Volume = 264 cm³

Radius (r) = 6 cm

π = $\frac{22}{7}$

264 = $\frac{22}{7}$ x 6² x $\frac{\mathrm{h}}{3}$

264 = $\frac{\mathrm{22\; x\; 36\; x\; h}}{\mathrm{7\; x\; 3}}$

Get rid of the fraction by multiplying both sides by the denominator 7 x 3.

264 x 7 x 3 = 22 x 36 x h

Divide both sides by 22 x 36

h = $\frac{\mathrm{264\; x\; 7\; x\; 3}}{\mathrm{22\; x\; 36}}$ = 7 cm

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

c² = 6² + 8²

c² = 100

Take square root of both sides.

c = $\sqrt{\mathrm{100}}$ = 10 cm

Perimeter is the sum of all the sides of a shape.

The perimeter of the rhombus = c + c + c + c

The perimeter of the rhombus = 4c

The perimeter of the rhombus = 4 x 10 cm

The perimeter of the rhombus = 40 cm

Radius (r) = 3 cm

π = $\frac{22}{7}$

Height (h) = 4 cm

Total surface area = 2 x $\frac{22}{7}$ x 3² + 2 x $\frac{22}{7}$ x 3 x 4 cm²

Total surface area = 132 cm²

Distance covered in one complete revolution = the circumference of the wheel

Distance covered in the 8 complete revolutions = 8 x the circumference of the wheel

Circumference = 2 x π x radius

Distance covered in the 8 complete revolutions = 8 x 2 x π x radius

Radius = $\frac{\mathrm{Diameter}}{2}$ = $\frac{\mathrm{21\; cm}}{2}$ = 10.5 cm

π = $\frac{22}{7}$

Distance covered in the 8 complete revolutions = 8 x 2 x $\frac{22}{7}$ x 10.5 cm

Distance covered in the 8 complete revolutions = 528 cm

Perimeter is the sum of all the sides of a shape.

A square has the same lengths (L) for all 4 sides.

Perimeter of a square = L + L + L + L = 4L

Area of a square = Length (L) x Length (L) = L²

Area of the square parcel of land = 256 m²

L² = 256

Take square root of both sides.

L = $\sqrt{\mathrm{256}}$ = 16 m

Perimeter of the square parcel of land = 4L

Perimeter of the square parcel of land = 4 x 16 m = 64 m

The rectangular field has the same perimeter.

Perimeter of the rectangular field = 64 m

Rectangle has two of the lengths equal and two of the breadth equal.

Perimeter of a rectangle = Length(L) + Length(L) + Breadth(B) + Breadth (B)

Perimeter of a rectangle = L + L + B + B

Perimeter of a rectangle = 2L + 2B

Perimeter of the rectangular field = 64 m

The length of the rectangular field = 20 m

2 x 20 m + 2B = 64 m

40 m + 2B = 64 m

2B = 64 m - 40 m

2B = 24 m

Divide both sides by 2.

B = $\frac{\mathrm{24\; m}}{2}$ = 12 m

The breadth of the rectangular field = 12 m

Area of a rectangle = Length x Breadth

Area of the rectangular field = 20 m x 12 m = 240 m²

Total surface area = 2(12 x 7 + 7 x 5 + 12 x 5) cm²

Total surface area = 358 cm²

Volume of a cylinder = π x radius x radius x height

1 Litre = 1000 cm³

85 litres = 85 x 1000 cm³

85 litres = 85000 cm³

Volume of kerosene = 85000 cm³

π = $\frac{22}{7}$

Radius = 7 cm

Let d = depth/height of kerosene in the container

85000 cm³ = $\frac{22}{7}$ x 7 cm x 7 cm x d

The denominator 7 cancels out one of the 7s multiplying.

85000 cm³ = 22 x 1 cm x 7 cm x d

85000 cm³ = 154 cm² x d

Divide both sides by 154 cm²

d = $\frac{85000{\mathrm{cm}}^3}{154{\mathrm{cm}}^2}$

d = 551.948 cm ≈ 552 cm

S = $\frac{\mathrm{4\; +\; 9\; +\; 11}}{2}$

S = $\frac{24}{2}$ = 12

Area = $\sqrt{\mathrm{12(12\; -\; 4)(12\; -\; 9)(12\; -\; 11)}}$

Area = $\sqrt{\mathrm{12(8)(3)(1)}}$

Area = $\sqrt{\mathrm{12\; x\; 8\; x\; 3\; x\; 1}}$

Area = $\sqrt{\mathrm{288}}$

Area = 16.970562748 ≈ 17 cm²

Radius = 8 cm
Slant height = 14 cm

We have to calculate for the vertical height to calculate the volume.

The vertical height, base radius and slant height forms a right-angle triangle and pythagoras theorem could be applied to calculate the height.

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

The slant height is the hypotenuse = 14 cm

Let h = the vertical height.

8² + h² = 14²

h² = 14² - 8²

h² = 196 - 64

h² = 132

Take the square root of both sides.

h = $\sqrt{\mathrm{132}}$ = 11.489

Vertical height = 11.489

π = $\frac{22}{7}$

Volume = $\frac{1}{3}x\frac{22}{7}x{8}^{2}x11.489$

Volume = 770.318 cm³ ≈ 770.32 cm³

Radius = 5 cm
Height = 10 cm

π = $\frac{22}{7}$

Total surface area of a closed cylinder = 2 x $\frac{22}{7}$ x 5² + 2 x $\frac{22}{7}$ x 5 x 10

Total surface area of a closed cylinder = 157.143 cm² + 314.286 cm²

Total surface area of a closed cylinder = 471.429 ≈ 471.43 cm²

Perimeter is the sum of all the sides of a shape.

The total perimeter before the sector was cut is the circumference of the circular plate.

Circumference of a circle = 2πr

The remaining perimeter will be the circumference - the length of the arc of the sector cut + the two radii.

