Another Test

Solution

= \( \normalsize 3.679 \: \times \: 10^5\)

Solution

⇒ \(\normalsize 7.823 \: \times \: 10^{-4}\)

Solution

Simplify \(\Large \frac{0.012 \: \times \: 0.00009}{0.06} \)

convert to standard form

⇒ \(\Large \frac{1.2 \: \times \: 10^{-2} \: \times \: 9\: \times \: 10^{-5}}{6 \: \times \: 10^{-2}} \)

⇒ \(\Large \frac{\not{1.2} \: \times \: 10^{-2} \: \times \: 9\: \times \: 10^{-5}}{\not{6} \: \times \: 10^{-2}} \)

⇒ \(\normalsize 0.2 \: \times \: 10^{-2 \: - \:(-2)} \: \times \: 9\: \times \: 10^{-5} \)

⇒ \(\normalsize 0.2 \: \times \: 10^{-2 \: + \:2} \: \times \: 9\: \times \: 10^{-5} \)

⇒ \(\normalsize 0.2 \: \times \: 10^{0} \: \times \: 9\: \times \: 10^{-5} \)

⇒ \(\normalsize 0.2 \: \times \: 1 \: \times \: 9\: \times \: 10^{-5} \)

⇒ \(\normalsize 0.2  \: \times \: 9\: \times \: 10^{-5} \)

= \(\normalsize 1.8\: \times \: 10^{-5} \)

Explanation:

Simplify \( \normalsize 2\frac{1}{2} \: + \: 5 \)

⇒ \( \Large \frac{5}{2} \: + \: \frac{5}{1} \)

⇒ \( \Large \frac{5 \: + \: 10}{2} \)

= \( \Large \frac{15}{2} \)

Solution

Simplify \( \normalsize 5 \frac{2}{3} \: - \: 3 \frac{1}{4} \)

Convert to improper fractions

⇒ \( \Large  \frac{17}{3} \: -\: \frac{13}{4} \)

⇒ \( \Large  \frac{68 \: - \: 39}{12}\)

= \( \Large  \frac{29}{12}\)

Solution

Simplify \( \normalsize 2\frac{3}{2} \: \times \: 1 \frac{3}{5} \: \div \: 1 \frac{2}{5} \)

convert to improper fractions

⇒ \( \Large \frac{7}{2} \: \times \:  \frac{8}{5} \: \div \:  \frac{7}{5} \)

Using BODMAS - divide first

⇒ \( \Large \frac{7}{2} \: \times \:  \left( \Large\frac{8}{5} \: \div \:  \frac{7}{5} \right) \)

⇒ \( \Large \frac{7}{2} \: \times \:  \left( \Large\frac{8}{5} \: \times \:  \frac{5}{7} \right) \)

⇒ \( \Large \frac{7}{2} \: \times \:  \left( \Large\frac{8}{\not{5}} \: \times \:  \frac{\not{5}}{7} \right) \)

⇒ \( \Large \frac{7}{2} \: \times \:  \left( \Large\frac{8}{7}  \right) \)

⇒ \( \Large \frac{7}{2} \: \times \:  \frac{8}{7}\)

⇒ \( \Large \frac{\not{7}}{2} \: \times \:  \frac{8}{\not{7}}\)

⇒ \( \Large \frac{1}{2} \: \times \:  \frac{8}{1}\)

⇒ \(\Large \frac{8}{2}\)

=  \(\normalsize 4\)

Solution

Simplify \( \normalsize 5 \frac{1}{4} \: - \: \left( \normalsize 1 \frac{2}{3} \: - \: \Large \frac{1}{2}\right) \)

Convert to improper fractions

⇒ \( \Large \frac{21}{4} \: - \: \left( \Large \frac{5}{3} \: - \: \frac{1}{2}\right) \)

⇒ \( \Large \frac{21}{4} \: - \: \left( \Large \frac{10 \: - \: 3}{6}\right) \)

⇒ \( \Large \frac{21}{4} \: - \: \left( \Large \frac{7}{6}\right) \)

⇒ \( \Large \frac{21}{4} \: - \:  \frac{7}{6} \)

⇒ \( \Large \frac{63 \: -\: 14}{12} \)

⇒ \( \Large \frac{49}{12} \)

Convert to proper fractions

= \( \normalsize 4 \frac{1}{12} \)

Solution

Convert 1.625 into a mixed fraction

1.625 = \( \Large \frac{1625}{1000} \)

Convert to proper fraction and simplify

= \( \normalsize 1 \frac{625}{1000} \)

Divide the fraction by 25

= \( \normalsize 1 \frac{25}{40} \)

Divide the fraction by 5

= \( \normalsize 1 \frac{5}{8} \)

Solution

27²

= 27 × 27

Solution

Step 1: Express 60 as a product of its prime factors.

