CBSE Class 7 Mathematics Perimeter And Area Worksheet Set 02

Question 1. Find the area of the shaded region in the following quadrants.
(a) A quadrant of a circle with a right triangle inscribed inside having base 2 and height 3.
(b) A quadrant of a circle with radius 4 and a right triangle inscribed inside having base 4 and height 4.
Answer:
(a) Area of quadrant = \(\frac{1}{4} \pi r^2\)
From the figure, the radius of the quadrant \(r = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}\)
Area of quadrant = \(\frac{1}{4} \times \frac{22}{7} \times 13 = \frac{286}{28} = \frac{143}{14} = 10.21 \, \text{sq. units}\)
Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 3 = 3 \, \text{sq. units}\)
Area of shaded region = Area of quadrant - Area of triangle = \(10.21 - 3 = 7.21 \, \text{sq. units}\)

(b) Radius of the quadrant \(r = 4\)
Area of quadrant = \(\frac{1}{4} \times \frac{22}{7} \times 4^2 = \frac{1}{4} \times \frac{22}{7} \times 16 = \frac{88}{7} = 12.57 \, \text{sq. units}\)
Area of triangle inside = \(\frac{1}{2} \times 4 \times 4 = 8 \, \text{sq. units}\)
Area of shaded region = \(12.57 - 8 = 4.57 \, \text{sq. units}\)
In simple words: The area of the shaded region is found by subtracting the area of the unshaded triangle from the total area of the quadrant.
Exam Tip: Always find the radius of the circle or quadrant first using Pythagoras theorem if the radius is not directly given.

Question 2. Find the area of the shaded region in each case.
(a) Rectangle of dimensions \(58\text{m} \times 45\text{m}\) with three identical inner rectangular paths/blocks.
(b) Outer rectangle with dimensions \(12\) and height containing inner cutout strips of width 3 and 4.
(c) Rectangle \(52.5\text{m} \times 86.5\text{m}\) with two intersecting paths of width 3.5m running through the middle.
(d) Outer rectangle \(75\text{m} \times 40\text{m}\) with inner rectangle \(65\text{m} \times 30\text{m}\).
(e) Trapezium-like shape with parallel sides 2 and 5, height 8, divided into sections.
(f) Rectangle of length 30 and breadth 12 with triangles inside.
Answer:
(a) Area of outer rectangle = \(58 \times 45 = 2610 \, \text{m}^2\)
Three inner unshaded rectangles each have dimensions \(8\text{m} \times 8\text{m}\) (or given by dimensions in figure):
Area of 3 unshaded rectangles = \(3 \times (8 \times 8) = 3 \times 64 = 192 \, \text{m}^2\)
Area of shaded region = \(2610 - 192 = 2418 \, \text{m}^2\)

(b) Total area minus unshaded cutout areas = \((12 \times \text{height}) - \text{unshaded parts} = 144 - 36 = 108 \, \text{sq. units}\)

(c) Area of rectangle = \(52.5 \times 86.5 = 4541.25 \, \text{m}^2\)
Area of two crossing paths of width 3.5m:
Horizontal path area = \(52.5 \times 3.5 = 183.75 \, \text{m}^2\)
Vertical path area = \(86.5 \times 3.5 = 302.75 \, \text{m}^2\)
Intersection area = \(3.5 \times 3.5 = 12.25 \, \text{m}^2\)
Total path area = \(183.75 + 302.75 - 12.25 = 474.25 \, \text{m}^2\)
Area of shaded region = \(4541.25 - 474.25 = 4067 \, \text{m}^2\)

(d) Outer dimensions = \(75\text{m} \times 40\text{m}\), Inner dimensions = \(65\text{m} \times 30\text{m}\)
Area of outer rectangle = \(75 \times 40 = 3000 \, \text{m}^2\)
Area of inner rectangle = \(65 \times 30 = 1950 \, \text{m}^2\)
Area of shaded pathway = \(3000 - 1950 = 1050 \, \text{m}^2\)