Remaining perimeter = Circumference - length of cut arc + the two radii

Length of an arc = $\frac{\mathrm{Angle}}{360}$ x circumference of the circle

Length of an arc = $\frac{\mathrm{Angle}}{360}$ x 2 x π x radius

Angle = 150°

π = $\frac{22}{7}$

Radius = 14 cm

Length of cut arc = $\frac{\mathrm{150°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 14 cm

Length of cut arc = 36.666666667 ≈ 36.7 cm

Circumference = 2 x x $\frac{22}{7}$ x 14 cm

Circumference = 88 cm

Remaining perimeter = 88 cm - 36.7 cm + 14 cm + 14 cm = 79.3 cm

Alternatively

The total angle for the circle = 360°

Remaining angle for the sector left = 360° - 150°

Remaining angle for the sector left = 210°

Remaining perimeter = Length of arc left + the two radii

Length of arc left = $\frac{\mathrm{210°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 14 cm = 51.3 cm

Remaining perimeter = 51.3 cm + 14 cm + 14 cm

Remaining perimeter = 79.3 cm

Volume of a square base pyramid = $\frac{\mathrm{Area\; of\; base\; x\; height}}{3}$

Area of a square = Length x Length = L²

Area of the base = 324 cm²

L² = 324

Take square root of both sides.

L = $\sqrt{\mathrm{324}}$ = 18 cm

The height of a square base pyramid is thrice the length of a side of its base

Height = 3 x Length of its base

Length of base = 18 cm

Height = 3 x 18 cm = 54 cm

Volume of the square base pyramid = $\frac{\mathrm{324\; x\; 54}}{3}$ = 5,832 cm³

Let r = the radius of the circle.

sin(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Hypothenuse}}$

sin 75 = $\frac{6}{\mathrm{r}}$

Multiply both sides by r

r x sin 75 = 6

Divide both sides by sin 75

r = $\frac{6}{\mathrm{sin\; 75}}$ = 6.2 cm

Volume of a cylinder = πr²h

Where r = radius and h = height of the cylinder.

Volume of a rectangular shape = Length x Breadth x Height

Volume of the rectangular tank = 4 m x 8 m x 11 m

Since the volumes of the rectangular tank and cylindrical tank are the same,

Volume of the cylindrical tank = 4 m x 8 m x 11 m

πr²h = 4 m x 8 m x 11 m

h = 7 cm

π = $\frac{22}{7}$

$\frac{22}{7}$ x r² x 7 = 4 x 8 x 11

The 7s cancel each other

22 x r² = 4 x 8 x 11

Divide both sides by 22

r² = $\frac{\mathrm{4\; x\; 8\; x\; 11}}{22}$

r² = 16

Take square root of both sides

r = $\sqrt{\mathrm{16}}$ = 4 m

The area of the trapezium is the summation of the area of the triangle and the rectangle.

Area of triangle = $\frac{1}{2}$ x base x height

Area of a rectangle = Length x Breadth

Base of the triangle = 24 - 16 = 8

Height of the triangle = x
Breadth of rectangle = x

300 = $\frac{1}{2}$ x 8 x x + 16 x x

2 divides itself once and 8, 4 times.

300 = 4x + 16x
300 = 20x

Divide both sides by 20

x = $\frac{300}{20}$ = 15

Perimeter is the summation of all the dimension of the boundaries of a shape.

Let w = the width of the rectangular lawn.

Length is 3 cm longer than the width → Length = w + 3

Perimeter = 2(w + 3) + 2w

But perimeter of the rectangular lawn = 42

2(w + 3) + 2w = 42
2 x w + 2 x 3 + 2w = 42
2w + 6 + 2w = 42
4w = 42 - 6
4w = 36

Divide both sides by 4

w = $\frac{36}{4}$

w = 9 cm

Let the length of the rectangle = 8 cm and b cm be the breadth.

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

The 10 cm is the hypothenuse, since it's facing the right-angle and also the longest side.

10² = 8² + b²

100 = 64 + b²

100 - 64 = b²

36 = b²

Take square root of both sides

$\sqrt{\mathrm{36}}$ = b

6 = b

Area of a rectangle = Length x Breadth

Length = 8 cm
Breadth = 6 cm

Area = 8 cm x 6 cm

Area = 48 cm²

Equilateral triangle has all three sides equal.

When you divide it, you will get two equal right angle triangles with base length being half of the side of the equilateral triangle.

Let x = the side of the equal lateral triangle.

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

(10$\sqrt{3}$)² + ${\left(\frac{\mathrm{x}}{2}\right)}^2$ = x²

Note: The square cancels the square root.

10² x 3 + $\frac{{\mathrm{x}}^2}{2^2}$ = x²

100 x 3 + $\frac{{\mathrm{x}}^2}{4}$ = x²

Get rid of the fraction by multiplying each sides by the denominator (4).

4 x 300 + x² = 4 x x²

1200 + x² = 4x²

1200 = 4x² - x²

1200 = 3x²

Divide both sides by 3

x² = $\frac{1200}{3}$

x² = 400

Take square root of both sides.

x = $\sqrt{\mathrm{400}}$ = 20

∴ each side of the equilateral triangle is 20 cm.

Perimeter is the sum of all the sides of a shape.

Perimeter of the equilateral = 20 cm + 20 cm + 20 cm

Perimeter of the equilateral = 60 cm

Area of △LMN = $\frac{1}{2}$ x |LM| x |MN|

Method I

Finding the angle x

sin x = $\frac{3}{5}$

Take sin inverse.

x = sin^-1($\frac{3}{5}$)

x = 36.9°

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

tan x = $\frac{\mathrm{|LM|}}{\mathrm{|MN|}}$

tan x = $\frac{6}{\mathrm{|MN|}}$

Multiply both sides by |MN|.

|MN| x tan x = 6

|MN| = $\frac{6}{\mathrm{tan\; x}}$

But x = 36.9°

|MN| = $\frac{6}{\mathrm{tan\; 36.9}}$ = 7.99 ≈ 8 cm

Area of △LMN = $\frac{1}{2}$ x |LM| x |MN|

|LM| = 6 cm

|MN| = 8 cm

Area of △LMN = $\frac{1}{2}$ x 6 cm x 8 cm = 24 cm²

Method II

Finding the other two sides.

sin(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Hypothenuse}}$

The opposite side of x = |LM|

The hypotenuse side is |LN|

|LM| = 6

sin x = $\frac{6}{\mathrm{|LN|}}$

sin x = $\frac{3}{5}$

$\frac{6}{\mathrm{|LN|}}$ = $\frac{3}{5}$

Cross multiply.

