Area & Perimeter of Plane Figures — Test Yourself

The base of a right triangle is 8 cm and its hypotenuse is 10 cm; the area of triangle is:

1. 24 cm²

2. 40 cm²

3. 48 cm²

4. 80 cm²

Answer

Given:

Base of the triangle = 8 cm

Hypotenuse of the triangle = 10 cm

Let h be the height of the triangle.

In a right-angled triangle, using the Pythagoras theorem,

⇒ Base² + Height² = Hypotenuse²

⇒ (8)² + h² = (10)²

⇒ 64 + h² = 100

⇒ h² = 100 - 64

⇒ h² = 36

⇒ h = ​$\sqrt{36}$

⇒ h = 6 cm.

By formula,

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

= $\dfrac{1}{2}$ x 8 x 6 cm²

= 24 cm².

Hence, option 1 is the correct option.

The area of an equilateral triangle is $4\sqrt{3}$ cm², its perimeter is :

1. 16 cm

2. 4 cm

3. 12 cm

4. 8 cm

Answer

Given,

Area of equilateral triangle = $4\sqrt{3}$ cm²

By formula,

Area of the equilateral triangle = $\dfrac{\sqrt{3}}{4}$ x side²

$\Rightarrow 4\sqrt{3} = \dfrac{\sqrt{3}}{4} \times \text{side}^2\\[1em] \Rightarrow \text{side}^2 = \dfrac{4 \times 4\sqrt{3}}{\sqrt{3}}\\[1em] \Rightarrow \text{side}^2 = 4 \times 4\\[1em] \Rightarrow \text{side}^2 = 16\\[1em] \Rightarrow \text{side} = \sqrt{16}\\[1em] \Rightarrow \text{side} = 4\text{ cm}.$

By formula,

Perimeter of equilateral triangle = 3 x side

= 3 x 4 cm

= 12 cm.

Hence, option 3 is the correct option.

If the perimeter of a square is 80 cm, its area is 80 cm².

1. true

2. false

3. none of these two

Answer

Given,

Perimeter of a square = 80 cm

⇒ 4 x side = 80

⇒ side = $\dfrac{80}{4}$

⇒ side = 20 cm

By formula,

Area of square = side²

= 20² cm²

= 400 cm².

So, given area is false.

Hence, option 2 is the correct option.

If the area of a trapezium is 32 cm² and distance between its parallel sides is 8 cm; the sum of length of its parallel side is :

1. 4 cm

2. 16 cm

3. 8 cm

4. 12 cm

Answer

Given, area of a trapezium = 32 cm²

Distance between its parallel sides = 8 cm

By formula,

Area of trapezium = $\dfrac{1}{2}$ x (sum of parallel sides) x distance between the parallel sides

⇒ 32 = $\dfrac{1}{2}$ x (Sum of parallel sides) x 8

⇒ 32 = Sum of parallel sides x 4

⇒ Sum of parallel sides = $\dfrac{32}{4}$

⇒ Sum of parallel sides = 8 cm.

Hence, option 3 is the correct option.

Statement 1: The side of a triangular board are 8 cm, 6 cm and 10 cm; the cost of painting it at the rate of ₹ 10 per square cm is $\dfrac{1}{2}$ x 6 x 8 x ₹ 10.

Statement 2: 10² = 8² + 6²

⇒ Sides of the triangle are 8 cm and 6 cm.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given,

Sides of a triangular board are 8 cm, 6 cm and 10 cm.

⇒ 10² = 100, 8² = 64 and 6² = 36.

⇒ 100 = 64 + 36

⇒ 10² = 8² + 6²

This confirms the triangle with sides 6 cm, 8 cm, and 10 cm is a right triangle, where 10 cm is the hypotenuse.

Thus, sides of the triangle are 8 cm and 6 cm.

So, statement 2 is true.

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

= $\dfrac{1}{2}$ x 8 x 6

Rate of painting = ₹ 10 per square cm

Cost of painting = Area x rate of painting

= $\dfrac{1}{2}$ x 8 x 6 x ₹ 10

So, statement 1 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Statement 1: A rhombus shaped sheet with perimeter 40 cm has one diagonal 12 cm and the other diagonal is 16 cm.

Statement 2: If the other diagonal of this rhombus = x cm, x = 10² - 6².

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given,

The perimeter of a rhombus = 40 cm.

One diagonal = 12 cm.

As we know, the perimeter of the rhombus = 4 x side

⇒ 4 x side = 40 cm

⇒ side = $\dfrac{40}{4}$ cm

⇒ side = 10 cm

Let ABCD be a rhombus. From figure,

AB = 10 cm

Let diagonal AC = 12 cm

We know that,

Diagonals of rhombus bisect each other.

Then, OA = OC = $\dfrac{12}{2}$ = 6 cm

Let OB be a cm.

Since the diagonal of a rhombus bisect at 90°.

Applying pythagoras theorem in triangle AOB, we get:

⇒ AB² = OA² + OB²

⇒ (10)² = (6)² + a²

⇒ a² = (10)² - (6)²

⇒ a² = 100 - 36

⇒ a² = 64

⇒ a = $\sqrt{64}$

⇒ a = 8

So, OB = 8 cm

BD = 2 x OB = 2 x 8 cm = 16 cm.

∴ Statement 1 is true.

As solved above,

If the other diagonal of this rhombus = x cm,

Then $\dfrac{x}{2}$ will be equal to 10² - 6².

∴ Statement 2 is false.

∴ Statement 1 is true, and statement 2 is false.

Hence, option 3 is the correct option.

Assertion (A): The perimeter of the adjoining figure is (32 + x) cm.

Reason (R): x² = 13² - 5² = 144 and x = 12 cm.

Perimeter = (32 + 12) cm

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Perimeter of the figure = sum of all the sides

= 7 + 13 + 12 + x

= (32 + x) cm

So, assertion (A) is true.