(e) Area of outer figure = Area of two triangles and central rectangle = \(\left(\frac{1}{2} \times 2 \times 8\right) + (5 \times 8) + \left(\frac{1}{2} \times 3 \times 8\right) = 8 + 40 + 12 = 60 \, \text{sq. units}\)

(f) Area of outer rectangle = \(30 \times 12 = 360 \, \text{sq. units}\)
Unshaded triangles inside have bases 12 and 18, height 12.
Area of unshaded triangles = \(\frac{1}{2} \times 30 \times 12 = 180 \, \text{sq. units}\)
Area of shaded region = \(360 - 180 = 180 \, \text{sq. units}\)
In simple words: Subtract the unshaded inner parts or paths from the total outer area to find the shaded region.
Exam Tip: For crossing paths inside a rectangle, remember to subtract the area of the overlapping square intersection in the middle once.

Question 3. A piece of wire is bent into the shape of an equilateral triangle of side 6.6 cm. It is rebent to form a circular ring. Find the diameter of the ring.
Answer:
Perimeter of the equilateral triangle = Length of the wire
Side of triangle = 6.6 cm
Perimeter = \(3 \times 6.6 = 19.8 \, \text{cm}\)
When the wire is bent into a circle, the circumference of the circle equals the length of the wire.
Circumference \(C = 19.8 \, \text{cm}\)
\(\pi d = 19.8\)
\(\frac{22}{7} \times d = 19.8\)
\(d = \frac{19.8 \times 7}{22} = \frac{138.6}{22} = 6.3 \, \text{cm}\)
In simple words: The total length of wire remains the same. Use the perimeter of the triangle to find the wire length, then use the circle circumference formula to find the diameter.
Exam Tip: Whenever a wire or shape is reshaped, the total length (perimeter or circumference) remains constant.

Question 4. Find the perimeter of a rectangular swimming pool whose length is 22m and area is 352 \(\text{m}^2\).
Answer:
Area of rectangle = \(\text{Length} \times \text{Breadth}\)
\(352 = 22 \times \text{Breadth}\)
Breadth = \(\frac{352}{22} = 16 \, \text{m}\)
Perimeter of rectangular pool = \(2 \times (\text{Length} + \text{Breadth})\)
Perimeter = \(2 \times (22 + 16) = 2 \times 38 = 76 \, \text{m}\)
In simple words: First find the breadth by dividing the area by the length, then plug both length and breadth into the perimeter formula.
Exam Tip: Always include proper units (like meters for perimeter and square meters for area) in your final answer.

Question 5. Find the ratio of radii of two circles whose circumference are in the ratio 2:5.
Answer:
Let the radii of the two circles be \(r_1\) and \(r_2\).
Circumferences are \(C_1 = 2\pi r_1\) and \(C_2 = 2\pi r_2\)
Given ratio of circumferences = \(\frac{C_1}{C_2} = \frac{2}{5}\)
\(\frac{2\pi r_1}{2\pi r_2} = \frac{2}{5}\)
\(\frac{r_1}{r_2} = \frac{2}{5}\)
Ratio of radii = \(2:5\)
In simple words: The ratio of the circumferences of two circles is directly equal to the ratio of their radii because \(2\pi\) cancels out.
Exam Tip: For similar figures, linear dimensions (radius, side, perimeter) share the same ratio.

Question 6. What is the ratio of areas of squares of sides 3cm and 4cm?
Answer:
Side of first square \(s_1 = 3 \, \text{cm}\)
Side of second square \(s_2 = 4 \, \text{cm}\)
Area of first square \(A_1 = s_1^2 = 3^2 = 9 \, \text{cm}^2\)
Area of second square \(A_2 = s_2^2 = 4^2 = 16 \, \text{cm}^2\)
Ratio of areas = \(\frac{A_1}{A_2} = \frac{9}{16} = 9:16\)
In simple words: Square the side lengths of both squares to find their areas, then write them as a ratio.
Exam Tip: The ratio of areas of two similar 2D figures is equal to the square of the ratio of their corresponding linear sides.