3 x |LN| = 6 x 5

Divide both sides by 3

|LN| = $\frac{\mathrm{6\; x\; 5}}{3}$ = 10 cm

From pythagorean theorem,

|LM|² + |MN|² = |LN|²

|LN| = 10

|LM| = 6

6² + |MN|² = 10²

|MN|² = 10² - 6²

|MN|² = 64

Take the square root of both sides.

|MN| = $\sqrt{\mathrm{64}}$ = 8 cm

Area of △LMN = $\frac{1}{2}$ x |LM| x |MN|

|LM| = 6 cm

|MN| = 8 cm

Area of △LMN = $\frac{1}{2}$ x 6 cm x 8 cm = 24 cm²

12 cm = 6 cm + 6 cm (Diagonals meet at the center)

5 cm = 2.5 cm + 2.5 cm

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

c² = 2.5² + 6²

c² = 42.25

Take square root of both sides.

c = $\sqrt{\mathrm{42.25}}$

c = 6.5 cm

Perimeter is the sum of all the sides of a shape.

Perimeter of the rhombus = c + c + c + c

Perimeter of the rhombus = 4 x c

Perimeter of the rhombus = 4 x 6.5 cm

Perimeter of the rhombus = 26 cm

The volume of both the cube and the cone formed are the same.

Volume of a cube = Length x Length x Length = L³

Height of the cube = Length of the cube = h

Volume of the cube = h³

Volume of a cone = $\frac{\mathrm{\pi }{\mathrm{r}}^2\mathrm{h}}{3}$

h³ = $\frac{\mathrm{\pi }{\mathrm{r}}^2\mathrm{h}}{3}$

The h at the right cancels one of the h at the left to produce h² at the left.

h² = $\frac{\mathrm{\pi }{\mathrm{r}}^2}{3}$

Multiply both sides by 3.

3 x h² = πr²

3h² = πr²

Divide both sides by π

r² = $\frac{3{\mathrm{h}}^2}{\mathrm{\pi }}$

Take the square root of both sides.

r = $\sqrt{\frac{3{\mathrm{h}}^2}{\mathrm{\pi }}}$

The square root of a square is the number itself.

$\sqrt{{\mathrm{h}}^2}$ = h

r = h x $\sqrt{\frac{3}{\mathrm{\pi }}}$

r = h$\sqrt{\frac{3}{\mathrm{\pi }}}$

Area of region not covered by the trapezium = Area of the circle - Area of the trapezium

Area of a circle = πr².

Where r is the radius.

π = $\frac{22}{7}$

Radius = 7 cm

r² = r x r

Area of the circle = $\frac{22}{7}$ x 7 cm x 7 cm = 154 cm²

Area of a trapezium = $\frac{1}{2}$ x (a + b) x h

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x height

Area of the trapezium = $\frac{1}{2}$ x (10 cm + 21 cm) x 8 cm

Area of the trapezium = $\frac{1}{2}$ x 31 cm x 8 cm = 124 cm²

Area of region not covered by the trapezium = 154 cm² - 124 cm²

Area of region not covered by the trapezium = 30 cm²

A cone is produced from a sector of a circle.

Let l = the radius of the circle which produces the cone and L = the length of the arc of the sector.

Figure 1

Let h = height of the cone and r = base radius of the cone.

Figure 2

From Pythagorean theorem,

l² = h² + r²

l = $\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2}$

h = 11 cm
r = 8 cm

l = $\sqrt{{11}^2+8^2}$

l = 13.6014705087 cm

Curved surface area = 8 cm x $\frac{22}{7}$ x 13.6014705087 cm

Curved surface area = 341.979829934 ≈ 341.98 cm²

Explantion to the formula curved surface area of a cone = r x π x l

The area of the sector of the circle which the cone is made from is the same as the curved surface area of the cone.

Area of a sector of a circle = $\frac{\mathrm{\theta }}{360}$ x π x r²

But the radius of the circle that produces the cone = l

Area of the sector the circle = $\frac{\mathrm{\theta }}{360}$ x π x l² ------------- eqn 1

The length of the arc of the circle becomes the circumference of the circular base of the cone.

Length of an arc of a circle = $\frac{\mathrm{\theta }}{360}$ x 2 x π x r

But the radius of the circle that produces the cone = l

Length of the arc of the circle = $\frac{\mathrm{\theta }}{360}$ x 2 x π x l

The length of the arc is the same as the circumference of the base of the cone.

Circumference of a circle = 2 x π x r

Circumference of the circular base of the cone = 2 x π x r

$\frac{\mathrm{\theta }}{360}$ x 2 x π x l = 2 x π x r

2 and π cancels each other.

$\frac{\mathrm{\theta }}{360}$ x l = r

Make $\frac{\mathrm{\theta }}{360}$ the subject.

$\frac{\mathrm{\theta }}{360}$ = $\frac{\mathrm{r}}{\mathrm{l}}$

But from ---------- eqn 1,

Area of the sector of the circle = $\frac{\mathrm{\theta }}{360}$ x π x l²

Subtitute $\frac{\mathrm{\theta }}{360}$ = $\frac{\mathrm{r}}{\mathrm{l}}$ into ------------ eqn 1

Area of the sector of the circle = $\frac{\mathrm{r}}{\mathrm{l}}$ x π x l²

The denominator, l cancels out one of the l in the l².

Area of the sector of the circle = r x π x l

∴ the curved surface area of the cone = r x π x l

Total surface area = 2(7 x 5 + 7 x 4 + 5 x 4) cm²

Total surface area = 2(35 + 28 + 20) cm²

Total surface area = 2 x 83 cm²

Total surface area = 166 cm²

Base area = Length x Breadth

Let h = the height of the pyramid.

Volume = $\frac{1}{3}$ x 9 x 5 x h

Volume = 15h

But the volume = 105

15h = 105

Divide both sides by 15

h = $\frac{105}{15}$ = 7

The height is 7 cm

PTRS is a parallelogram.

Since line PU is parallel to line SR the height of PTRS is the same as the height of ∆SUR

Let h = the height of ∆SUR

Base of ∆SUR = 8 cm

Area of ∆SUR = 24 cm²

$\frac{1}{2}$ x 8 cm x h = 24 cm²

4 cm x h = 24 ²

Divide both sides by 4 cm

h = $\frac{\mathrm{24\; }{\mathrm{cm}}^2}{\mathrm{4\; cm}}$ = 6 cm

Height of the parallelogram = 6 cm

Base of the parallelogram = 8 cm

Area of the parallelogram = 8 cm x 6 cm = 48 cm²

π = $\frac{22}{7}$

Total surface area cylinder = 165 cm²

2 x πr² + 2πrh = 165

Radius (r) = $\frac{\mathrm{Diameter}}{2}$ = $\frac{7}{2}$ = 3.5 cm

2 x $\frac{22}{7}$ x 3.5² + 2 x $\frac{22}{7}$ x 3.5 x h = 165

Get rid of the fractions by multiplying both sides by the denominator 7.