From figure,

AB = AC - BC = 12 - 7 = 5 cm

Since, ABE is a right angled triangle. Using pythagoras theorem,

⇒ AE² = AB² + BE²

⇒ 13² = 5² + x²

⇒ 169 = 25 + x²

⇒ x² = 169 - 25

⇒ x² = 144

⇒ x = $\sqrt{144}$

⇒ x = 12 cm

Perimeter = (32 + x) = (32 + 12) cm.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

Assertion (A): If BC = 14 cm, AB = 14 x 4 cm

Reason (R): AB = 4 x 2r = 4 x 14 cm

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given, BC = 14 cm

From figure,

Diameter (d) = 2r = 14 cm

AB has 4 circles (3 circles and 2 semi-circles).

AB = 4 x diameter of one circle = 4 x 2r = 4 x 14 cm = 56 cm.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

AD is altitude of an isosceles triangle ABC in which AB = AC = 30 cm and BC = 36 cm. A point O is marked on AD in such a way that ∠BOC = 90°. Find the area of quadrilateral ABOC.

Answer

ΔABC is shown in the figure below:

Area of isosceles triangle ABC =

$= \dfrac{b}{4} \sqrt{4a^2 - b^2}\\[1em] = \dfrac{36}{4} \sqrt{4 \times 30^2 - 36^2}\\[1em] = 9 \sqrt{4 \times 30^2 - 36^2}\\[1em] = 9 \sqrt{4 \times 900 - 1,296}\\[1em] = 9 \sqrt{3600 - 1,296}\\[1em] = 9 \sqrt{2,304}\\[1em] = 9 \times 48 \\[1em] = 432 \text{ cm}^2$

∠ BOC = ∠ COD = 45° (∵ AD divide ∠ BOC in 2 equal halves)

Let OB = OC = x.

In Δ BOC, by using the Pythagoras theorem,

OB² + OC² = BC²

⇒ x² + x² = (36)²

⇒ 2x² = 1,296

⇒ x² = $\dfrac{1,296}{2}$

⇒ x² = 648

⇒ x = $\sqrt{648}$

⇒ x = 18 $\sqrt{2}$

Now the area of triangle BOC = $\dfrac{1}{2}$ x base x height

$= \dfrac{1}{2} \times 18 \sqrt{2} \times 18 \sqrt{2}\\[1em] = \dfrac{1}{2} \times 18 \times 18 \times 2\\[1em] = 18 \times 18\\[1em] = 324 \text{ cm}^2$

Area of quadrilateral ABOC = Area of Δ ABC - Area of Δ BOC

= 432 - 324 cm²

= 108 cm²

Hence, the area of quadrilateral ABOC is 108 cm².

A footpath of uniform width runs all around the outside of a rectangular field 30 m long and 24 m wide. If the path occupies an area of 360 m², find its width.

Answer

Given:

The length of the rectangular field = 30 m

The breadth of the rectangular field = 24 m

Area of path = 360 m²

Let the width of the path be x m.

The length of the smaller rectangular field = 30 m - x m - x m

= 30 - 2x m

The breadth of the smaller rectangular field = 24 m - x m - x m

= 24 - 2x m

As we know, the area of a rectangle = length x breadth

⇒ Area of the larger rectangular field = 30 x 24 m²

= 720 m²

⇒ Area of the smaller rectangular field = (30 - 2x) x (24 - 2x) m²

= (720 - 108x - 4x²) m²

Area of the path = Area of larger rectangular field - Area of smaller rectangular field

⇒ 360 = 720 - (720 - 108x - 4x²)

⇒ 360 = 720 - 720 + 108x + 4x²

⇒ 4x² + 108x - 360 = 0

⇒ x² + 27x - 90 = 0

⇒ x² + 30x - 3x - 90 = 0

⇒ x(x + 30) - 3(x + 30) = 0

⇒ (x + 30)(x - 3) = 0

⇒ x = - 30 or 3

Since the width of the path cannot be negative,

x = 3 m

Hence, the width of the path is 3 m.

A wire when bent in the form of a square encloses an area of 484 m². Find the largest area enclosed by the same wire when bent to form :

(i) an equilateral triangle.

(ii) a rectangle of breadth 16 m.

Answer

(i) Area of the square = 484 m²

Let a be the length of side of the square.

⇒ a² = 484

⇒ a = $\sqrt{484}$

⇒ a = 22 m

Total length of the wire = Perimeter of the square = 4 x 22m = 88 m

Perimeter of the square = Perimeter of equilateral triangle.

⇒ 3 x side = 88 m

⇒ side = $\dfrac{88}{3}$ m

⇒ side = 29.3 m

Area of equilateral triangle = $\dfrac{\sqrt{3}}{4}$ x side²

= $\dfrac{\sqrt{3}}{4}$ x (29.3)² m²

= $\dfrac{\sqrt{3}}{4}$ x 858.49 m²

= 372.57 m²

Hence, the area of the equilateral triangle is 372.57 sq. m.

(ii) Given:

Length of the rectangle = 16 m

Let b be the breadth of the rectangle.

Perimeter of the rectangle = Perimeter of the square

⇒ 2(l + b) = 88 m

⇒ 2(16 + b) = 88 m

⇒ 16 + b = $\dfrac{88}{2}$ m

⇒ 16 + b = 44 m

⇒ b = 44 - 16 m

⇒ b = 28 m

Area of the rectangle = l x b

= 16 x 28 m²

= 448 m²

Hence, the area of the rectangle is 448 sq. m.

For the trapezium given below; find its area.

Answer

Draw CE parallel to DA which meets AB at point E.

Since, AB || DC thus AE || DC and DA || CE.

Since, opposite sides are parallel, thus AECD is a parallelogram.

Opposite sides of a parallelogram are equal, thus CE = AD = 10 cm and AE = DC = 12 cm.