Step 2: Pick the product of prime factors in pairs of the same digits. Then multiply digits without pairs (stand-alone digits) together. This gives the required number. i.e.

‘3  and 5’ stand-alone

The required number = 3 × 5

= 15

Check

60 × 15 = 900

900 is a perfect square

⇒ \( \normalsize \sqrt{900} = 30 \)

Solution

Step 1: Calculate the L.C.M

L.C.M = 2 × 2 × 2 × 2 × 3 × 3 = 144

Step 2: Calculate the H.C.F

H.C.F = 2 × 2 × 3 × 3  = 36

Divide the L.C.M of 36, 72 and 144 by their H.C.F

i.e. \( \Large \frac{144}{36} \\ = \normalsize 4 \)

Solution

Find the H.C.F. of 18, 40 and 48

Short division method

The steps are as follows:

• Write the numbers in individual rows.
• Divide the numbers by common prime factors.
• Factorisation stops when we reach prime numbers which cannot be further divided.
• HCF is the product of all the common factors.

H.C.F = 2 × 3 = 6

Prime Factorization Method

Express each number as a product of its prime factors and multiply their common prime factors.

18 = 2 × 3 × 3

40 = 2³ × 5

48 = 2⁴ × 3

Compare powers of the same base. The one with the lowest power is a common factor.

∴ 2 × 3 = 6

Solution

To find the value of ( x ), we start with the given information that the highest common factor (H.C.F) of 48, ( x ), and 88 is 8.

Firstly, let's determine the prime factorization of 48 and 88:

48 = \( \normalsize 2^4 \: \times \: 3\)

88 = \( \normalsize 2^3 \: \times \: 11\)

Common prime factors:

H.C.F = \( \normalsize 2^3 = 8\)

Now, we know that the H.C.F of 48, x, and 88 is 8.

We need x to be a number such that its H.C.F. with both 48 and 88 is 8.

Since 8 is a factor of 48 and 88, x must also be divisible by 8.

The only option that is divisible by 8 is 56

i.e. 56 ÷ 8 = 7

Therefore, the correct value of x is 56.

Solution

0.025 = \( \Large \frac{25}{1000} \\ = \Large \frac{1}{40} \)

Solution

Express 15 as a percentage of 20.

⇒ \(\Large \frac{15}{20} \normalsize \: \times \: 100 \\ = \Large \frac{3}{4} \normalsize \: \times \: 100\\ = \normalsize 3 \: \times \: 25 \\ =\normalsize 75.00 \% \)

Solution

Find 7.5% of 4 kg 300 g, correct to 3 significant figures.

convert kg to g

1 kg = 1000 g

∴ 4 kg = 4 × 1000 g = 4000 g

4 kg 300 g = 4000 g + 300 g = 4300 g

7.5% of 4300 g

⇒ \( \Large \frac{7.5}{100} \normalsize \: \times \: 4300 \\ =\normalsize 7.5 \: \times \: 43 \\=\normalsize 322.5  \)

To 3.s.f ≈ 323 g

Solution

Express 68 as a binary number

= 1000100₂

Solution

= \( \normalsize 2^4 \: \times \: 3^2 \)

Solution

= (1 × 2²) + (1 × 2¹) + (1 × 2⁰)

= 4 + 2 + 1

= 7

Explanation:

• Simple interest, I = ₦2,400.00
• Time, T = 2 years
• Rate, R = 5%
• Principal = ?

Using the formula:

I = \( \Large \frac{PRT}{100}\)

make P the subject

P = \( \Large \frac{100I}{RT} \)

substitute the given values into the equation

P = \( \Large \frac{100 \: \times \: 2400}{5 \: \times \: 2} \)

P = \( \Large \frac{100 \: \times \: 2400}{10} \)

P = \( \normalsize 10 \: \times \: 2400 \)

P = ₦24,000.00

Solution

Solution

• Principal, P= ₦15,000
• Time, T = 2½ years  = \( \Large \frac{5}{2} \normalsize \: years\)
• Rate = 5½% = \( \Large \frac{11}{2} \normalsize \: \%\)
• Simple Interest, I = ?