2 x 22 x 3.5² + 2 x 22 x 3.5 x h = 165 x 7

539 + 154h = 1155

154h = 1155 - 539

154h = 616

Divide both sides by 154

h = $\frac{616}{154}$ = 4 cm

Length = 15 cm

Width = x cm

If the sides are double

New length = 15 cm X 2 = 30 cm

New width = 2 x x = 2x

Area of a rectangle = Length x Width

Area of the new rectangular board = 30 cm x 2xcm = 60x cm²

Base area is a square.

Area of a square = Length x Length

Base area of the pyramid = 10 cm x 10 cm = 100 cm²

Volume = 1400 cm³

Let h = the height of the pyramid.

1400 = $\frac{1}{3}$ x 100 x h

Get rid of the fraction by multiplying both sides of the equation by the denominator 3

1400 x 3 = $\frac{1}{3}$ x 100 x h x 3

1400 x 3 = 100 x h

Divide both sides by 100

h = $\frac{\mathrm{1400\; x\; 3}}{100}$ = 42 cm

Let h = height of triangle XYZ

h² + 4² = 5²

h² + 16 = 25

h² = 25 - 16

h² = 9

Take square root of both sides.

h = $\sqrt{9}$ = 3 cm

Base = 8 cm

Height = 3 cm

Area of triangle XYZ = $\frac{1}{2}$ x 8 cm x 3 cm = 12 cm²

Total surface ara = 2(12 x 8 + 12 x 3 + 3 x 8) cm²

Total surface ara = 2 x 156 cm² = 312 cm²

Radius = $\frac{\mathrm{Diameter}}{2}$ = $\frac{\mathrm{6\; cm}}{2}$ = 3 cm

π = $\frac{22}{7}$

Volume = $\frac{4}{3}$ x $\frac{22}{7}$ x 3³ = 113.143 cm³ ≈ 113 cm³

Let h = height of the water

Volume of a cuboid = Length x Breadth x Height

Since the length and breadth are in cm, convert the litres to cm³

1 litre = 1000 cm³

180 litres = 180 x 1000 cm³

Length = 250 cm, breadth = 120 cm

Volume = 180 litres = 180 x 1000 cm³

250 cm x 120 cm x h cm = 180 x 1000 cm³

Divide both sides by 250 cm x 120 cm

h = $\frac{\mathrm{180\; x\; 1000}}{\mathrm{250\; x\; 120}}$ cm = 6 cm

Let h = distance between the parallel sides

a = 6 cm and b = 7 cm

$\frac{1}{2}$ x (6 cm + 7 cm) x h = 49 cm²

$\frac{1}{2}$ x 13 cm x h = 49 cm²

Get rid of the fraction by multiplying both sides by the denominator 2.

2 x $\frac{1}{2}$ x 13 cm x h = 2 x 49 cm²

13 cm x h = 98 cm²

Divide both sides by 13 cm.

h = $\frac{98}{13}$ cm = 7.53846 cm ≈ 7.5 cm (one decimal place)

Base is square. Area of a square is length x length

Let h = the height of the pyramid.

Length of the base = height of the pyramid.

Area of the base = Length x Length = h x h

Volume = $\frac{1}{3}$ x h x h x h

But volume = 114$\frac{1}{3}$ = $\frac{\mathrm{3\; x\; 114\; +\; 1}}{3}$ = $\frac{343}{3}$

$\frac{343}{3}$ = $\frac{1}{3}$ x h x h x h

Get rid of the fractions by multiplying both sides by the denominator, 3

3 x $\frac{343}{3}$ = 3 x $\frac{1}{3}$ x h x h x h

343 = 1 x h x h x h

h³ = 343

Take the cube root of both sides

h = $\sqrt[3]{\mathrm{343}}$ or ${343}^{\frac{1}{3}}$ = 7 cm

Both methods of finding the area requires that we solve for |QT|.

Considering the triangle

Apply pythagoras theorem

|QT|² + |RT|² = |QR|²

|QT|² + 3.3² = 5.5²

|QT|² + 10.89 = 30.25

|QT|² = 30.25 - 10.89

|QT|² = 19.36

|QT| = √19.36

|QT| = 4.4 cm

Method I

Find area of trapezium

Area of trapezium = $\frac{1}{2}$ x (Sum of parallel sides) x Height

Area of trapezium = $\frac{1}{2}$ x (|PQ| + [|PQ| + |RT|]) x |QT|

Area of trapezium = $\frac{1}{2}$ x (6.7 cm + [6.7 cm + 3.3 cm]) x 4.4 cm

Area of trapezium = $\frac{1}{2}$ x (6.7 cm + 10.0 cm) x 4.4 cm

Area of trapezium = $\frac{1}{2}$ x 16.7 cm x 4.4 cm

Area of trapezium = 36.74 cm²

Method II

Find area of rectangle and triangle

Area of rectangle = |PQ| x |QT|

Area of rectangle = 6.7 cm x 4.4 cm = 29.48 cm²

Area of triangle = $\frac{1}{2}$ x Base x Height

Area of triangle = $\frac{1}{2}$ x |RT| x |QT|

Area of triangle = $\frac{1}{2}$ x 3.3 cm x 4.4 cm

Area of triangle = 7.26 cm²

Total area = Area of rectangle + Area of triangle

Total area = 29.48 cm² + 7.26 cm²

Total area = 36.74 cm²

Area of field = Area of the rectangle - Area of the semicircle

Area of rectangle = |PQ| x |QR|

Area of rectangle = 8 cm x 4 cm = 32 cm²

Area of semicircle = $\frac{1}{2}$ x π x r²

Radius = $\frac{1}{2}$ x (8 cm - [2 cm + 2 cm])

Radius = $\frac{1}{2}$ x (8 cm - 4 cm)

Radius = $\frac{1}{2}$ x 4 cm

Radius = 2 cm

Area of semicircle = $\frac{1}{2}$ x $\frac{22}{7}$ x (2 cm)²

Area of semicircle = $\frac{1}{2}$ x $\frac{22}{7}$ x 4 cm²

Area of semicircle = $\frac{88}{14}$ cm²

Area of semicircle = 6.29 cm²

Area of field = Area of rectangle - Area of semicircle

Area of field = 32 cm² - 6.29 cm²

Area of field = 25.71 cm²

Area of field ≈ 25.7 cm²

Volume = 264 cm³

Radius (r) = 6 cm

π = $\frac{22}{7}$

264 = $\frac{22}{7}$ x 6² x $\frac{\mathrm{h}}{3}$

264 = $\frac{\mathrm{22\; x\; 36\; x\; h}}{\mathrm{7\; x\; 3}}$

Get rid of the fraction by multiplying both sides by the denominator 7 x 3.

