Sasi Math (Revised for 1024)

Scroll:Measurement >> Area & Perimeter of Composite Figure >> ps (1743)

Written Instructions:

For each Multiple Choice Question (MCQ), four options are given. One of them is the correct answer. Make your choice (1,2,3 or 4). Write your answers in the brackets provided..

For each Short Answer Question(SAQ) and Long Answer Question(LAQ), write your answers in the blanks provided.

Leave your answers in the simplest form or correct to two decimal places.

The figure below is formed by two right-angled triangles, a circle and a circle. Given that WX = XY = XZ = 18cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

18cm 18cm

___cm²

Answer:_______________

The figure below is formed by two right-angled triangles, a circle and a circle. Given that WX = XY = XZ = 28cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

28cm 28cm

___cm²

Answer:_______________

The figure below is formed by two right-angled triangles, a circle and a circle. Given that WX = XY = XZ = 10cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

10cm 10cm

___cm²

Answer:_______________

The figure below is formed by two right-angled triangles, a circle and a circle. Given that WX = XY = XZ = 20cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

20cm 20cm

___cm²

Answer:_______________

The figure below is formed by two right-angled triangles, a circle and a circle. Given that WX = XY = XZ = 34cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

34cm 34cm

___cm²

Answer:_______________

The figure below is formed by two right-angled triangles, a circle and a circle. Given that WX = XY = XZ = 24cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

24cm 24cm

___cm²

Answer:_______________

The figure below is formed by two right-angled triangles, a circle and a circle. Given that WX = XY = XZ = 40cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

40cm 40cm

___cm²

Answer:_______________

The figure below is formed by two right-angled triangles, a circle and a circle. Given that WX = XY = XZ = 26cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

26cm 26cm

___cm²

Answer:_______________

The figure below is formed by two right-angled triangles, a circle and a circle. Given that WX = XY = XZ = 36cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

36cm 36cm

___cm²

Answer:_______________

10)

The figure below is formed by two right-angled triangles, a circle and a circle. Given that WX = XY = XZ = 14cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

14cm 14cm

___cm²

Answer:_______________

Π = 3.142

18cm 18cm

Answer: 208cm²

SOLUTION 1 :

18cm 18cm

Step 1: Diameter = 18cm

Radius = 18 ÷ 2

= 9cm

Step 2: $\frac{1}{2}$ rugby → quadrant - triangle

→ ( $\frac{1}{4}$ x π x r² ) - ( $\frac{1}{2}$ x b x h )

→ ( $\frac{1}{4}$ x π x 9 x 9 ) - ( $\frac{1}{2}$ x 9 x 9 )

= 23.126cm²

Step 3: shaded area → big circle - rugby

→ (π x r²) - (2 x 23.126)

→ (π x 9 x 9) - (2 x 23.126)

= 208.250cm²

≈ 208cm²

Π = 3.142

28cm 28cm

Answer: 504cm²

SOLUTION 1 :

28cm 28cm

Step 1: Diameter = 28cm

Radius = 28 ÷ 2

= 14cm

Step 2: $\frac{1}{2}$ rugby → quadrant - triangle

→ ( $\frac{1}{4}$ x π x r² ) - ( $\frac{1}{2}$ x b x h )

→ ( $\frac{1}{4}$ x π x 14 x 14 ) - ( $\frac{1}{2}$ x 14 x 14 )

= 55.958cm²

Step 3: shaded area → big circle - rugby

→ (π x r²) - (2 x 55.958)

→ (π x 14 x 14) - (2 x 55.958)

= 503.916cm²

≈ 504cm²

Π = 3.142

10cm 10cm

Answer: 64cm²

SOLUTION 1 :

10cm 10cm

Step 1: Diameter = 10cm

Radius = 10 ÷ 2

= 5cm

Step 2: $\frac{1}{2}$ rugby → quadrant - triangle

→ ( $\frac{1}{4}$ x π x r² ) - ( $\frac{1}{2}$ x b x h )

→ ( $\frac{1}{4}$ x π x 5 x 5 ) - ( $\frac{1}{2}$ x 5 x 5 )

= 7.138cm²

Step 3: shaded area → big circle - rugby

→ (π x r²) - (2 x 7.138)

→ (π x 5 x 5) - (2 x 7.138)

= 64.274cm²

≈ 64cm²

Π = 3.142

20cm 20cm

Answer: 257cm²

SOLUTION 1 :

20cm 20cm

Step 1: Diameter = 20cm

Radius = 20 ÷ 2

= 10cm

Step 2: $\frac{1}{2}$ rugby → quadrant - triangle

→ ( $\frac{1}{4}$ x π x r² ) - ( $\frac{1}{2}$ x b x h )