In isosceles triangle EBC,

Draw CF ⊥ EB.

In an isosceles triangle, the perpendicular from the common vertex to the base, bisects it.

Thus, EF = $\dfrac{EB}{2} = \dfrac{8}{2}$ = 4 cm.

In right triangle CEF,

By pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

EC² = CF² + EF²

10² = CF² + 4²

100 = CF² + 16

CF² = 100 - 16

CF² = 84

CF = $\sqrt{84}$ = 9.16 cm

By formula,

Area of trapezium = $\dfrac{1}{2}$ x sum of parallel sides x distance between them

Area of trapezium ABCD = $\dfrac{1}{2} \times (AB + DC) \times CF$

$= \dfrac{1}{2} \times (20 + 12) \times 9.16 \\[1em] = \dfrac{1}{2} \times 32 \times 9.16 \\[1em] = 146.56 \text{ cm}^2.$

Hence, area of trapezium ABCD = 146.56 cm².

For the trapezium given below; find its area.

Answer

Draw CO such that CO is perpendicular to AB.

AO = DC = 8 cm

AB = AO + OB

⇒ 14 = 8 + OB

⇒ OB = 14 - 8 = 6 cm

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

In triangle BCO, by using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ (6)² + Height² = 10²

⇒ 36 + Height² = 100

⇒ Height² = 100 - 36

⇒ Height² = 64

⇒ Height = $\sqrt{64}$

⇒ Height = 8 cm

Area of trapezium ABCD = $\dfrac{1}{2}$ x (8 + 14) x 8

= $\dfrac{1}{2}$ x 22 x 8 sq. cm

= 11 x 8 sq. cm

= 88 sq. cm

Hence, the area of trapezium ABCD is 88 sq. cm.

For the trapezium given below; find its area.

Answer

Draw CE parallel to DA which meets AB at point E.

AE = DC = 20 cm

AB = AE + EB

⇒ 32 = 20 + EB

⇒ EB = 32 - 20 = 12 cm

And, DA = CE = 10 cm

For the Δ EBC,

Let the sides of the triangle be:

a = 10 cm, b = 12 cm and c = 16 cm.

The semi-perimeter s:

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{10 + 12 + 16}{2}\\[1em] = \dfrac{38}{2}\\[1em] = 19$

∵ Area of triangle EBC = $\sqrt{s(s - a)(s - b)(s - c)}$

= $\sqrt{19(19 - 10)(19 - 12)(19 - 16)}$ cm²

= $\sqrt{19 \times 9 \times 7 \times 3}$ cm²

= $\sqrt{3,591}$ cm²

= 59.9 cm²

Let h be the height of Δ EBC,

Area of Δ EBC = $\dfrac{1}{2}$ x base x height

⇒ $\dfrac{1}{2}$ x 12 x height = 59.9

⇒ 6 x height = 59.9

⇒ height = $\dfrac{59.9}{6}$

⇒ height = 9.98 cm

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

= $\dfrac{1}{2}$ x (20 + 32) x 9.98

= $\dfrac{1}{2}$ x 52 x 9.98 sq. cm

= 26 x 9.98 sq. cm

= 259.65 sq. cm

Hence, the area of trapezium ABCD is 259.65 sq. cm.

For the trapezium given below; find its area.

Answer

Draw DE and CF perpendicular to AB.

ABCD is an isosceles trapezium.

Let EF = FB = x cm

DC = EF = 18 cm

AB = AE + EF + FB

⇒ 30 = x + 18 + x

⇒ 30 = 2x + 18

⇒ 2x = 30 - 18

⇒ 2x = 12

⇒ x = $\dfrac{12}{2}$

⇒ x = 6 cm

As EDA is a right angled triangle, by using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ (6)² + height² = 12²

⇒ 36 + height² = 144

⇒ height² = 144 - 36

⇒ height² = 108 cm

⇒ height = $\sqrt{108}$ cm

⇒ height = 10.39 cm

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

= $\dfrac{1}{2}$ x (18 + 30) x 10.39

= $\dfrac{1}{2}$ x 48 x 10.39 sq. cm

= 24 x 10.39 sq. cm

= 249.41 sq. cm

Hence, the area of trapezium ABCD is 249.41 sq. cm.

The perimeter of a rectangular board is 70 cm. Taking its length as x cm, find its width in terms of x.

If the area of the rectangular board is 300 cm²; find its dimensions.

Answer

Given:

Perimeter of the rectangular board = 70 cm

Length = x cm

Let b be the width of the rectangular board.

Perimeter of a rectangle = 2(l + b)

⇒ 2(x + b) = 70

⇒ x + b = $\dfrac{70}{2}$

⇒ x + b = 35

⇒ b = 35 - x

Area = l x b

⇒ x $\times$ (35 - x) = 300

⇒ 35x - x² = 300

⇒ 35x - x² - 300 = 0

⇒ x² - 35x + 300 = 0

⇒ x² - 20x - 15x + 300 = 0

⇒ x(x - 20) - 15(x - 20) = 0

⇒ (x - 20)(x - 15) = 0

⇒ x = 20 or 15

Therefore, b = (35 - x)

b = 15 or 20

Hence, the width of rectangular board is (35 - x) and the dimensions of the board are 20 cm and 15 cm.

The area of a rectangle is 640 m². Taking its length as x m; find, in terms of x, the width of the rectangle. If the perimeter of the rectangle is 104 m; find its dimensions.

Answer

Given:

Area of the rectangle = 640 m²

Length = x meters

Let b be the width of the rectangle.