264 x 7 x 3 = 22 x 36 x h

Divide both sides by 22 x 36

h = $\frac{\mathrm{264\; x\; 7\; x\; 3}}{\mathrm{22\; x\; 36}}$ = 7 cm

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

c² = 6² + 8²

c² = 100

Take square root of both sides.

c = $\sqrt{\mathrm{100}}$ = 10 cm

Perimeter is the sum of all the sides of a shape.

The perimeter of the rhombus = c + c + c + c

The perimeter of the rhombus = 4c

The perimeter of the rhombus = 4 x 10 cm

The perimeter of the rhombus = 40 cm

Radius (r) = 3 cm

π = $\frac{22}{7}$

Height (h) = 4 cm

Total surface area = 2 x $\frac{22}{7}$ x 3² + 2 x $\frac{22}{7}$ x 3 x 4 cm²

Total surface area = 132 cm²

Distance covered in one complete revolution = the circumference of the wheel

Distance covered in the 8 complete revolutions = 8 x the circumference of the wheel

Circumference = 2 x π x radius

Distance covered in the 8 complete revolutions = 8 x 2 x π x radius

Radius = $\frac{\mathrm{Diameter}}{2}$ = $\frac{\mathrm{21\; cm}}{2}$ = 10.5 cm

π = $\frac{22}{7}$

Distance covered in the 8 complete revolutions = 8 x 2 x $\frac{22}{7}$ x 10.5 cm

Distance covered in the 8 complete revolutions = 528 cm

Perimeter is the sum of all the sides of a shape.

A square has the same lengths (L) for all 4 sides.

Perimeter of a square = L + L + L + L = 4L

Area of a square = Length (L) x Length (L) = L²

Area of the square parcel of land = 256 m²

L² = 256

Take square root of both sides.

L = $\sqrt{\mathrm{256}}$ = 16 m

Perimeter of the square parcel of land = 4L

Perimeter of the square parcel of land = 4 x 16 m = 64 m

The rectangular field has the same perimeter.

Perimeter of the rectangular field = 64 m

Rectangle has two of the lengths equal and two of the breadth equal.

Perimeter of a rectangle = Length(L) + Length(L) + Breadth(B) + Breadth (B)

Perimeter of a rectangle = L + L + B + B

Perimeter of a rectangle = 2L + 2B

Perimeter of the rectangular field = 64 m

The length of the rectangular field = 20 m

2 x 20 m + 2B = 64 m

40 m + 2B = 64 m

2B = 64 m - 40 m

2B = 24 m

Divide both sides by 2.

B = $\frac{\mathrm{24\; m}}{2}$ = 12 m

The breadth of the rectangular field = 12 m

Area of a rectangle = Length x Breadth

Area of the rectangular field = 20 m x 12 m = 240 m²

Total surface area = 2(12 x 7 + 7 x 5 + 12 x 5) cm²

Total surface area = 358 cm²

Volume of a cylinder = π x radius x radius x height

1 Litre = 1000 cm³

85 litres = 85 x 1000 cm³

85 litres = 85000 cm³

Volume of kerosene = 85000 cm³

π = $\frac{22}{7}$

Radius = 7 cm

Let d = depth/height of kerosene in the container

85000 cm³ = $\frac{22}{7}$ x 7 cm x 7 cm x d

The denominator 7 cancels out one of the 7s multiplying.

85000 cm³ = 22 x 1 cm x 7 cm x d

85000 cm³ = 154 cm² x d

Divide both sides by 154 cm²

d = $\frac{85000{\mathrm{cm}}^3}{154{\mathrm{cm}}^2}$

d = 551.948 cm ≈ 552 cm

S = $\frac{\mathrm{4\; +\; 9\; +\; 11}}{2}$

S = $\frac{24}{2}$ = 12

Area = $\sqrt{\mathrm{12(12\; -\; 4)(12\; -\; 9)(12\; -\; 11)}}$

Area = $\sqrt{\mathrm{12(8)(3)(1)}}$

Area = $\sqrt{\mathrm{12\; x\; 8\; x\; 3\; x\; 1}}$

Area = $\sqrt{\mathrm{288}}$

Area = 16.970562748 ≈ 17 cm²

Radius = 8 cm
Slant height = 14 cm

We have to calculate for the vertical height to calculate the volume.

The vertical height, base radius and slant height forms a right-angle triangle and pythagoras theorem could be applied to calculate the height.

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

The slant height is the hypotenuse = 14 cm

Let h = the vertical height.

8² + h² = 14²

h² = 14² - 8²

h² = 196 - 64

h² = 132

Take the square root of both sides.

h = $\sqrt{\mathrm{132}}$ = 11.489

Vertical height = 11.489

π = $\frac{22}{7}$

Volume = $\frac{1}{3}x\frac{22}{7}x{8}^{2}x11.489$

Volume = 770.318 cm³ ≈ 770.32 cm³

Radius = 5 cm
Height = 10 cm

π = $\frac{22}{7}$

Total surface area of a closed cylinder = 2 x $\frac{22}{7}$ x 5² + 2 x $\frac{22}{7}$ x 5 x 10

Total surface area of a closed cylinder = 157.143 cm² + 314.286 cm²

Total surface area of a closed cylinder = 471.429 ≈ 471.43 cm²

Perimeter is the sum of all the sides of a shape.

The total perimeter before the sector was cut is the circumference of the circular plate.

Circumference of a circle = 2πr

The remaining perimeter will be the circumference - the length of the arc of the sector cut + the two radii.

