Samacheer Kalvi Class 8 Maths Solutions Chapter 2 Measurements Exercise 2.2

Question 1. Find the perimeter and area of the figures given below. (\(\pi = \frac{22}{7}\))
(i)
(ii)
Answer:
(i) For the given figure:
Radius of the semicircle \( r = \frac{7}{2} \) m \( = 3.5 \) m
Length of the arc of the semicircle \( = \frac{1}{2} \times 2\pi r = \pi r = \frac{22}{7} \times \frac{7}{2} = 11 \) m
The perimeter of the figure is found by adding all the lengths of the outer boundary.
\( \therefore \) Perimeter \( P = 11 + 10 + 7 + 10 = 38 \) m
Area of the figure \( = \) Area of the rectangle \( - \) Area of the semicircle
\( = (l \times b) - \frac{1}{2}\pi r^2 \) sq. units
\( = (10 \times 7) - \frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \)
\( = 70 - \frac{11 \times 7}{2 \times 2} = 70 - \frac{77}{4} \)
\( = \frac{280 - 77}{4} = \frac{203}{4} = 50.75 \) m\(^2\)
Area of the figure \( = 50.75 \) m\(^2\) (approximately). This shape looks like a rectangle with a semi-circle cut out from one side.

(ii) For the given figure:
Radius of the quadrant circle \( r = 3.5 \) cm
Length of the arc of one quadrant circle \( = \frac{1}{4} \times 2\pi r = \frac{1}{2}\pi r \)
\( = \frac{1}{2} \times \frac{22}{7} \times 3.5 = 11 \times 0.5 = 5.5 \) cm
Length of arcs of 2 such quadrant circles \( = 2 \times 5.5 = 11 \) cm
The perimeter is the sum of all outside lengths.
\( \therefore \) Perimeter \( P = 11 + 6 + 3.5 + 6 + 3.5 = 30 \) cm
Area of the figure \( = \) Area of 2 quadrant circles \( + \) Area of a rectangle
\( = (2 \times \frac{1}{4}\pi r^2) + (l \times b) \) sq. units
\( = (\frac{1}{2}\pi r^2) + (l \times b) \)
\( = (\frac{1}{2} \times \frac{22}{7} \times 3.5 \times 3.5) + (6 \times 3.5) \)
\( = (11 \times 3.5 \times 0.5) + 21 \)
\( = 19.25 + 21 = 40.25 \) cm\(^2\)
\( \therefore \) Area \( = 40.25 \) cm\(^2\) (approximately). Always ensure to include all external boundary segments when calculating perimeter.
In simple words: For the first figure, we find the perimeter by adding all outside edges, including the curved part. The area is found by taking the rectangle's area and subtracting the semicircle's area. For the second figure, we add the lengths of the two curved ends, the two long straight sides, and the two short straight sides to get the perimeter. The total area is the rectangle's area plus the area of two quarter circles (which is one semicircle).

🎯 Exam Tip: Remember to clearly identify the individual shapes that make up the combined figure, and ensure you use the correct formulas for each shape's perimeter and area.

Question 2. Find the area of the shaded part in the following figures. (\(\pi = 3.14\))
(i)
(ii)
Answer:
(i) For the given figure:
The radius of each quadrant circle is half the side of the square. So, \( r = \frac{10}{2} = 5 \) cm.
Area of the shaded part \( = \) Area of 4 quadrant circles
\( = 4 \times \frac{1}{4} \times \pi r^2 = \pi r^2 \)
\( = 3.14 \times 5 \times 5 = 3.14 \times 25 = 78.5 \) cm\(^2\)
Area of the shaded part \( = 78.5 \) cm\(^2\). This calculation represents the area of the four leaves.
Area of the unshaded part \( = \) Area of the square \( - \) Area of the shaded part
\( = (10 \times 10) - 78.5 \)
\( = 100 - 78.5 = 21.5 \) cm\(^2\)
Area of the unshaded part \( = 21.5 \) cm\(^2\) (approximately).

