I.G.C.S.E. Area. You can access the solutions from the end of each question

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Marlene Clark
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1 I.G.C.S.E. Area Index: Please click on the question number you want Question Question Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 You can access the solutions from the end of each question

2 Question For each of the questions, find the area of each shape. Decide which information to use: you may not need all it. a. b 5.3 cm 3.6 m 3.8 cm 5.4 cm 8.3 m c. d. 7.5 cm 7. mm 6.4 mm 60 cm 3.4 mm 0 cm Click here to read the solution to this question

3 Solution to question a. Rectangle b Trapezium 5.3 cm 3.6 m 3.8 cm 5.4 cm Area = length width = = 9.88 = 9.9 m 8.3 m Area = ( a+ ) 7.5 cm b h = ( ) 3.8 = 4.3 = 4.3cm c. Parallelogram d. Kite 7. mm 6.4 mm 60 cm 3.4 mm 0 cm Area = length height = = = mm Area = length of the product diagonals = 60 0 = 3600 cm Click here to read the question again

4 Question Find the shaded area of each of the following. a. b 0 m 6.5 m 8 cm 6 m 8 cm Click here to read the solution to this question

5 Solution to question a. 0 m 6.5 m Shaded area = area of rectangle area of trapezium = ( 0 + 6) 6.5 = 65 5 = 3 m 6 m b. 8 cm Shaded area = area of rectangle area of kite = = 44 7 = 7cm 8 cm ( ) Click here to read the question again

6 Question 3 A trapezium of area of 0cm has parallel sides 6 cm apart and one of these sides is 0 cm long. Find the length of the other parallel side. Click here to read the solution to this question

7 Solution to question 3 A trapezium of area of 0cm has parallel sides 6 cm apart and one of these sides is 0 cm long. Find the length of the other parallel side Drawing a diagram 0 cm 6 cm a b h 0 = ( 0 + b) 6 0 = b Area of a trapezium = ( + ) ( ) 40 = 0 + b b = 30cm b Click here to read the question again

8 Question 4 A kite of area 6cm has one diagonal 4 cm shorter than the other. Find the length of each diagonal. Click here to read the solution to this question

9 Solution to question 4 A kite of area 6cm has one diagonal 4 cm shorter than the other. Find the length of each diagonal. First draw a diagram x 4 x Let the length of the longer diagonal be x cm. Therefore the length of the shorter diagonal is x 4 cm. Area of kite = the product of the diagonals 6 = x( x 4) = x 4x 0 = x 4x product = sum = 4 factors = 6, 0 = x 6x + x x ( x 6) + ( x 6) ( x )( x + ) 0 = 0 = 6 either x 6 = 0 or x + = 0 x = 6 x = (not possible) Therefore the lengths of the diagonals are x = 6 cm or x 4 = 6 4 = cm. Click here to read the question again

10 Question 5 A floor 6 m by m is covered by square tiles with side 0 cm. How many tiles are needed? Click here to read the solution to this question

11 Solution to question 5 A floor 6 m by m is covered by square tiles with side 0 cm. How many tiles are needed? The area of the floor in Each tile has area cm is 0cm 0cm = 400cm 6m m = 600cm 00cm = 70000cm Number of tiles needed = = 800 tiles 400 Click here to read the question again

12 Question 6 Find the area of the following shapes a. b. 5 m 6 m 4.7 cm 8.4 cm m 5 cm Click here to read the solution to this question

13 Solution to question 6 a. 5 m 6 m m Area of a triangle = base = 5 = 7.5 m height b. 4.7 cm 8.4 cm Area of a triangle = base = =.75 =.8 cm 5 cm height Click here to read the question again

14 Question 7 Find the perimeter and area of the following shapes a. b. 9 cm cm 0 cm Click here to read the solution to this question

15 Solution to question 7 a. b. a 9 cm cm 0 cm C = πd = π 9 = 8.3cm A = π r = π ( 4.5) = 63.6cm Note a = 0 r = 0 6 = 4cm Perimeter = π = 0 + 6π = 38.8cm = 4 + π 6 Area = π = 96.5cm Click here to read the question again

16 Question 8 A circle radius 8 cm is inscribed inside a square as shown. Find the area shaded. 8 cm Click here to read the solution to this question

17 Solution to question 8 8 cm 6 cm Note the side of the square = r = 8 = 6cm Shaded area = area of square area of circle = 6 6 π 8 = 56 64π = 54.9cm Click here to read the question again

18 Question 9 Find the minor arc length AB and the area of the minor sector AOB. A O 35! 4.5 cm B Click here to read the solution to this question

19 Solution to question 9 A O 35! 4.5 cm B θ Minor arc length AB = π r! 360! 35 = π 4.5! 360 = 0.875π =.75cm Area of minor sector θ AB = πr! 360! 35 = π 4.5! 360 = 6.9cm Click here to read the question again

I.G.C.S.E. Volume & Surface Area. You can access the solutions from the end of each question

I.G.C.S.E. Volume & Surface Area Index: Please click on the question number you want Question 1 Question Question Question 4 Question 5 Question 6 Question 7 Question 8 You can access the solutions from