Remaining perimeter = Circumference - length of cut arc + the two radii

Length of an arc = $\frac{\mathrm{Angle}}{360}$ x circumference of the circle

Length of an arc = $\frac{\mathrm{Angle}}{360}$ x 2 x π x radius

Angle = 150°

π = $\frac{22}{7}$

Radius = 14 cm

Length of cut arc = $\frac{\mathrm{150°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 14 cm

Length of cut arc = 36.666666667 ≈ 36.7 cm

Circumference = 2 x x $\frac{22}{7}$ x 14 cm

Circumference = 88 cm

Remaining perimeter = 88 cm - 36.7 cm + 14 cm + 14 cm = 79.3 cm

Alternatively

The total angle for the circle = 360°

Remaining angle for the sector left = 360° - 150°

Remaining angle for the sector left = 210°

Remaining perimeter = Length of arc left + the two radii

Length of arc left = $\frac{\mathrm{210°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 14 cm = 51.3 cm

Remaining perimeter = 51.3 cm + 14 cm + 14 cm

Remaining perimeter = 79.3 cm

Volume of a square base pyramid = $\frac{\mathrm{Area\; of\; base\; x\; height}}{3}$

Area of a square = Length x Length = L²

Area of the base = 324 cm²

L² = 324

Take square root of both sides.

L = $\sqrt{\mathrm{324}}$ = 18 cm

The height of a square base pyramid is thrice the length of a side of its base

Height = 3 x Length of its base

Length of base = 18 cm

Height = 3 x 18 cm = 54 cm

Volume of the square base pyramid = $\frac{\mathrm{324\; x\; 54}}{3}$ = 5,832 cm³

Let r = the radius of the circle.

sin(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Hypothenuse}}$

sin 75 = $\frac{6}{\mathrm{r}}$

Multiply both sides by r

r x sin 75 = 6

Divide both sides by sin 75

r = $\frac{6}{\mathrm{sin\; 75}}$ = 6.2 cm

Volume of a cylinder = πr²h

Where r = radius and h = height of the cylinder.

Volume of a rectangular shape = Length x Breadth x Height

Volume of the rectangular tank = 4 m x 8 m x 11 m

Since the volumes of the rectangular tank and cylindrical tank are the same,

Volume of the cylindrical tank = 4 m x 8 m x 11 m

πr²h = 4 m x 8 m x 11 m

h = 7 cm

π = $\frac{22}{7}$

$\frac{22}{7}$ x r² x 7 = 4 x 8 x 11

The 7s cancel each other

22 x r² = 4 x 8 x 11

Divide both sides by 22

r² = $\frac{\mathrm{4\; x\; 8\; x\; 11}}{22}$

r² = 16

Take square root of both sides

r = $\sqrt{\mathrm{16}}$ = 4 m

Length of arc = $\frac{\mathrm{\theta }}{360}$ x 2πr

Where r = radius of the circle.

Length of the arc = 50 cm
θ = 75

π = $\frac{22}{7}$

$\frac{75}{360}$ x 2 x $\frac{22}{7}$ x r = 50

$\frac{\mathrm{75\; x\; 2\; x\; 22}}{\mathrm{360\; x\; 7}}$ x r = 50

$\frac{3300}{2520}$ x r = 50

Get rid of the fraction by multiplying both sides of the equation by the denominator (2520)

3300 x r = 50 x 2520
3300 x r = 126000

Divide both sides of the equation by 3300

r = $\frac{126000}{3300}$

r = 38.1818 ≈ 38.2 cm (three significant figures)

Method I

The angle for the entire circle is 360°

If 140° = 40 cm²
360° = $\frac{\mathrm{40\; x\; 360}}{140}$ = 102.86 cm²

Method II

Area of a sector = $\frac{\mathrm{\theta }}{360}$ x πr²

But πr² = area of the circle

Area of a sector = $\frac{\mathrm{\theta }}{360}$ x area of circle

Area of a sector = 40 cm²
θ = 140

40 = $\frac{140}{360}$ x area of circle

Make area of circle the subject.

Multiply both sides by 360

40 x 360 = 140 x area of circle

Divide both sides by 140

$\frac{\mathrm{40\; x\; 360}}{140}$ = area of circle

Area of circle = 102.86 cm²

The area of the trapezium is the summation of the area of the triangle and the rectangle.

Area of triangle = $\frac{1}{2}$ x base x height

Area of a rectangle = Length x Breadth

Base of the triangle = 24 - 16 = 8

Height of the triangle = x
Breadth of rectangle = x

300 = $\frac{1}{2}$ x 8 x x + 16 x x

2 divides itself once and 8, 4 times.

300 = 4x + 16x
300 = 20x

Divide both sides by 20

x = $\frac{300}{20}$ = 15

Perimeter is the summation of all the dimension of the boundaries of a shape.

Let w = the width of the rectangular lawn.

Length is 3 cm longer than the width → Length = w + 3

Perimeter = 2(w + 3) + 2w

But perimeter of the rectangular lawn = 42

2(w + 3) + 2w = 42
2 x w + 2 x 3 + 2w = 42
2w + 6 + 2w = 42
4w = 42 - 6
4w = 36

Divide both sides by 4

w = $\frac{36}{4}$

w = 9 cm

Let the length of the rectangle = 8 cm and b cm be the breadth.

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

The 10 cm is the hypothenuse, since it's facing the right-angle and also the longest side.

10² = 8² + b²

100 = 64 + b²

100 - 64 = b²

36 = b²

Take square root of both sides

$\sqrt{\mathrm{36}}$ = b

6 = b

Area of a rectangle = Length x Breadth

Length = 8 cm
Breadth = 6 cm

Area = 8 cm x 6 cm

Area = 48 cm²

1. Find the number of times the cyclist complete one cycle by dividing the distance overed by the circumference.

Number of completed cycles = $\frac{\mathrm{302\; km}}{\mathrm{9\; km}}$ = 33.555

Number of completed cycles = 33

Note: completed cycle is when the cyclist reaches the starting point.

2. Distance from starting point = Distance covered - Number of completed cycles x the circumference

Distance from starting point = 302 km - 33 x 9 km

Distance from starting point = 302 km - 297 km

Distance from starting point = 5 km

Equilateral triangle has all three sides equal.