(ii) For the given figure:
Area of the shaded part \( = \) Area of semicircle \( - \) Area of the triangle
Radius of the semicircle \( r = \frac{14}{2} = 7 \) cm
Base of the triangle \( b = 14 \) cm, Height of the triangle \( h = 7 \) cm
\( = (\frac{1}{2}\pi r^2) - (\frac{1}{2}bh) \) cm\(^2\)
\( = (\frac{1}{2} \times 3.14 \times 7 \times 7) - (\frac{1}{2} \times 14 \times 7) \)
\( = (\frac{153.86}{2}) - 49 \)
\( = 76.93 - 49 = 27.93 \) cm\(^2\)
Area of the shaded part \( = 27.93 \) cm\(^2\) (approximately). The semicircle has a uniform curve, making its area calculation straightforward.
In simple words: For the first figure, we calculate the area of a full circle using the quadrant's radius, which is given as the shaded area. Then we subtract this from the square's area to get the unshaded part. For the second figure, we find the area of the semicircle and then take away the area of the triangle inside it to get the shaded area.

🎯 Exam Tip: Always pay close attention to which part of the figure is shaded and ensure your calculation directly corresponds to that area. Double-check your use of radius vs. diameter.

Question 3. Find the area of the combined figure given which is got by joining of two parallelograms
Answer:
The given figure is made of two parallelograms joined together.
For one parallelogram: Base \( b = 8 \) cm, Height \( h = 3 \) cm
Area of the figure \( = \) Area of 2 parallelograms
\( = 2 \times (b \times h) \) sq. units
\( = 2 \times 8 \times 3 \) cm\(^2\)
\( = 48 \) cm\(^2\)
\( \therefore \) Area of the given figure \( = 48 \) cm\(^2\). A parallelogram's area is simply base multiplied by its perpendicular height.
In simple words: The shape is made of two identical parallelograms. We find the area of one parallelogram (base times height) and then multiply it by two to get the total area.

🎯 Exam Tip: Remember that the height of a parallelogram must be the perpendicular distance between its parallel bases, not the slanted side length.

Question 4. Find the area of the combined figure given, formed by joining a semicircle of diameter 6 cm with a triangle of base 6 cm and height 9 cm. (\(\pi = 3.14\))
Answer:
The figure is a combination of a semicircle and a triangle.
Radius of the semicircle \( r = \frac{6}{2} = 3 \) cm
Area of the figure \( = \) Area of the semicircle of radius 3 cm \( + \) Area of triangle with base 9 cm and height 3 cm
\( = (\frac{1}{2}\pi r^2) + (\frac{1}{2}bh) \) sq. units
\( = (\frac{1}{2} \times 3.14 \times 3 \times 3) + (\frac{1}{2} \times 9 \times 3) \)
\( = (\frac{28.26}{2}) + (\frac{27}{2}) \)
\( = 14.13 + 13.5 = 27.63 \) cm\(^2\)
\( \therefore \) Area of the figure \( = 27.63 \) cm\(^2\) (approximately). This calculation sums the individual areas to find the total.
In simple words: To find the total area, we add the area of the semicircle (half of pi times radius squared) to the area of the triangle (half of base times height).

🎯 Exam Tip: For combined figures, clearly break down the figure into simpler shapes and calculate their areas separately before adding them up. Pay careful attention to the specific dimensions used for each component shape.

Question 5. The door mat which is in a hexagonal shape has the following measures as given in the figure. Find its area.
Answer:
The hexagonal doormat can be divided into two identical trapeziums.
Height of each trapezium \( h = \frac{70}{2} = 35 \) cm
The parallel sides of each trapezium are \( a = 90 \) cm and \( b = 70 \) cm.
Area of the doormat \( = \) Area of 2 trapeziums
\( = 2 \times \frac{1}{2}h(a + b) \) sq. units
\( = 2 \times \frac{1}{2} \times 35 \times (90 + 70) \)
\( = 35 \times 160 = 5600 \) cm\(^2\)
\( \therefore \) Area of the door mat \( = 5600 \) cm\(^2\). This is an efficient way to calculate the area of a complex symmetrical polygon by breaking it into simpler, identical parts.
In simple words: We split the doormat into two identical trapezium shapes. We find the area of one trapezium by using its parallel sides and height, then multiply that by two to get the total area of the doormat.