Area of the rectangle = l x b

⇒ x × b = 640

⇒ b = $\dfrac{640}{x}$ m

By formula,

Perimeter of the rectangle = 2(l + b)

⇒ $2\Big(x + \dfrac{640}{x}\Big)$ = 104

⇒ $\Big(x + \dfrac{640}{x}\Big) = \dfrac{104}{2}$

⇒ $\Big(x + \dfrac{640}{x}\Big)$ = 52

⇒ x² + 640 = 52x

⇒ x² - 52x + 640 = 0

⇒ x² - 32x - 20x + 640 = 0

⇒ x(x - 32) - 20(x - 32) = 0

⇒ (x - 20)(x - 32) = 0

⇒ x - 20 = 0 or x - 32 = 0

⇒ x = 20 m or x = 32 m.

Now,

⇒ b = $\dfrac{640}{x}$

Case 1 : x = 20 m

⇒ b = $\dfrac{640}{20}$ = 32 m.

Case 2 : x = 32 m

⇒ b = $\dfrac{640}{32}$ = 20 m.

Hence, the width of rectangle is $\dfrac{640}{x}$ and the dimensions of the rectangle are 20 m and 32 m.

The length of a rectangle is twice the side of a square and its width is 6 cm greater than the side of the square. If area of the rectangle is three times the area of the square; find the dimensions of each.

Answer

Given:

Let s be the side of the square.

Length of the rectangle = 2 x Side of the square

⇒ l = 2s

Width of the rectangle = Side of the square + 6

⇒ w = s + 6

Area of the rectangle = 3 x area of the square

⇒ l x w = 3s²

⇒ 2s x (s + 6) = 3s²

⇒ 2s² + 12s = 3s²

⇒ 2s² + 12s - 3s² = 0

⇒ - s² + 12s = 0

⇒ s² - 12s = 0

⇒ s(s - 12) = 0

⇒ s = 0 or 12

Since the side cannot be zero, s = 12 cm.

The dimensions of the rectangle:

l = 2 x s = 2 x 12 cm = 24 cm

w = s + 6 = 12 + 6 = 18 cm

Hence, the length and width of the rectangle are 24 cm and 18 cm, and the side of the square is 12 cm.

ABCD is a square with each side 12 cm. P is a point on BC such that area of Δ ABP : area of trapezium APCD = 1 : 5. Find the length of CP.

Answer

Square ABCD is shown in the figure below:

Given:

$\dfrac{\text{Area of Δ ABP}}{\text{Area of trapezium APCD}} = \dfrac{1}{5}$

Area of Δ ABP = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x 12 x (12 - CP)

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

= $\dfrac{1}{2}$ x (12 + CP) x 12

$⇒ \dfrac{\dfrac{1}{2} \times 12 \times (12 - CP)}{\dfrac{1}{2} \times (12 + CP) \times 12} = \dfrac{1}{5}\\[1em] ⇒ \dfrac{\cancel{\dfrac{1}{2}} \times 12 \times (12 - CP)}{\cancel{\dfrac{1}{2}} \times (12 + CP) \times 12} = \dfrac{1}{5}\\[1em] ⇒ \dfrac{\cancel{12} \times (12 - CP)}{ (12 + CP) \times \cancel{12}} = \dfrac{1}{5}\\[1em] ⇒ \dfrac{(12 - CP)}{ (12 + CP)} = \dfrac{1}{5}\\[1em] ⇒ 5 \times (12 - CP) = 1 \times (12 + CP)\\[1em] ⇒ 60 - 5CP = 12 + CP\\[1em] ⇒ 60 - 12 = 5CP + CP\\[1em] ⇒ 48 = 6CP\\[1em] ⇒ CP = \dfrac{48}{6}\\[1em] ⇒ CP = 8 \text { cm}$

Hence, the length of CP is 8 cm.

A rectangular plot of land measures 45 m x 30 m. A boundary wall of height 2.4 m is built all around the plot at a distance of 1 m from the plot. Find the area of the inner surface of the boundary wall.

Answer

Given:

The boundary wall is built all around the rectangular plot at a distance of 1 meter, which means the wall forms a larger rectangle around the plot as shown in the figure below:

Dimensions of the outer rectangle (including the boundary wall):

Length of the outer rectangle = (45 + 1 + 1) m = 47 m

Width of the outer rectangle = (30 + 1 + 1) m = 32 m

Perimeter of the outer rectangle = 2(length + breadth)

= 2(47 + 32)

= 2 x 79 = 158 m

Area of the inner surface of the boundary wall is the perimeter of the outer rectangle multiplied by the height of the wall.

Inner surface area of boundary wall = Perimeter of outer rectangle x height of wall

= 158 m x 2.4 m

= 379.2 m²

Hence, the area of the inner surface of the boundary wall is 379.2 m².

A wire when bent in the form of a square encloses an area = 576 cm². Find the largest area enclosed by the same wire when bent to form:

(i) an equilateral triangle.

(ii) a rectangle whose adjacent sides differ by 4 cm.

Answer

(i) Area of the square = 576 cm²

Let a be the length of side of the square.

⇒ a² = 576

⇒ a = $\sqrt{576}$

⇒ a = 24 cm

Total length of the wire = Perimeter of the square = 4 x 24 cm = 96 cm

Perimeter of the square = Perimeter of equilateral triangle.

⇒ 3 x side = 96 cm

⇒ side = $\dfrac{96}{3}$ cm

⇒ side = 32 cm

Area of equilateral triangle = $\dfrac{\sqrt{3}}{4}$ x side²

= $\dfrac{\sqrt{3}}{4}$ x (32)² cm²

= $\dfrac{\sqrt{3}}{4}$ x 1,024 cm²

= 256 $\sqrt{3}$ cm²

Hence, the area of equilateral triangle is 256 $\sqrt{3}$ cm².