WORKSHEET 4 PAGE-2

Question 7. The perimeter of both square and rectangle are equal. Find the missing dimensions if needed. (Note: Text from page header continuation).
Answer:
Since the complete numeric values for this specific problem are cut off by the page boundary, the general relation is:
\(\text{Perimeter of square} = 4s\)
\(\text{Perimeter of rectangle} = 2(l + b)\)
Equating both: \(4s = 2(l + b) \implies 2s = l + b\)
In simple words: Equate the perimeter formula of the square to the perimeter formula of the rectangle to solve for any unknown side.

Exam Tip: Always write down the given formulas before substituting known values.

Question 1. Name any 5 objects around you that have:
(a) only one vertical line of symmetry
(b) only one horizontal line of symmetry
(c) four lines of symmetry
(d) three lines of symmetry
(e) no line of symmetry
Answer:
(a) Only one vertical line of symmetry: Human face, Butterfly, Letter 'A', Vase, Leaf
(b) Only one horizontal line of symmetry: Letter 'B', Letter 'C', Letter 'E', Letter 'D', Arrow pointing up/down
(c) Four lines of symmetry: Square, Rhombus, A window pane (square), A four-pointed star
(d) Three lines of symmetry: Equilateral triangle, Triskelion symbol, Mercedes-Benz logo
(e) No line of symmetry: Scalene triangle, Letter 'F', Letter 'G', Shoe, Cloud
In simple words: Line of symmetry is a line that divides an object into two identical, mirror-image halves.
Exam Tip: Visualize folding the object along the line to check if both halves overlap perfectly.

Question 2. Write clearly all those capital letters of English which have:
(a) only vertical line of symmetry
(b) only horizontal line of symmetry
(c) both vertical and horizontal lines of symmetry
(d) no line of symmetry
(e) order of rotation 1
(f) order of rotation 2
(g) order of rotation 4
(h) reflectional symmetry about a vertical mirror
(i) reflectional symmetry about a horizontal mirror
(j) reflectional symmetry about both vertical and horizontal mirrors
Answer:
(a) Only vertical line of symmetry: A, M, T, U, V, W, Y
(b) Only horizontal line of symmetry: B, C, D, E, K
(c) Both vertical and horizontal lines of symmetry: H, I, O, X
(d) No line of symmetry: F, G, J, L, N, P, Q, R, S, Z
(e) Order of rotation 1 (no rotational symmetry other than 360 degrees): F, G, J, L, P, Q, R, S, Z
(f) Order of rotation 2: H, I, N, S, X, Z (Note: S and Z have rotational order 2)
(g) Order of rotation 4: None of the standard English capital letters have rotational symmetry of order 4 (Square/cross shapes do, but letters do not).
(h) Reflectional symmetry about a vertical mirror: A, M, T, U, V, W, Y, H, I, O, X
(i) Reflectional symmetry about a horizontal mirror: B, C, D, E, K, H, I, O, X
(j) Reflectional symmetry about both vertical and horizontal mirrors: H, I, O, X
In simple words: Vertical symmetry means left and right halves match; horizontal symmetry means top and bottom halves match.

Question 3. State the number of lines of symmetry for the following figures:
(a) scalene triangle
(b) isosceles triangle
(c) equilateral triangle
(d) square
(e) rectangle
(f) rhombus
(g) parallelogram
(h) quadrilateral (irregular)
(i) trapezium
(j) circle
Answer:
(a) Scalene triangle: 0 lines of symmetry
(b) Isosceles triangle: 1 line of symmetry
(c) Equilateral triangle: 3 lines of symmetry
(d) Square: 4 lines of symmetry
(e) Rectangle: 2 lines of symmetry
(f) Rhombus: 2 lines of symmetry
(g) Parallelogram: 0 lines of symmetry
(h) Quadrilateral (irregular): 0 lines of symmetry
(i) Trapezium (general): 0 lines of symmetry (an isosceles trapezium has 1)
(j) Circle: Infinite lines of symmetry
In simple words: Regular shapes have more lines of symmetry, while irregular or asymmetric shapes have zero.
Exam Tip: A circle has an infinite number of lines of symmetry because every line passing through its center is a line of symmetry.