When you divide it, you will get two equal right angle triangles with base length being half of the side of the equilateral triangle.

Let x = the side of the equal lateral triangle.

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

(10$\sqrt{3}$)² + ${\left(\frac{\mathrm{x}}{2}\right)}^2$ = x²

Note: The square cancels the square root.

10² x 3 + $\frac{{\mathrm{x}}^2}{2^2}$ = x²

100 x 3 + $\frac{{\mathrm{x}}^2}{4}$ = x²

Get rid of the fraction by multiplying each sides by the denominator (4).

4 x 300 + x² = 4 x x²

1200 + x² = 4x²

1200 = 4x² - x²

1200 = 3x²

Divide both sides by 3

x² = $\frac{1200}{3}$

x² = 400

Take square root of both sides.

x = $\sqrt{\mathrm{400}}$ = 20

∴ each side of the equilateral triangle is 20 cm.

Perimeter is the sum of all the sides of a shape.

Perimeter of the equilateral = 20 cm + 20 cm + 20 cm

Perimeter of the equilateral = 60 cm

Area of △LMN = $\frac{1}{2}$ x |LM| x |MN|

Method I

Finding the angle x

sin x = $\frac{3}{5}$

Take sin inverse.

x = sin^-1($\frac{3}{5}$)

x = 36.9°

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

tan x = $\frac{\mathrm{|LM|}}{\mathrm{|MN|}}$

tan x = $\frac{6}{\mathrm{|MN|}}$

Multiply both sides by |MN|.

|MN| x tan x = 6

|MN| = $\frac{6}{\mathrm{tan\; x}}$

But x = 36.9°

|MN| = $\frac{6}{\mathrm{tan\; 36.9}}$ = 7.99 ≈ 8 cm

Area of △LMN = $\frac{1}{2}$ x |LM| x |MN|

|LM| = 6 cm

|MN| = 8 cm

Area of △LMN = $\frac{1}{2}$ x 6 cm x 8 cm = 24 cm²

Method II

Finding the other two sides.

sin(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Hypothenuse}}$

The opposite side of x = |LM|

The hypotenuse side is |LN|

|LM| = 6

sin x = $\frac{6}{\mathrm{|LN|}}$

sin x = $\frac{3}{5}$

$\frac{6}{\mathrm{|LN|}}$ = $\frac{3}{5}$

Cross multiply.

3 x |LN| = 6 x 5

Divide both sides by 3

|LN| = $\frac{\mathrm{6\; x\; 5}}{3}$ = 10 cm

From pythagorean theorem,

|LM|² + |MN|² = |LN|²

|LN| = 10

|LM| = 6

6² + |MN|² = 10²

|MN|² = 10² - 6²

|MN|² = 64

Take the square root of both sides.

|MN| = $\sqrt{\mathrm{64}}$ = 8 cm

Area of △LMN = $\frac{1}{2}$ x |LM| x |MN|

|LM| = 6 cm

|MN| = 8 cm

Area of △LMN = $\frac{1}{2}$ x 6 cm x 8 cm = 24 cm²

12 cm = 6 cm + 6 cm (Diagonals meet at the center)

5 cm = 2.5 cm + 2.5 cm

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

c² = 2.5² + 6²

c² = 42.25

Take square root of both sides.

c = $\sqrt{\mathrm{42.25}}$

c = 6.5 cm

Perimeter is the sum of all the sides of a shape.

Perimeter of the rhombus = c + c + c + c

Perimeter of the rhombus = 4 x c

Perimeter of the rhombus = 4 x 6.5 cm

Perimeter of the rhombus = 26 cm

The volume of both the cube and the cone formed are the same.

Volume of a cube = Length x Length x Length = L³

Height of the cube = Length of the cube = h

Volume of the cube = h³

Volume of a cone = $\frac{\mathrm{\pi }{\mathrm{r}}^2\mathrm{h}}{3}$

h³ = $\frac{\mathrm{\pi }{\mathrm{r}}^2\mathrm{h}}{3}$

The h at the right cancels one of the h at the left to produce h² at the left.

h² = $\frac{\mathrm{\pi }{\mathrm{r}}^2}{3}$

Multiply both sides by 3.

3 x h² = πr²

3h² = πr²

Divide both sides by π

r² = $\frac{3{\mathrm{h}}^2}{\mathrm{\pi }}$

Take the square root of both sides.

r = $\sqrt{\frac{3{\mathrm{h}}^2}{\mathrm{\pi }}}$

The square root of a square is the number itself.

$\sqrt{{\mathrm{h}}^2}$ = h

r = h x $\sqrt{\frac{3}{\mathrm{\pi }}}$

r = h$\sqrt{\frac{3}{\mathrm{\pi }}}$

Area of region not covered by the trapezium = Area of the circle - Area of the trapezium

Area of a circle = πr².

Where r is the radius.

π = $\frac{22}{7}$

Radius = 7 cm

r² = r x r

Area of the circle = $\frac{22}{7}$ x 7 cm x 7 cm = 154 cm²

Area of a trapezium = $\frac{1}{2}$ x (a + b) x h

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x height

Area of the trapezium = $\frac{1}{2}$ x (10 cm + 21 cm) x 8 cm

Area of the trapezium = $\frac{1}{2}$ x 31 cm x 8 cm = 124 cm²

Area of region not covered by the trapezium = 154 cm² - 124 cm²

Area of region not covered by the trapezium = 30 cm²

A cone is produced from a sector of a circle.

Let l = the radius of the circle which produces the cone and L = the length of the arc of the sector.

Figure 1

Let h = height of the cone and r = base radius of the cone.

Figure 2

From Pythagorean theorem,

l² = h² + r²

l = $\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2}$

h = 11 cm
r = 8 cm

l = $\sqrt{{11}^2+8^2}$

l = 13.6014705087 cm

Curved surface area = 8 cm x $\frac{22}{7}$ x 13.6014705087 cm

Curved surface area = 341.979829934 ≈ 341.98 cm²

Explantion to the formula curved surface area of a cone = r x π x l

The area of the sector of the circle which the cone is made from is the same as the curved surface area of the cone.

Area of a sector of a circle = $\frac{\mathrm{\theta }}{360}$ x π x r²

But the radius of the circle that produces the cone = l

Area of the sector the circle = $\frac{\mathrm{\theta }}{360}$ x π x l² ------------- eqn 1

The length of the arc of the circle becomes the circumference of the circular base of the cone.