(ii) Given:

Let l be the length and b be the breadth of the rectangle.

l - b = 4 ...............(1)

Perimeter of rectangle = Perimeter of square

⇒ 2(l + b) = 96 cm

⇒ 2(l + b) = 96 cm

⇒ l + b = $\dfrac{96}{2}$ cm

⇒ l + b = 48 cm ...............(2)

Add equation (1) and (2), we get

⇒ (l - b) + (l + b) = 4 + 48

⇒ l - b + l + b = 52

⇒ 2l = 52

⇒ l = $\dfrac{52}{2}$

⇒ l = 26 cm

So, b = l - 4 = 26 - 4 = 22 cm

Area of rectangle = l x b

= 26 x 22 cm²

= 572 cm²

Hence, the area of rectangle is 572 cm².

The area of a parallelogram is y cm² and its height is h cm. The base of another parallelogram is x cm more than the base of the first parallelogram and its area is twice the area of the first. Find, in terms of y, h and x, the expression for the height of the second parallelogram.

Answer

Given:

Area of the first parallelogram = y cm²

Height of the first parallelogram = h cm

Area of the first parallelogram = base x height

⇒ y = base x h

⇒ base of the first parallelogram = $\dfrac{y}{h}$ cm

The base of the second parallelogram = $\Big(\dfrac{y}{h} + x\Big)$ cm

Area of the second parallelogram = base x height

⇒ $2y = \Big(\dfrac{y}{h} + x\Big)$ x height

⇒ height = $\dfrac{2hy}{y + hx}$

Hence, the height of the second parallelogram is $\dfrac{2hy}{y + hx}$.

The distance between parallel sides of a trapezium is 15 cm and the length of the line segment joining the mid-points of its non-parallel sides is 26 cm. Find the area of the trapezium.

Answer

Given:

Let the given trapezium be as shown in the figure below:

The distance between parallel sides of a trapezium, (i.e., height) = 15 cm

The length of the line segment joining the mid-points of its non-parallel sides, EF = 26 cm

Since EF is half of the sum of the lengths of the parallel sides.

EF = $\dfrac{1}{2}$ x (AD + BC)

⇒ $\dfrac{1}{2}$ x (AD + BC) = 26

⇒ AD + BC = 26 x 2 cm = 52 cm

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

= $\dfrac{1}{2}$ x (AD + BC) x height

= $\dfrac{1}{2}$ x 52 x 15

= 26 x 15

= 390 cm²

Hence, the area of the trapezium is 390 cm².

The diagonal of a rectangular plot is 34 m and its perimeter is 92 m. Find its area.

Answer

Given:

Diagonal of the rectangular plot = 34 m

Perimeter of the rectangular plot = 92 m

Let l be the length and b be the breadth of the rectangular plot.

Perimeter = 2(l + b)

⇒ 2(l + b) = 92

⇒ l + b = $\dfrac{92}{2}$

⇒ l + b = 46

Squaring both side,

⇒ (l + b)² = 46²

⇒ l² + b² + 2lb = 2,116 ...............(1)

As all the angles of rectangle are 90°, we can use the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ l² + b² = 34²

⇒ l² + b² = 1,156

Using equation (1), we get

⇒ 2,116 - 2lb = 1,156

⇒ 2,116 - 1,156 = 2lb

⇒ 960 = 2lb

⇒ lb = $\dfrac{960}{2}$

⇒ lb = 480

Hence, the area of the rectangular plot is 480 m².

The cost of fencing a circular field at the rate of ₹ 240 per metre is ₹ 52,800. The field is to be ploughed at the rate of ₹ 12.50 per m². Find the cost of ploughing the field.

Answer

Given:

Rate of fencing = ₹ 240 per metre

Total cost = ₹ 52,800

Let r be the radius of the field.

Total cost = Circumference of field x Rate of fencing

$⇒ 52,800 = 2πr \times 240\\[1em] ⇒ 2πr = \dfrac{52,800}{240}\\[1em] ⇒ 2πr = 220\\[1em] ⇒ 2 \times \dfrac{22}{7} \times r = 220\\[1em] ⇒ \dfrac{44}{7} \times r = 220\\[1em] ⇒ r = \dfrac{220 \times 7}{44}\\[1em] ⇒ r = \dfrac{5 \times 7}{1}\\[1em] ⇒ r = 35 \text{ m}$

Area of the field = πr²

$= \dfrac{22}{7} \times 35^2\\[1em] = \dfrac{22}{7} \times 1225\\[1em] = 22 \times 175\\[1em] = 3,850 \text{ m}^2$

Rate of ploughing = ₹ 12.50 per m²

Total cost of ploughing = Area of the field x Rate of ploughing

= 3,850 x 12.50

= ₹ 48,125

Hence, the total cost of ploughing is ₹ 48,125.

Two circles touch each other externally. The sum of their areas is 58π cm² and the distance between their centres is 10 cm. Find the radii of the two circles.

Answer

Let r₁ and r₂ be the radii of the two circles.

Sum of the areas = 58π

⇒ A₁ + A₂ = 58π

⇒ πr₁² + πr₂² = 58π

⇒ π(r₁² + r₂²) = 58π

⇒ $\cancel{π}$ (r₁² + r₂²) = 58 $\cancel{π}$

⇒ r₁² + r₂² = 58 ...............(1)

Also, since the circles touch externally, the sum of their radii equals the distance between their centers.

r₁ + r₂ = 10

⇒ r₁ = 10 - r₂

Substituting the value of r₁ in equation (1),

⇒ (10 - r₂)² + r₂² = 58

⇒ 10² + r₂² - 2 x 10 x r₂ + r₂² = 58

⇒ 100 + 2r₂² - 20r₂ = 58

⇒ 100 + 2r₂² - 20r₂ - 58 = 0

⇒ 2r₂² - 20r₂ - 42 = 0

⇒ r₂² - 10r₂ - 21 = 0

⇒ r₂² - 3r₂ - 7r₂ - 21 = 0

⇒ r₂(r₂ - 3) - 7(r₂ - 3) = 0

⇒ (r₂ - 3)(r₂ - 7) = 0

⇒ r₂ = 3 cm or 7 cm

Using equation (1), we find r₁:

If r₂ = 3, then r₁ = 10 - 3 = 7 cm

If r₂ = 7, then r₁ = 10 - 7 = 3 cm

Hence, the radii of the two circles are 7 cm and 3 cm.