Using the formula:

I = \( \Large\frac{PRT}{100} \)

substitute the given values into the formula:

I = \( \Large\frac{15000 \: \times \: 11 \: \times \: 5}{100 \: \times \: 2 \: \times \: 2} \)

I = \(\Large \frac{150 \: \times \: 11 \: \times \: 5}{4} \)

I = \( \Large\frac{8250}{4} \)

I = ₦2062.5

Solution

Felix bought a car at the rate of ₦3,088,144.00. He made the cash payment of 32% and paid the remaining through bank transfer. How much was paid through transfer?

He made the cash payment of: \( \Large \frac{32}{100}\normalsize \: \times \: 3,088,144\\ \normalsize = 0.32 \: \times \: 3,088,144\\ \normalsize = 988,206.08 \)

Therefore he paid ₦3,088,144.00 - ₦988,206.08 through transfer

= ₦2,099,937.92

Solution

Round up the numbers to whole numbers;

= ₦11 + ₦6 +  ₦125 +  ₦16 + ₦15 = ₦173.00

Solution

Interior angle = 165º

Exterior angle = 180 - 165 = 15º

Exterior angle = \( \Large \frac{360}{n}\)

15 = \( \Large \frac{360}{n}\)

15n = 360

n = \( \Large \frac{360}{15} \\ = \normalsize  24\)

Solution

M \( \normalsize \propto \Large \frac{1}{N^2} \)

introducing constant k

M =  \( \normalsize  k \Large \frac{1}{N^2} \)

M = \(  \Large \frac{k}{N^2} \)

2 = \(  \Large \frac{k}{\frac{1}{2}^2} \)

2 = \(  \Large \frac{k}{\frac{1}{4}} \)

2 = \(  \normalsize k \: \times \: 4 \)

⇒ k = \( \Large \frac{2}{4} \)

∴ k = \( \Large \frac{1}{2} \)

substituting k into M = \(  \Large \frac{k}{N^2} \)

we have:

M = \(  \Large \frac{\frac{1}{2}}{N^2} \)

∴ M = \(  \Large \frac{1}{2N^2} \)

Explanation: 140º angle cannot be constructed using only a ruler and compass; this is because angles that can be constructed with these tools are limited to multiples of 15 degrees, and 140 does not fall into that category

Solution

The formula for sum of interior angles is:

(2n - 4)90

where n = number of sides.

when n= 10

⇒ ([2 × 10] - 4) × 90

⇒ (20 - 4) × 90

⇒ 16 × 90

= 1440º

Explanation

Solution

Area of circle = \( \normalsize \pi r^2 \)

diameter of circle = 42 cm

radius = \( \Large \frac{42}{2} \\ \normalsize = 21 cm \)

⇒ \( \normalsize \pi = \Large \frac{22}{7}\)

∴ Area of circle = \( \Large \frac{22}{7} \normalsize \: \times \:  21^2\\ =\Large \frac{22}{7} \normalsize \: \times \:  21 \: \times \: 21 \\ =\normalsize 22 \: \times \: 3 \: \times \:  21\\ = \normalsize 1386 \: cm^2 \)

Solution

From the diagram, we are given the adjacent (y) and hypotenuse (25 cm) sides, therefore we can use the formula for cosine to find y.

Remember: SOHCAHTOA

cos θ = \( \Large \frac{adj}{hyp}\)

cos 50 = \( \Large \frac{y}{25}\)

y = cos 50 × 25

y = 0.643 × 25

y = 16.075 cm

y ≅ 16.1 cm (to 1.d.p)

Solution

Circumference of a circle = 2πr

∴ 21 cm = 2πr

make radius, r the subject of the formula

r = \(\Large \frac{21}{2 \pi} \normalsize \: cm \)

Solution

• base radius = 3½ cm
• height = 10 cm
• π = \( \Large \frac{22}{7}\)