🎯 Exam Tip: When dealing with symmetrical polygons, look for ways to divide them into fewer, identical, simpler shapes to make calculations faster and less error-prone.

Question 6. A rocket drawing has the measures as given in the figure. Find its area.
Answer:
The rocket figure can be divided into three basic shapes: a triangle, a rectangle, and a trapezium.
For the rectangle in the middle:
Length \( l = 120 - 20 - 20 = 80 \) cm
Breadth \( b = 30 \) cm
For the triangle at the top:
Base \( = 30 \) cm (same as rectangle's breadth)
Height \( = 20 \) cm
For the trapezium at the bottom:
Height \( h = 20 \) cm
Parallel sides are \( a = 50 \) cm and \( b = 30 \) cm
Area of the figure \( = \) Area of rectangle \( + \) Area of triangle \( + \) Area of trapezium
\( = (l \times b) + (\frac{1}{2} \times \text{base} \times \text{height}) + (\frac{1}{2} \times h \times (a + b)) \) sq. units
\( = (80 \times 30) + (\frac{1}{2} \times 30 \times 20) + (\frac{1}{2} \times 20 \times (50 + 30)) \)
\( = 2400 + 300 + (10 \times 80) \)
\( = 2400 + 300 + 800 = 3500 \) cm\(^2\)
Area of the figure \( = 3500 \) cm\(^2\). Breaking complex shapes into simpler ones makes calculation manageable.
In simple words: We find the area of the rectangle, the triangle, and the trapezium parts separately. Then we add all these areas together to get the total area of the rocket shape.

🎯 Exam Tip: When given a complex figure, carefully identify all the simple geometric shapes that compose it. List their dimensions and use the appropriate area formulas for each before summing them up.

Question 7. Find the area of the irregular polygon shaped fields given below.
Answer:
The irregular polygon can be divided into smaller, recognizable shapes: a trapezium and several triangles.
Area of the field \( = \) Area of trapezium FBCH \( + \) Area of \( \triangle \) DHC \( + \) Area of \( \triangle \) EGD \( + \) Area of \( \triangle \) EGA \( + \) Area of \( \triangle \) ABF
Formula for area of a triangle \( = \frac{1}{2}bh \) sq. units
Formula for area of a trapezium \( = \frac{1}{2} \times h \times (a + b) \) sq. units

1. Area of trapezium FBCH:
Parallel sides \( a = 10 \) m, \( b = 8 \) m
Height \( h = (8 + 3) = 11 \) m
\( = \frac{1}{2} \times (10 + 8) \times 11 = \frac{1}{2} \times 18 \times 11 = 9 \times 11 = 99 \) m\(^2\)

2. Area of \( \triangle \) DHC:
Base \( = 5 \) m, Height \( = 8 \) m
\( = \frac{1}{2} \times 5 \times 8 = 20 \) m\(^2\)

3. Area of \( \triangle \) EGD:
Base \( = (10 + 5) = 15 \) m, Height \( = 8 \) m
\( = \frac{1}{2} \times 15 \times 8 = 60 \) m\(^2\)

4. Area of \( \triangle \) EGA:
Base \( = (8 + 6) = 14 \) m, Height \( = 8 \) m
\( = \frac{1}{2} \times 14 \times 8 = 56 \) m\(^2\)

5. Area of \( \triangle \) ABF:
Base \( = 6 \) m, Height \( = 3 \) m
\( = \frac{1}{2} \times 6 \times 3 = 9 \) m\(^2\)

\( \therefore \) Total Area of the field \( = 99 + 20 + 60 + 56 + 9 = 244 \) m\(^2\)
Area of the field \( = 244 \) m\(^2\). This method is useful for finding the area of any complex polygon by breaking it down into simple shapes.
In simple words: To find the area of this uneven field, we divide it into one trapezium and several triangles. We calculate the area of each small shape using its own dimensions and then add all these areas together to get the total area of the field.

🎯 Exam Tip: When dealing with irregular polygons on a grid or with given segment lengths, always break them down into basic shapes like triangles, rectangles, and trapeziums. Carefully identify the base and height for each, ensuring they are perpendicular for triangles and trapeziums.