The given figure shows a rectangle ABCD inscribed in a circle as shown alongside.

If AB = 28 cm and BC = 21 cm, find the area of the shaded portion of the given figure.

Answer

ABCD is a rectangle. So, ∠ABC = 90°.

By using the Pythagoras theorem,

Let h be the diagonal of the rectangle.

Base² + Height² = Diagonal²

⇒ 28² + 21² = h²

⇒ 784 + 441 = h²

⇒ 1,225 = h²

⇒ h = $\sqrt{1,225}$

⇒ h = 35 cm

Diagonal of rectangle = Diameter of circle.

Radius = $\dfrac{d}{2}$ = $\dfrac{35}{2}$ = 17.5 cm

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

= πr² - lb cm²

= $\dfrac{22}{7}$ x 17.5² - 28 x 21 cm²

= $\dfrac{22}{7}$ x 306.25 - 28 x 21 cm²

= 22 x 43.75 - 588 cm²

= 962.5 - 588 cm²

= 374.5 cm²

Hence, area of the shaded portion = 374.5 cm².

A square is inscribed in a circle of radius 7 cm. Find the area of the square.

Answer

Diagonal of the square = Diameter of the circle.

Radius of circle = 7 cm

Diameter of circle = 2 x 7 cm = 14 cm

Diagonal of square = 14 cm

Let s be the side of square.

By using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ s² + s² = 14²

⇒ 2s² = 196

⇒ s² = $\dfrac{196}{2}$

⇒ s² = 98 cm

⇒ s = $\sqrt{98}$ cm

Area of square = Side²

= $(\sqrt{98})^2$ cm²

= 98 cm²

Hence, the area of the square is 98 cm².

A metal wire, when bent in the form of an equilateral triangle of largest area, encloses an area of $484{\sqrt3} \text{ cm}^2$. If the same wire is bent into the form of a circle of largest area, find the area of this circle.

Answer

Given:

Area of the equivalent triangle = $484{\sqrt3} \text{ cm}^2$

Let a be the length of the equivalent triangle.

Area of equivalent triangle = $\dfrac{\sqrt{3}}{4} \times a^2$

$⇒ \dfrac{\sqrt{3}}{4} \times a^2 = 484{\sqrt3} \text{ cm}^2\\[1em] ⇒ \dfrac{\cancel{\sqrt{3}}}{4} \times a^2 = 484 \cancel{\sqrt{3}} \text{ cm}^2\\[1em] ⇒ \dfrac{1}{4} \times a^2 = 484 \text{ cm}^2\\[1em] ⇒ a^2 = 484 \times 4 \text{ cm}^2\\[1em] ⇒ a^2 = 1,936 \text{ cm}^2\\[1em] ⇒ a = \sqrt{1,936} \text{ cm}^2\\[1em] ⇒ a = 44 \text{ cm}$

Perimeter of the triangle = 3 x side

= 3 x 44 cm = 132 cm

Perimeter of triangle = Circumference of circle

$⇒ 2πr = 132\\[1em] ⇒ 2 \times \dfrac{22}{7} \times r = 132\\[1em] ⇒ \dfrac{44}{7} \times r = 132\\[1em] ⇒ r = \dfrac{7 \times 132}{44}\\[1em] ⇒ r = \dfrac{7 \times 3}{1}\\[1em] ⇒ r = 21$

Area of circle = πr²

$= \dfrac{22}{7} \times 21^2\\[1em] = \dfrac{22}{7} \times 441\\[1em] = 22 \times 63\\[1em] = 1,386 \text{ cm}^2$

Hence, the area of the circle is 1,386 cm².

The perimeter of a triangle is 450 m and its sides are in the ratio 12 : 5 : 13. Find the area of the triangle.

Answer

Given:

Perimeter of the triangle = 450 m

Ratio of the sides = 12 : 5 : 13.

Let the sides of triangle be 12a, 5a and 13a.

Perimeter of the triangle = Sum of sides of the triangle

⇒ 12a + 5a + 13a = 450

⇒ 30a = 450

⇒ a = $\dfrac{450}{30}$

⇒ a = 15

Thus, sides are 12 x 15, 5 x 15 and 13 x 15 m

= 180 m, 75 m and 195 m

The sides of the triangle are:

a = 180 m, b = 75 m and c = 195 m.

The semi-perimeter s:

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{180 + 75 + 195}{2}\\[1em] = \dfrac{450}{2}\\[1em] = 225$

∵ Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

= $\sqrt{225(225 - 180)(225 - 75)(225 - 195)}$ m²

= $\sqrt{225 \times 45 \times 150 \times 30}$ m²

= $\sqrt{45,562,500}$ m²

= 6,750 m²

Hence, the area of the triangle is 6,750 m².

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Answer

Let the sides of the triangle be:

a = 26 cm, b = 28 cm and c = 30 cm.

The semi-perimeter s of the triangle is:

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{26 + 28 + 30}{2}\\[1em] = \dfrac{84}{2}\\[1em] = 42$

∵ Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

= $\sqrt{42(42 - 26)(42 - 28)(42 - 30)}$ cm²

= $\sqrt{42 \times 16 \times 14 \times 12}$ cm²

= $\sqrt{112,896}$ cm²

= 336 cm²

Area of triangle = Area of parallelogram

(∵ Area of parallelogram = base x height)

Let h be the height of parallelogram.

⇒ 336 = 28 x h

⇒ h = $\dfrac{336}{28}$

⇒ h = 12 cm

Hence, the height of the parallelogram is 12 cm.

Using the information in the following figure, find the area of the trapezium.