Volume of a cylinder = \( \normalsize \pi r^2 h \\ = \Large \frac{22}{7} \normalsize \: \times \: \left(\Large \frac{7}{2}\right)^2 \normalsize \: \times \: 10\\ =  \Large \frac{22}{7} \normalsize \: \times \: \Large \frac{7}{2}  \normalsize \: \times \: \Large \frac{7}{2}  \normalsize \: \times \: 10  \\ = \Large \frac{22}{2} \normalsize \: \times \: \Large \frac{7}{2}  \normalsize \: \times \: 10 \\ =  \normalsize 11 \: \times \: 7 \: \times \: 5 \\ =  \normalsize 385 \: cm^3\)

Solution

Using Pythagoras theorem

x² = 4² + 3²

x² = 16 + 9

x² = 25

take the square root of both sides

x = \( \normalsize \sqrt{25} \)

∴ x = 5

Solution

Area of a circle = 154 cm² = πr²

make r the subject of the formula and solve for r

r² = \( \Large \frac{154}{\pi} \)

r = \(\sqrt{ \Large \frac{154}{\frac{22}{7}} }\)

r = \(\sqrt{ \normalsize 154 \: \times \: \Large \frac{7}{22}}\)

r = \(\sqrt{ \Large \frac{\not{154}}{1} \: \times \: \Large \frac{7}{\not{22}}}\)

r = \(\sqrt{ \normalsize 7 \: \times \: 7}\)

r = \(\sqrt{ \normalsize 49}\)

r = 7 cm

Perimeter of circle = 2πr

but r = 7 cm

∴ Perimeter of circle = \( \normalsize 2 \: \times \: \pi \: \times \: 7  \\ = \normalsize 2 \: \times \: \Large \frac{22}{\not{7}} \normalsize \: \times \: \Large \frac{\not{7}}{1} \\ = \normalsize 2 \: \times \:22\\= \normalsize 44 \: cm \)

Solution

First thing we need to do is find x before finding the perimeter.

Using Pythagoras theorem:

x² = 15² + 20²

x² = 225 + 400

x² = 625

x = \( \sqrt{\normalsize 625} \)

x = 25 cm

The perimeter of the triangle = 15 + 20 + x

= 15 + 20 + 25

= 60 cm

Solution

• opposite = x
• hypotenuse = 10 m

Since we are given the opposite and hypotenuse sides we can use the sine formula to find x

SOHCAHTOA

sin θ = \( \Large \frac{opposite}{hypotenuse}\)

sin 45 = \( \Large \frac{x}{10}\)

x = sin 45 × 10

x = 0.707 × 10

x = 7.07 m

x ≅ 7.1 m (to 1 d.p.)

Solution

The shape is a trapezium

Area of trapezium = \( \Large \frac{1}{2} \left(\normalsize a \: + \: b \right) \normalsize  h\)

where:

• h = 4 cm
• a = 12 cm
• b = EF

EF = ED + DC + CF

To find ED, consider ΔADE

AE² = AD² + ED²

i.e. 5² = 4² + ED²

ED² = 5² - 4²

ED² = 25 - 16

ED² = 9

ED = \( \normalsize \sqrt{9} \)

ED = 3 cm

Similarly CF = 3 cm

and DC = 12 cm (since ABDC is a rectangle)

We can now redraw the trapezium

b = EF = ED + DC + CF

b = EF = 3 + 12 + 3

∴ b = 18 cm

Area of trapezium = \( \Large \frac{1}{2} \left(\normalsize a \: + \: b \right) \normalsize  h\)

Area of trapezium = \( \Large \frac{1}{2} \left(\normalsize 12 \: + \: 18 \right) \normalsize \: \times \:  4 \\ = \Large \frac{1}{2}  \normalsize\: \times\:  30 \: \times\: 4 \\ = \normalsize 60 \: cm^2 \)

2nd method

Find the sum of the area of ΔADE, rectangle ABDE and ΔBCF

area of ΔADE = \( \Large \frac{1}{2} \normalsize bh \)

• b = ED = 3 cm
• h = 4 cm

area of ΔADE = \( \Large \frac{1}{2} \normalsize \: \times \:  3 \: \times \: 4 \\ \normalsize = 6 \: cm^2 \)

∴ area of ΔBCF = 6 cm²

area of rectangle = 12 × 4 = 48 cm²

Area of shape = 6 + 48 + 6 = 60 cm²

Solution

x = adjacent

15 m = opposite

SOHCAHTOA

Use the tan formula to find x

tan θ = \( \Large \frac{opposite}{adjacent}\)

tan 50 = \( \Large \frac{15}{x}\)

x = \( \Large \frac{15}{tan\:50}\)

x = \( \Large \frac{15}{1.1918}\)

x = 12.586 m

x ≅ 12.59 m (to 2 d.p.)

