Divide each side into segments of lengths $a, b$, and $c$.

• Three Squares: You get three squares with areas $a^2, b^2$, and $c^2$ along the main diagonal.
• Six Rectangles: The remaining space consists of two rectangles of area $a b$, two of area $b c$, and two of area $c a$.
• Total Area: Summing all nine individual regions perfectly matches the algebraic expansion of the identity.

Identity: $(a+b)(a-b)=a^2-b^2$
This model starts with a square of area $a^2$ and involves removing a smaller square of area $b^2$ from one corner.

• The L-shape: After removing $b^2$, you are left with an L-shaped region with a total area of $a^2-b^2$.
• Rearrangement: Cut this L-shape into two rectangles. One has dimensions $a$ by $(a-b)$, and the other is $b$ by $(a-b)$.
• Final Rectangle: Placing these side-by-side forms a single large rectangle with a length of $(a+b)$ and a width of $(a-b)$.

Q.39:

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

Solution:

To find the area, we first identify all side lengths of the triangle.
Since the perimeter is 40 cm and the two equal sides are 15 cm each, the third side (base) is $40-(15+15)=10 {~cm}$.
Now, we use Heron’s formula. The semi-perimeter $s$ is half the perimeter: $s=40 / 2=20 {~cm}$.
The area formula is $\sqrt{s(s-a)(s-b)(s-c)}$.
Plugging in the values, we get $\sqrt{20(20-15)(20-15)(20-10)}$, which simplifies to $\sqrt{20 \times 5 \times 5 \times 10}=\sqrt{5000}$.
This calculates to $50 \sqrt{2}$, which is approximately 70.71 square ${c m}$.

Q.40:

An isosceles triangle has base 10 cm, and its area is $60 \ {cm}^2$. What are the lengths of the equal sides?

Solution:

We start with the base of 10 cm and an area of 60 square cm.
Using the area formula Area $= \frac{1}{2} \times$ base $\times$ height, we solve $60=\frac{1}{2} \times 10 \times h$, which gives a height of 12 cm.
In an isosceles triangle, the height from the vertex to the base bisects the base into two equal parts of 5 cm each.
This creates a right-angled triangle with a base of 5 cm and a height of 12 cm.
Applying the Pythagorean theorem, the equal side $L$ is $\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}$.
Therefore, the equal sides are ${1 3 ~ c m}$ each.

Q.41:

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

Solution:

Given the area is 54 square cm and one leg is 12 cm, we find the second leg (b) using Area $= \frac{1}{2} \times a \times b$.
So, $54=\frac{1}{2} \times 12 \times b$, which means $b=9 {~cm}$.
Now we have two legs, 12 cm and 9 cm. To find the perimeter, we first need the hypotenuse $(c)$ using $c=\sqrt{12^2+9^2}$.
This equals $\sqrt{144+81}=\sqrt{225}=15 {~cm}$.
Adding all three sides together $(12+9+15)$, we find that the total perimeter of the triangle is ${3 6 ~ c m}$.

Q.42:

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

Solution:

The sides are in the ratio $2: 3: 4$, so let them be $2 x, 3 x$, and $4 x$.
Their sum is the perimeter: $2 x+3 x+4 x=45$, so $9 x=45$, which gives $x=5$.
The sides are therefore $10 {~cm}, 15 {~cm}$, and 20 cm.
The semi-perimeter $s$ is $45 / 2=22.5 {~cm}$.
Applying Heron’s formula: $\sqrt{22.5(22.5-10)(22.5-15)(22.5-20)}$, which is $\sqrt{22.5 \times 12.5 \times 7.5 \times 2.5}$.
Calculating this product gives $\sqrt{5273.4375}$, resulting in a final area of approximately 72.62

Q.43:

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

Solution:

The sides are 7, 24, and 25 cm.
Method 1: Using Heron’s Formula

$s=\frac{7+24+25}{2}=28$
$\text { Area }=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{28(28-7)(28-24)(28-25)}=\sqrt{28 \times 21 \times 4 \times 3}$
$=\sqrt{7056}=84 \ {cm}^2$

Method 2: Using Right Triangle Property
Check:

$7^2+24^2=49+576=625=25^2$

$7^2+24^2=25^2$

So, the triangle is right-angled.
Area:

$\text { Area }=\frac{1}{2} \times 7 \times 24=84\ {cm}^2$

Q.44:

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.

Solution:

Distance travelled $=$ (number of rotations) × (circumference of wheel)
Diameter $=60 {~cm}$
Circumference $=\pi d=60 \pi {~cm}$

For 100 rotations:
Distance $=100 \times 60 \pi=6000 \pi {~cm}$

Convert to metres:

$6000 \pi {~cm}=60 \pi {~m}$

Q.45:

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

Solution:

Circumference of the circle $=66 {~cm}$

$2 \pi r=66$

$r=\frac{66}{2 \pi}=\frac{33}{\pi}$

Using $\pi=\frac{22}{7}$,

$r=\frac{33}{22 / 7}=10.5 {~cm}$

Area of a quadrant:

$\text { Area }=\frac{1}{4} \pi r^2$
$=\frac{1}{4} \times \frac{22}{7} \times(10.5)^2=86.625 {~cm}^2$

Q.46:

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.