Answer

Draw a perpendicular line CP on AB, such that CP = 9 m and AP = 23 m.

Let PB = a m

In Δ CPB, by using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ a² + (9)² = (15)²

⇒ a² + 81 = 225

⇒ a² = 225 - 81

⇒ a² = 144

⇒ a² = 144

⇒ a = $\sqrt{144}$

⇒ a = 12 m

So, AB = AP + PB = 23 + 12 = 35 m

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

= $\dfrac{1}{2}$ (23 + 35) x 9

= $\dfrac{1}{2}$ x 58 x 9

= 29 x 9

= 261 m²

Hence, the area of the trapezium is 261 m².

Sum of the areas of two squares is 400 cm². If the difference of their perimeters is 16 cm, find the sides of the two squares.

Answer

Given:

Sum of the areas of two squares = 400 cm².

The difference of their perimeters = 16 cm

Let a and b be the sides of 2 square.

Sum of Areas,

⇒ a² + b² = 400 ...............(1)

Difference of Perimeters,

⇒ 4a - 4b = 16

⇒ 4(a - b) = 16

⇒ a - b = $\dfrac{16}{4}$

⇒ a - b = 4

⇒ a = 4 + b

Substituting the value of a in equation (1), we get

⇒ (4 + b)² + b² = 400

⇒ 4² + b² + 2 x 4 x b + b² = 400

⇒ 16 + b² + 8b + b² = 400

⇒ 16 + 2b² + 8b - 400 = 0

⇒ 2b² + 8b - 384 = 0

⇒ b² + 4b - 192 = 0

⇒ b² + 16b - 12b - 192 = 0

⇒ b(b + 16) - 12(b + 16) = 0

⇒ (b + 16)(b - 12) = 0

⇒ b = - 16 or 12

Since the side of a square cannot be negative, b = 12 cm.

So, a = 4 + b = 16 cm

Hence, the sides of the squares are 16 cm and 12 cm.

Find the area and the perimeter of a square with diagonal 24 cm.

[Take $\sqrt{2}$ = 1.41].

Answer

Diagonal of the square = 24 cm

Let a be the side of square.

By using the Pythagoras theorem,

AB² + BC² = AC²

⇒ a² + a² = (24)²

⇒ 2a² = 576

⇒ a² = $\dfrac{576}{2}$

⇒ a² = 288

⇒ a = $\sqrt{288}$

⇒ a = 12 $\sqrt{2}$

⇒ a = 12 x 1.41 = 16.92 cm

Area of the square = side²

= ($\sqrt{288}$)² cm²

= 288 cm²

Perimeter of the square = 4 x side

= 4 x 16.92 cm

= 67.68 cm

Hence, the area of the square is 288 cm² and the perimeter is 67.68 cm.

A steel wire, when bent in the form of a square, encloses an area of 121 cm². The same wire is bent in the form of a circle. Find area of the circle.

Answer

Given:

Area of the square = 121 cm²

Let s be the side of the square.

As we know, the area of the square = side²

⇒ s² = 121

⇒ s = $\sqrt{121}$

⇒ s = 11

Total length of the wire = Perimeter of the square

As we know, the perimeter of the square = 4 x side

= 4 x 11

= 44 cm

Perimeter of the square = Circumference of the circle

Let r be the radius of the circle.

⇒ 2πr = 44

$⇒ 2 \times \dfrac{22}{7} \times r = 44\\[1em] ⇒ \dfrac{44}{7} \times r = 44\\[1em] ⇒ r = \dfrac{7 \times 44}{44}\\[1em] ⇒ r = \dfrac{308}{44}\\[1em] ⇒ r = 7$

Area of the circle = πr²

$= \dfrac{22}{7} \times 7^2\\[1em] = \dfrac{22}{7} \times 49\\[1em] = \dfrac{1,078}{7}\\[1em] = 154$

Hence, the area of the circle is 154 cm².

The perimeter of a semicircular plate is 108 cm, find its area.

Answer

Given:

Perimeter of the semicircular plate = 108 cm

Let r be the radius of the plate.

The perimeter of the semicircular plate includes the curved part (half the circumference of a circle) and the diameter. So,

⇒ πr + 2r = Perimeter

⇒ πr + 2r = 108

⇒ r(π + 2) = 108

⇒ r $\Big(\dfrac{22}{7} + 2\Big)$ = 108

⇒ r $\Big(\dfrac{22}{7} + \dfrac{14}{7}\Big)$ = 108

⇒ r $\Big(\dfrac{22 + 14}{7} \Big)$ = 108

⇒ r $\Big(\dfrac{36}{7}\Big)$ = 108

⇒ r = $\Big(\dfrac{7 \times 108}{36}\Big)$

⇒ r = 7 x 3 = 21 cm

Area of semicircular plate = $\dfrac{1}{2}$ πr²

$= \dfrac{1}{2} \times \dfrac{22}{7} \times 21^2\\[1em] = \dfrac{22}{14} \times 441\\[1em] = \dfrac{9,702}{14} \\[1em] = 693 \text{ cm}^2$

Hence, the area of the semicircular plate is 693 cm².

Two circles touch externally. The sum of their areas is 130π sq. cm and the distance between their centres is 14 cm. Find the radii of the circles.

Answer

Given:

Sum of areas of two circles = 130π cm²

The distance between their centres = 14 cm

Let $r_1$ and $r_2$ be the radii of 2 circles.