Solution

Area of parallelogram = base × height

• base = 15 cm
• height = 2.5 cm

Area of parallelogram = 15 × 2.5 = 37.5 cm²

Solution

Volume of cuboid = l × b × h

• length = 25 cm
• breadth or width = 15 cm
• height  = 10 cm

∴ Volume of cuboid = 25 × 15 × 10 = 3750 cm³

Solution

Volume of cone = \( \Large \frac{1}{3} \normalsize \pi r^2 h\)

• height is \( \Large \frac{14}{3}\normalsize \: cm \)
• base radius \( \Large \frac{3}{2}\normalsize\: cm \)
• \(\normalsize \pi = \Large \frac{22}{7}\)

Volume of cone = \( \Large \frac{1}{3}\: \times\: \Large \frac{22}{7} \: \times\: \left( \Large \frac{3}{2}\right)^2 \: \times\: \Large \frac{14}{3} \)

Volume of cone = \( \Large \frac{1}{3}\: \times\: \Large \frac{22}{7} \: \times\:  \Large \frac{9}{4}\: \times\: \Large \frac{14}{3} \)

Volume of cone = \( \Large \frac{1}{3}\: \times\: \Large \frac{22}{7} \: \times\:  \Large \frac{\not{9}}{\not{4}}\: \times\: \Large \frac{\not{14}}{\not{3}} \)

Volume of cone = \( \Large \frac{1}{3}\: \times\: \Large \frac{22}{7} \: \times\:  \Large \frac{3}{2}\: \times\: \Large \frac{7}{1} \)

Volume of cone = \( \Large \frac{1}{\not{3}}\: \times\: \Large \frac{22}{\not{7}} \: \times\:  \Large \frac{\not{3}}{2}\: \times\: \Large \frac{\not{7}}{1} \)

Volume of cone = \(\normalsize 22 \: \times \:  \Large \frac{1}{2} \)

Volume of cone = \(\normalsize 11\: cm^3 \)

Solution

Area of parallelogram = base × height

• Area = 28 cm²
• base = 7 cm
• height, h = ?

28 = 7 × h

h = \( \Large \frac{28}{7} \)

h = 4 cm

Solution

Bearing of Y from X = 90° + 90° + 90° + 55° = 325°

Solution

S83ºW to a three-figure bearing = 90º + 90º + 83º = 263º

Solution

the three-figure bearing of X from Y = 040°

Solution

• hypotenuse = 16 m
• opposite = x

SOHCAHTOA

sin θ = \(\Large \frac{opposite}{hypotenuse}\)

sin 30 = \(\Large \frac{x}{16}\)

x = sin 30 × 16

x = 0.5 × 16

x = 8 m

Solution

Sum of angles in a triangle = 180

∴ x + 2 + 2x + 23 + 3x + 5 = 180

collect like terms

x + 2x + 3x + 2 + 23 + 5 = 180

6x + 30 = 180

6x = 180 - 30

6x = 150

x = \( \Large \frac{150}{6} \)

x = 25°

Solution

scale factor = 3 : 4

area factor = 3² : 4²

area factor = 9 : 16

area factor = \( \Large \frac{area \: of \: smaller \: shape}{area \: of \: bigger \: shape} \)

i.e. \( \Large \frac{9}{16} = \frac{81}{x} \)

cross multiply

9 × x = 81 × 16

x = \( \Large \frac{81\:\times\:16}{9} \)

x = \( \Large \frac{\not{81}\:\times\:16}{\not{9}} \)

x = 9 × 16 = 144 cm²

Solution

⇒ \( \Large \frac{420}{6800} \normalsize \: \times \: 100 \)

⇒ \( \Large \frac{420}{68\not{0}\not{0}} \normalsize \: \times \: 1\not{0}\not{0} \)

⇒ \( \Large \frac{420}{68}\)

= 6.176 %

≅ 6.2 % (to 1 d.p.)