Solution:

Outer radius $r=28 {~cm}$

1. Distance travelled in one complete turn
   Distance $=$ circumference of the wheel
   $\text { Circumference }=2 \pi r=2 \pi \times 28=56 \pi {~cm}$
   Using $\pi=\frac{22}{7}$:
   $56 \pi=56 \times \frac{22}{7}=176 {~cm}$
2. Number of turns in 1 km
   Convert distance:
   $1 {~km}=100000 {~cm}$
   $\text { Number of turns }=\frac{100000}{176} \approx 568.18$

Q.47:

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

Solution:

Let the two rectangles have sides $\left(l_1, w_1\right)$ and $\left(l_2, w_2\right)$. If perimeters are equal, $l_1+w_1= l_2+w_2$.

If areas are equal, $l_1 w_1=l_2 w_2$.

These two conditions mean that $l$ and $w$ are roots of the same quadratic equation $x^2-($Sum$) x+($Product$)=0$.

Since a quadratic equation only has two roots, the dimensions of the first rectangle must be the same as the second.

This proves the rectangles are congruent.

Q.48:

You know that the area of a parallelogram is base $\times$ height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides $\times$ height, i.e., $\frac{1}{2}(a+b) h$.

Solution:

Step 1: Area of the Parallelogram
The area of the parallelogram part is calculated as base × height, which gives us ah.
Step 2: Area of the Triangle
The triangle sharing the same height ${h}$ has a base equal to the difference of the parallel sides, ${b}-{a}$. Its area is $\frac{1}{2} \times(b-a) \times h$.

Step 3: Combine and Simplify
Adding these two areas together gives $a h+\frac{1}{2}(b-a) h$. Factoring out the height, we get $h\left(a+\frac{1}{2} b-\frac{1}{2} a\right)$, which simplifies to $h\left(\frac{1}{2} a+\frac{1}{2} b\right)$. Finally, we reach the identity: $\frac{1}{2}(a+b) h$.

Q.49:

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).

Solution:

A trapezium can be analyzed by drawing a diagonal that splits it into two distinct triangles which share the same vertical height.

Step 1: Divide the Shape
Consider a trapezium with parallel sides $a$ and $b$ and height $h$. Let us draw a diagonal from one corner to the opposite corner. This action divides the entire trapezium into two separate triangles.

Step 2: Area of the First Triangle
The first triangle has one of the parallel sides, $a$, as its base. Since the height of the trapezium is $h$, the area of this triangle is calculated as $\frac{1}{2} \times a \times h$.

Step 3: Area of the Second Triangle
The second triangle uses the other parallel side, $b$, as its base. Because the parallel lines remain a constant distance apart, this triangle also has the same height $h$. Its area is $\frac{1}{2} \times b \times h$.

Step 4: Sum the Areas
To find the total area of the trapezium, we simply add the areas of these two triangles together:

$\text { Total Area }=\left(\frac{1}{2} \times a \times h\right)+\left(\frac{1}{2} \times b \times h\right)$

Step 5: Factor the Equation
By factoring out the common terms $\frac{1}{2}$ and $h$ from the expression, we arrive at the final formula:

$\text { Total Area }=\frac{1}{2} \times(a+b) \times h$

This demonstrates that the area is indeed half the sum of the parallel sides multiplied by the perpendicular height.

Q.50:

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?

Solution:

Take a trapezium with parallel sides a and b, and height h. Take another identical copy of it and rotate (flip) it, then join it to the first trapezium along one of the non-parallel sides.

The new figure formed is a parallelogram whose:

• Base $=a+b$ (sum of parallel sides)
• Height $=h$

So,

$\text { Area of parallelogram }=(a+b) \times h$

But this parallelogram is made of two identical trapeziums, so:

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

Q.51:

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

Solution:

1. Algebraic Method Let diagonals $A C=d_1$ and $B D=d_2$ intersect at $O$.
   In a kite, diagonals are perpendicular and one bisects the other.
   Area $=$ sum of 4 right triangles: $\text { Area }=\frac{1}{2}(A O \cdot B O+O C \cdot B O+O C \cdot D O+A O \cdot D O)$
   $=\frac{1}{2}(A O+O C)(B O+D O)=\frac{1}{2}(A C)(B D)$
   $=\frac{1}{2} d_1 d_2$
2. Geometric Method The diagonals divide the kite into 4 right triangles. Their combined area equals half the product of the diagonals.