Area of the first circle + Area of the second circle = 130π

⇒ A₁ + A₂ = 130π

⇒ πr₁² + πr₂² = 130π

⇒ π(r₁² + r₂²) = 130π

⇒ $\cancel{π}$ (r₁² + r₂²) = 130 $\cancel{π}$

⇒ r₁² + r₂² = 130 ...............(1)

Also, since the circles touch externally, the sum of their radii equals the distance between their centers.

r₁ + r₂ = 14

⇒ r₁ = 14 - r₂

Putting the value of r₁ in equation (1),

⇒ (14 - r₂)² + r₂² = 130

⇒ 14² + r₂² - 2 x 14 x r₂ + r₂² = 130

⇒ 196 + 2r₂² - 28r₂ = 130

⇒ 196 + 2r₂² - 28r₂ - 130 = 0

⇒ 2r₂² - 28r₂ + 66 = 0

⇒ r₂² - 14r₂ + 33 = 0

⇒ r₂² - 11r₂ - 3r₂ + 33 = 0

⇒ r₂(r₂ - 11) - 3(r₂ - 11) = 0

⇒ (r₂ - 11)(r₂ - 3) = 0

⇒ r₂ = 11 cm or 3 cm

Using equation (1), we find r₁:

If r₂ = 11, then r₁ = 14 - 11 = 3cm

If r₂ = 3, then r₁ = 14 - 3 = 11cm

Hence, the radii of the two circles are 11 cm and 3 cm.

The diameters of the front and the rear wheels of a tractor are 63 cm and 1.54 m respectively. The rear wheel is rotating at $24\dfrac{6}{11}$ revolutions per minute. Find :

(i) the revolutions per minute made by the front wheel.

(ii) the distance travelled by the tractor in 40 minutes.

Answer

(i) Given:

Diameter of the rear wheel = 1.54 m

Diameter of the front wheel = 0.63 m

Radius of the rear wheel = $\dfrac{d}{2}$ = $\dfrac{1.54}{2}$ = 0.77 m

Radius of the front wheel = $\dfrac{d}{2}$ = $\dfrac{0.63}{2}$ = 0.315 m

Distance travelled by the tractor in 1 revolution of the rear wheel = Circumference of the rear wheel

= 2πr

$= 2 \times \dfrac{22}{7} \times 0.77\\[1em] = \dfrac{44}{7} \times 0.77\\[1em] = 44 \times 0.11\\[1em] = 4.84 \text{ m}$

The number of revolutions per minute for the rear wheel is = $24\dfrac{6}{11}$ = $\dfrac{270}{11}$

So, the distance traveled by the rear wheel in one minute is = 4.84 x $\dfrac{270}{11}$

= 0.44 x 270

= 118.8 m

Let x be the number of revolutions made by the front wheel.

The total distance travelled by tractor in 1 min = Number of revolutions made by the front wheel in 1 min x Circumference of wheel

$⇒ x \times 2 \times \dfrac{22}{7} \times 0.315 = 118.8\\[1em] ⇒ x \times 44 \times 0.045 = 118.8\\[1em] ⇒ x \times 1.98 = 118.8\\[1em] ⇒ x = \dfrac{118.8}{1.98}\\[1em] ⇒ x = 60$

Hence, the number of revolutions made by the front wheel is 60.

(ii) Distance travelled by the tractor in 40 minutes = Number of revolutions made by the rear wheel in 40 min x Circumference of the rear wheel

$= \dfrac{270}{11} \times 40 \times 4.84 \\[1em] = 270 \times 40 \times 0.44 \\[1em] = 4,752 \text{ m}$

Hence, the distance travelled by the tractor in 40 minutes is 4,752 m = 4.752 km.

Two circles touch each other externally. The sum of their areas is 74π cm² and the distance between their centres is 12 cm. Find the diameters of the circle.

Answer

Given:

Sum of area = 74π cm²

The distance between their centres = 12 cm

Let r₁ and r₂ be the radius of 2 circles.

⇒ A₁ + A₂ = 74π

⇒ πr₁² + πr₂² = 74π

⇒ π(r₁² + r₂²) = 74π

⇒ $\cancel{π}$ (r₁² + r₂²) = 74 $\cancel{π}$

⇒ r₁² + r₂² = 74 ...............(1)

And, r₁ + r₂ = 12

⇒ r₁ = 12 - r₂

Putting the value of r₁ in equation (1),

⇒ (12 - r₂)² + r₂² = 74

⇒ 12² + r₂² - 2 x 12 x r₂ + r₂² = 74

⇒ 144 + 2r₂² - 24r₂ = 74

⇒ 144 + 2r₂² - 24r₂ - 74 = 0

⇒ 2r₂² - 24r₂ + 70 = 0

⇒ r₂² - 12r₂ + 35 = 0

⇒ r₂² - 7r₂ - 5r₂ + 35 = 0

⇒ r₂(r₂ - 7) - 5(r₂ - 7) = 0

⇒ (r₂ - 7)(r₂ - 5) = 0

⇒ r₂ = 7 cm or 5 cm

⇒ r₁ = 12 - r₂ = 12 - 7 or 12 - 5 = 5 cm or 7 cm

Hence, the radius of 2 circles = 7 cm and 5 cm.

If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.

Answer

Let a be the side of the square.

By using the Pythagoras theorem,

AB² + BC² = AC²

⇒ a² + a² = AC²

⇒ AC² = 2a²

⇒ AC = a $\sqrt{2}$

Diagonal of the square = Diameter of the circle

d = a $\sqrt{2}$

Radius, r = $\dfrac{d}{2}$ = $\dfrac{a \sqrt{2}}{2}$

Now, the ratio of the area of the circle to the area of the square is:

$= \dfrac{\text{Area of circle}}{\text{Area of square}}\\[1em] = \dfrac{πr^2}{\text{side}^2}\\[1em] = \dfrac{π \times \Big(\dfrac{a \sqrt{2}}{2}\Big)^2}{a^2}\\[1em] = \dfrac{π \times a^2 \times 2}{4a^2}\\[1em] = \dfrac{π \times a^2}{2a^2}\\[1em] = \dfrac{π}{2}\\[1em] = \dfrac{22}{14}\\[1em] = \dfrac{11}{7}\\[1em]$

Hence, the ratio of the area of the circle to the area of the square is 11 : 7.