Length of an arc of a circle = $\frac{\mathrm{\theta }}{360}$ x 2 x π x r

But the radius of the circle that produces the cone = l

Length of the arc of the circle = $\frac{\mathrm{\theta }}{360}$ x 2 x π x l

The length of the arc is the same as the circumference of the base of the cone.

Circumference of a circle = 2 x π x r

Circumference of the circular base of the cone = 2 x π x r

$\frac{\mathrm{\theta }}{360}$ x 2 x π x l = 2 x π x r

2 and π cancels each other.

$\frac{\mathrm{\theta }}{360}$ x l = r

Make $\frac{\mathrm{\theta }}{360}$ the subject.

$\frac{\mathrm{\theta }}{360}$ = $\frac{\mathrm{r}}{\mathrm{l}}$

But from ---------- eqn 1,

Area of the sector of the circle = $\frac{\mathrm{\theta }}{360}$ x π x l²

Subtitute $\frac{\mathrm{\theta }}{360}$ = $\frac{\mathrm{r}}{\mathrm{l}}$ into ------------ eqn 1

Area of the sector of the circle = $\frac{\mathrm{r}}{\mathrm{l}}$ x π x l²

The denominator, l cancels out one of the l in the l².

Area of the sector of the circle = r x π x l

∴ the curved surface area of the cone = r x π x l

Total surface area = 2(7 x 5 + 7 x 4 + 5 x 4) cm²

Total surface area = 2(35 + 28 + 20) cm²

Total surface area = 2 x 83 cm²

Total surface area = 166 cm²

Base area = Length x Breadth

Let h = the height of the pyramid.

Volume = $\frac{1}{3}$ x 9 x 5 x h

Volume = 15h

But the volume = 105

15h = 105

Divide both sides by 15

h = $\frac{105}{15}$ = 7

The height is 7 cm

Length of the arc = 1$\frac{19}{36}$ cm = $\frac{\mathrm{1\; x\; 36\; +\; 19}}{36}$ cm = $\frac{55}{36}$ cm

π = $\frac{22}{7}$

Radius = 3.5 cm

$\frac{55}{36}$ = $\frac{\mathrm{\theta }}{360}$ x 2 x $\frac{22}{7}$ x 3.5

$\frac{55}{36}$ = $\frac{\mathrm{\theta \; x\; 2\; x\; 22\; x\; 3.5}}{\mathrm{360\; x\; 7}}$

Cross multiply.

36 x θ x 2 x 22 x 3.5 = 55 x 360 x 7

Divide both sides by 36 x 2 x 22 x 3.5

θ = $\frac{\mathrm{55\; x\; 360\; x\; 7}}{\mathrm{36\; x\; 2\; x\; 22\; x\; 3.5}}$

θ = 25°

PTRS is a parallelogram.

Since line PU is parallel to line SR the height of PTRS is the same as the height of ∆SUR

Let h = the height of ∆SUR

Base of ∆SUR = 8 cm

Area of ∆SUR = 24 cm²

$\frac{1}{2}$ x 8 cm x h = 24 cm²

4 cm x h = 24 ²

Divide both sides by 4 cm

h = $\frac{\mathrm{24\; }{\mathrm{cm}}^2}{\mathrm{4\; cm}}$ = 6 cm

Height of the parallelogram = 6 cm

Base of the parallelogram = 8 cm

Area of the parallelogram = 8 cm x 6 cm = 48 cm²

π = $\frac{22}{7}$

Total surface area cylinder = 165 cm²

2 x πr² + 2πrh = 165

Radius (r) = $\frac{\mathrm{Diameter}}{2}$ = $\frac{7}{2}$ = 3.5 cm

2 x $\frac{22}{7}$ x 3.5² + 2 x $\frac{22}{7}$ x 3.5 x h = 165

Get rid of the fractions by multiplying both sides by the denominator 7.

2 x 22 x 3.5² + 2 x 22 x 3.5 x h = 165 x 7

539 + 154h = 1155

154h = 1155 - 539

154h = 616

Divide both sides by 154

h = $\frac{616}{154}$ = 4 cm

Length of the arc = $\frac{\mathrm{72°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 3.5 cm

Length of the arc = 4.4 cm

Length = 15 cm

Width = x cm

If the sides are double

New length = 15 cm X 2 = 30 cm

New width = 2 x x = 2x

Area of a rectangle = Length x Width

Area of the new rectangular board = 30 cm x 2xcm = 60x cm²

Let c = length of the chord.

Method I

Using the cosine rule

c² = 5² + 5² - 2 x 5 x 5 cos70°

c² = 25 + 25 - 50cos70°

c² = 50 - 50cos70°

c² = 32.9

Take the square root of both sides to get rid of the square.

c = $\sqrt{\mathrm{32.9}}$ = 5.7359 cm ≈ 5.7 cm (one decimal place)

Method II

Using the sine rule

Since the radii are the same, the triangle the chord makes with the circle is an isosceles triangle (has two sides equal).

The base angles of an isosceles triangle are the same.

Let y = base angle of the isosceles triangle.

y + y + 70° = 180° (Sum of angles in a triangle).

2y = 180° - 70°

2y = 110°

Divide both sides by 2.

y = $\frac{\mathrm{110°}}{2}$ = 55°

Each of the base angles is 55°

Applying the sine rule,

$\frac{\mathrm{sin\; 70°}}{\mathrm{c}}$ = $\frac{\mathrm{sin\; 55°}}{5}$

Cross multiply.

c x sin 55° = 5 x sin 70°

Divide both sides by sin 55°

c = $\frac{\mathrm{5\; x\; sin\; 70°}}{\mathrm{sin\; 55°}}$ = 5.7358 cm ≈ 5.7 cm (one decimal place)

Base area is a square.

Area of a square = Length x Length

Base area of the pyramid = 10 cm x 10 cm = 100 cm²

Volume = 1400 cm³

Let h = the height of the pyramid.

1400 = $\frac{1}{3}$ x 100 x h

Get rid of the fraction by multiplying both sides of the equation by the denominator 3

1400 x 3 = $\frac{1}{3}$ x 100 x h x 3

1400 x 3 = 100 x h

Divide both sides by 100

h = $\frac{\mathrm{1400\; x\; 3}}{100}$ = 42 cm