COMBINED FIGURES WORKSHEET

Question 1 :

Find the perimeter and area of the shape shown below. (Use π ≈ 3.14).

Question 2 :

Jennifer has a key-chain which is in the form of an equilateral triangle and a semicircle attached to a square of side 5 cm as shown below. Find its area. (Use π ≈ 3.14 and √3 = 1.732).

Question 3 :

A 3-fold invitation card is given with measures as shown below. Find its area.

Question 4 :

A 3-fold invitation card is given with measures as shown below. Find its area.

Question 5 :

Find the area of the shaded region in the square of side 10 cm as shown below (Use π ≈ 3.14).

Question 6 :

Find the area of an irregular polygon field whose measures are as shown below.

1. Answer :

Radius of a circular quadrant, r = 3.5 cm and side of a square, a = 3.5 cm.

The given figure is formed by the joining of 4 quadrants of a circle with each side of a square. The boundary of the given figure consists of 4 arcs and 4 radii.

Perimeter of the given combined shape :

= (4 x 1/4 x 2πr) + 4r

= (1 x 2 x 3.14 x 3.5) + 4(3.5)

= 21.98 + 14

= 35.98 cm (approximately)

Area of the given combined shape :

= a² + (4 x 1/4 x πr²)

= (3.5 x 3.5) + (1 x 3.14 x 3.5 x 3.5)

= 12.25 + 38.465

= 50.715

= 50.72 cm² (approximately)

2. Answer :

Side of the square = 5 cm

Diameter of the semi-circle = 5 cm ⇒ Radius = 2.5 cm

Side of the equilateral triangle = 5 cm

Area of the key-chain :

= (1/2)πr² + a² + (√3/4)a²

= (0.5 x 3.14 x 2.5 x 2.5) + (5 x 5) + [(1.732/4) x 5 x 5]

= 9.8125 + 25 + 10.825

= 45.64 cm² (approximately)

3. Answer :

Figures I and II are trapeziums separately as well as combinedly.

The parallel sides of the combined trapezium (I and II) are 5 cm and 16 cm and its height, h = 8 + 8 = 16 cm, length of the rectangle (III) = 16 cm and its breadth = 8 cm.

Area of the combined invitation card :

= area of the combined trapezium + area of the rectangle

= (1/2) x h x (a + b) + l x w

= 0.5 x 16 x (5 + 16) + 16 x 8

= 8 x 21 + 128

= 168 + 128

 = 296 cm²

4. Answer :

The mat given in the figure can be split into two rectangles as follows :

Area of the entire mat :

= area of the I rectangle + area of the II rectangle

= l₁ x w₁ + l₂ x w₂

= 5 x 2 + 9 x 2

= 10 + 18

= 28 ft²

Cost per square foot = $20.

The total cost of the entire mat = 28 x $20 = $560.

5. Answer :

Mark the unshaded parts of the given figure as I, II, III and IV

Area of the I and III parts :

= Area of the square – Area of 2 semicircles

= a² - 2 x (1/2) x πr²

= 10 x 10 - 2 x 0.5 x 3.14 x 5 x 5

= 100 - 78.5

= 21.5 cm²

Similarly, the area of the II and IV parts = 21.5 cm².

Area of the unshaded parts (I, II, III and IV) :

= 2 x 21.5

= 43 cm² (approximately)

Area of the shaded part :

= area of the square – area of the unshaded parts

= 100 – 43

= 57 cm² (approximately)

6. Answer :

The given field has four triangles (I, III, IV and V) and a trapezium.

Area of triangle (I) :

= (1/2) x b x h

= 0.5 x 5 x 6

= 15 m²

Area of trapezoid (II) :

= (1/2) x h x (a + b)

= 0.5 x 13 x (6 + 4)

= 0.5 x 13 x 10

= 65 m²

Area of triangle (III) :

= (1/2) x b x h

= 0.5 x 8 x 4

= 16 m²

Area of triangle (IV) :

= (1/2) x b x h

= 0.5 x 13 x 10

= 65 m²

Area of triangle (V) :

= (1/2) x b h

= 0.5 x 13 x 10

= 65 m²

The total area of the field :

= 15 + 65 + 16 + 65 + 65

= 226 m²

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