→ ( $\frac{1}{4}$ x π x 10 x 10 ) - ( $\frac{1}{2}$ x 10 x 10 )

= 28.550cm²

Step 3: shaded area → big circle - rugby

→ (π x r²) - (2 x 28.550)

→ (π x 10 x 10) - (2 x 28.550)

= 257.100cm²

≈ 257cm²

Π = 3.142

34cm 34cm

Answer: 743cm²

SOLUTION 1 :

34cm 34cm

Step 1: Diameter = 34cm

Radius = 34 ÷ 2

= 17cm

Step 2: $\frac{1}{2}$ rugby → quadrant - triangle

→ ( $\frac{1}{4}$ x π x r² ) - ( $\frac{1}{2}$ x b x h )

→ ( $\frac{1}{4}$ x π x 17 x 17 ) - ( $\frac{1}{2}$ x 17 x 17 )

= 82.510cm²

Step 3: shaded area → big circle - rugby

→ (π x r²) - (2 x 82.510)

→ (π x 17 x 17) - (2 x 82.510)

= 743.018cm²

≈ 743cm²

Π = 3.142

24cm 24cm

Answer: 370cm²

SOLUTION 1 :

24cm 24cm

Step 1: Diameter = 24cm

Radius = 24 ÷ 2

= 12cm

Step 2: $\frac{1}{2}$ rugby → quadrant - triangle

→ ( $\frac{1}{4}$ x π x r² ) - ( $\frac{1}{2}$ x b x h )

→ ( $\frac{1}{4}$ x π x 12 x 12 ) - ( $\frac{1}{2}$ x 12 x 12 )

= 41.112cm²

Step 3: shaded area → big circle - rugby

→ (π x r²) - (2 x 41.112)

→ (π x 12 x 12) - (2 x 41.112)

= 370.224cm²

≈ 370cm²

Π = 3.142

40cm 40cm

Answer: 1028cm²

SOLUTION 1 :

40cm 40cm

Step 1: Diameter = 40cm

Radius = 40 ÷ 2

= 20cm

Step 2: $\frac{1}{2}$ rugby → quadrant - triangle

→ ( $\frac{1}{4}$ x π x r² ) - ( $\frac{1}{2}$ x b x h )

→ ( $\frac{1}{4}$ x π x 20 x 20 ) - ( $\frac{1}{2}$ x 20 x 20 )

= 114.200cm²

Step 3: shaded area → big circle - rugby

→ (π x r²) - (2 x 114.200)

→ (π x 20 x 20) - (2 x 114.200)

= 1028.400cm²

≈ 1028cm²

Π = 3.142

26cm 26cm

Answer: 434cm²

SOLUTION 1 :

26cm 26cm

Step 1: Diameter = 26cm

Radius = 26 ÷ 2

= 13cm

Step 2: $\frac{1}{2}$ rugby → quadrant - triangle

→ ( $\frac{1}{4}$ x π x r² ) - ( $\frac{1}{2}$ x b x h )

→ ( $\frac{1}{4}$ x π x 13 x 13 ) - ( $\frac{1}{2}$ x 13 x 13 )

= 48.250cm²

Step 3: shaded area → big circle - rugby

→ (π x r²) - (2 x 48.250)

→ (π x 13 x 13) - (2 x 48.250)

= 434.498cm²

≈ 434cm²

Π = 3.142

36cm 36cm

Answer: 833cm²

SOLUTION 1 :

36cm 36cm

Step 1: Diameter = 36cm

Radius = 36 ÷ 2

= 18cm

Step 2: $\frac{1}{2}$ rugby → quadrant - triangle

→ ( $\frac{1}{4}$ x π x r² ) - ( $\frac{1}{2}$ x b x h )

→ ( $\frac{1}{4}$ x π x 18 x 18 ) - ( $\frac{1}{2}$ x 18 x 18 )

= 92.502cm²

Step 3: shaded area → big circle - rugby

→ (π x r²) - (2 x 92.502)

→ (π x 18 x 18) - (2 x 92.502)

= 833.004cm²

≈ 833cm²

10)

Π = 3.142

14cm 14cm

Answer: 126cm²

SOLUTION 1 :

14cm 14cm

Step 1: Diameter = 14cm

Radius = 14 ÷ 2

= 7cm

Step 2: $\frac{1}{2}$ rugby → quadrant - triangle

→ ( $\frac{1}{4}$ x π x r² ) - ( $\frac{1}{2}$ x b x h )

→ ( $\frac{1}{4}$ x π x 7 x 7 ) - ( $\frac{1}{2}$ x 7 x 7 )

= 13.990cm²

Step 3: shaded area → big circle - rugby

→ (π x r²) - (2 x 13.990)

→ (π x 7 x 7) - (2 x 13.990)

= 125.978cm²

≈ 126cm²