Solution

• • The rhombus is a shape formed by 4 straight lines.
  • All sides are equal and, opposite sides are parallel to each other
  • The diagonals of a rhombus are not equal but perpendicular to each other.

Area of rhombus = 360 cm²

Length of diagonal = 24 cm

Length of second diagonal = x  (as shown in the diagram above)

Area of ABCD = 2 × Area of ΔACB

ΔACB = \( \Large \frac{1}{2} \normalsize bh \)

360 = \(\normalsize 2 \left( \Large \frac{1}{2} \normalsize \: \times \: 24 \: \times \:  \Large \frac{x}{ 2} \right)\)

360 = \(\normalsize 2  \: \times \: \left(  \normalsize 12 \: \times \:  \Large \frac{x}{ 2} \right)\)

360 = \(\normalsize \not{2}  \: \times \: \left(  \normalsize 12 \: \times \:  \Large \frac{x}{ \not{2}} \right)\)

360 = 12x

x = \( \Large \frac{360}{12} \)

x = 30 cm

Solution

Volume scale factor = \( \Large \frac{64}{125} \)

Length scale factor = \( \Large \frac{\sqrt[3]{64}}{\sqrt[3]{125}} \\ = \Large \frac{4}{5} \)

∴ Area scale factor = \( \Large \frac{4^2}{5^2} \\  = \Large \frac{16}{25} \\ = \normalsize 16 \: : \: 25\)

Solution

Area scale factor = \( \Large \frac{4^2}{5^2} \\  = \Large \frac{16}{25} \\ = \normalsize 16 \: : \: 25\)

Solution

Linear scale factor = 3 : 7

Volume scale factor = \( \Large \frac{3}{7} \\ = \Large \frac{3^3}{7^3} \\ = \Large \frac{3 \: \times \: 3 \: \times \: 3}{7\: \times \:7\: \times \:7}\\ = \Large \frac{27}{343} \)

Solution

Area scale factor of two similar shapes is 4 : 9

Linear scale factor = \( \Large \frac{\sqrt{4}}{\sqrt{9}}\\ = \Large \frac{2}{3} \)

∴ Volume scale factor = \( \Large \frac{2^3}{3^3} \\ = \Large \frac{8}{27} \)

Volume scale factor = \( \Large \frac{Volume\:of\:smaller\:shape}{Volume\:of\:bigger\:shape} \)

⇒ \( \Large \frac{8}{27} = \frac{x}{270} \)

x = \( \Large \frac{270 \: \times \: 8}{27} \)

x = 10 × 8 = 80 cm³

Solution

Area of parallelogram = b × h

• b = 18 cm
• h = ?

use the sin formula to calculate h

• h = opposite
• PQ = 15 cm = hypotenuse

sin 30º = \( \Large \frac{h}{15} \)

h = sin 30 × 15

h = 0.5 × 15 = 7.5 cm

Area of parallelogram = 18 × 7.5 = 135 cm²

Solution

Area of circle = \( \normalsize \pi r^2 \)

r = 7 cm

Area of circle = \( \Large \frac{22}{7} \: \times \: \normalsize  7^2 \\ =\Large \frac{22}{7} \: \times \: \normalsize  7 \: \times \: 7 \\  = \normalsize 22 \: \times\: 7 \\=\normalsize 154\: cm^2 \)

Solution

2y + 5y + y = 180 (angles on a straight line)

8y = 180

y = \( \Large \frac{180}{8}\)

y = 22.5º

Solution

Area of shaded portion = Area of rectangle  - Area of semi-circle

Area of rectangle = 15 × 20 = 300 cm²

Area of semi-circle = \( \Large \frac{\pi r^2}{2} \normalsize \: or \: \Large \frac{\pi d^2}{8} \)

where d = 15 cm, r = 15/2 = 7.5 cm

Area of semi-circle = \( \Large \frac{22}{7} \normalsize \: \times \: 7.5^2 \: \times \: \Large \frac{1}{2}\\ = \Large \frac{\not{22}}{7} \normalsize \: \times \: 56.25 \: \times \: \Large \frac{1}{\not{2}} \\ = \Large \frac{11}{7} \normalsize \: \times \: 56.25 \\ = \Large \frac{618.75}{7} \\ = 88.39 \: cm^2\)

⇒ Area of shaded portion = 300 - 88.39 = 211.61 cm²

∴ Area of shaded portion ≈ 212 cm²