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 = 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 = 14cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

14cm 14cm

___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 = 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 = 32cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

32cm 32cm

___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 = 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 = 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 = 22cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

22cm 22cm

___cm²

Answer:_______________

10)

The figure below is formed by two right-angled triangles, a circle and a circle. Given that WX = XY = XZ = 16cm, find the area of the shaded region rounding off your answer to the nearest whole number.

Π = 3.142

16cm 16cm

___cm²

Answer:_______________

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

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

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

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

32cm 32cm

Answer: 658cm²

SOLUTION 1 :

32cm 32cm

Step 1: Diameter = 32cm

Radius = 32 ÷ 2

= 16cm

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 16 x 16 ) - ( $\frac{1}{2}$ x 16 x 16 )

= 73.088cm²

Step 3: shaded area → big circle - rugby

→ (π x r²) - (2 x 73.088)

→ (π x 16 x 16) - (2 x 73.088)

= 658.176cm²

≈ 658cm²

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

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

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

22cm 22cm

Answer: 311cm²

SOLUTION 1 :

22cm 22cm

Step 1: Diameter = 22cm

Radius = 22 ÷ 2

= 11cm

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 11 x 11 ) - ( $\frac{1}{2}$ x 11 x 11 )

= 34.546cm²

Step 3: shaded area → big circle - rugby

→ (π x r²) - (2 x 34.546)

→ (π x 11 x 11) - (2 x 34.546)

= 311.090cm²

≈ 311cm²

10)

Π = 3.142

16cm 16cm

Answer: 165cm²

SOLUTION 1 :

16cm 16cm

Step 1: Diameter = 16cm

Radius = 16 ÷ 2

= 8cm

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 8 x 8 ) - ( $\frac{1}{2}$ x 8 x 8 )

= 18.272cm²

Step 3: shaded area → big circle - rugby

→ (π x r²) - (2 x 18.272)

→ (π x 8 x 8) - (2 x 18.272)

= 164.544cm²

≈ 165cm²