Area Of Plane Figures

Side of square ACDE, a = 4 cm

As Area of square, Ar (ACDE) = a² sq. cm

⇒ A(ACDE) = (4)² = 16 sq.cm

Height of Δ ABC, h = 6 – 4 = 2 cm

⇒ Ar (ABC) = 4 sq.cm

Area enclosed by the figure, Area = Ar(ABC) + Ar (ACDE)

⇒ Area = 4 + 16

⇒ Area = 20 sq.cm

Area of square ABCF, Ar(ABCF) = (side)²

⇒ Ar(ABCF) = 18² = 324 sq.cm

Given Height of trapezium EDCF, h = 8 cm

As we know that,

⇒ Ar(EDCF) = 100 sq.cm

Area of given figure, Req. area = Ar(EDCF) + Ar(ABCF)

⇒ Req. area = 100 + 324 = 424 sq.cm

∴ Area of given figure is 424 sq. cm

Area of rectangle ABCD, Ar(ABCD) = BC × CD

⇒ Ar(ABCD) = 20 × 15 = 300 sq.cm

Given height of trapezium, h = 8 cm

Sum of parallel sides of trapezium = EF + AD

⇒ 6 + 15 = 21 cm

⇒ Ar(EFAD) = 84 sq.cm

Now, required area = Ar(EFAD) + Ar(ABCD)

⇒ 84 + 300

⇒ 384 sq.cm

∴ Area of given figure is 384 sq.cm

Since the quadrilateral can be divided into two triangles, therefore the area of the figure is equal to the sum of these triangles.

As we know that,

⇒ A₁ = 25 sq.cm

Similarly,

⇒ A₂ = 30 sq.cm

Required Area = A₁ + A₂

⇒ 25 + 30

⇒ 55 sq.cm

∴ the area of quadrilateral ABCD is 55 sq.cm

We can see that the shaded region represents a trapezium, with the parallel sides are of length 20cm and 28 cm

Height of trapezium = h

As we know that,

⇒ Area = 96 sq.cm

∴ Area of the shaded part is 96 sq. cm

As we know that,

And we have 3 triangles here, namely, Δ DEI, Δ DHC, Δ FGA

Therefore,

⇒ Ar(Δ DEI) = 1200 sq. m

Similarly,

⇒ Ar(Δ DHC) = 1600 sq.m

And,

⇒ Ar(Δ FGA) = 1250 sq. m

Also, there are two trapeziums, namely, EIGF and ABCH

⇒ Ar(EIGF) = 3850 sq.m

Also,

⇒ Ar(ABCH) = 2800 sq.m

Now, the area of the given figure is equal to the sum of the individual areas of triangles and trapeziums.

⇒ Req. area = Ar(Δ DEI) + Ar(Δ DHC) + Ar(Δ FGA) + Ar(EIGF) + Ar(ABCH)

⇒ Req. area = 1200 + 1600 + 1250 + 3850 + 2800

⇒ Req. area = 10700 sq.m

∴ the area of the given figure is 10700 sq.m

As we know that,

And we have 3 triangles here, namely, Δ EFH, Δ AJG, Δ AKB

Therefore,

⇒ Ar(Δ EFH) = 400 sq. m

Similarly,

⇒ Ar(Δ AKB) = 750 sq.m

And,

⇒ Ar(Δ AJG) = 1400 sq. m

Also, there are three trapeziums, namely, EICD, ICBK and FHJG

⇒ Ar(EICD) = 3600 sq.m

Also,

⇒ Ar(ICBK) = 2100 sq.m

And,

⇒ Ar(FHJG) = 2400 sq.m

Now, the area of the given figure is equal to the sum of the individual areas of triangles and trapeziums.

⇒ Req. area = Ar(Δ EFH) + Ar(Δ AKB) + Ar(Δ AJG) + Ar(EICD) + Ar(ICBK) + Ar(FHJG)

⇒ Req. area = 400 + 750 + 1400 + 3600 + 2100 + 2400

⇒ Req. area = 10650 sq.m

∴ the area of the given figure is 10650 sq.m

Let the parallel sides of the trapezium be 5x and 3x.

Distance between the parallel sides, h = 16 cm

Area of trapezium, A = 960 sq. cm

As we know that,

⇒ 16 × 4x = 960

⇒ 4x = 60

⇒ x = 15

⇒ 5x = 75 and 3x = 45

∴ the length of the parallel sides is 75cm and 45cm

The diagonals of the rhombus given are 45 cm and 30 cm.

As we know that,

⇒ Area of one tile = 675 sq.cm

Total area of floor = (area of one tile) × (Total number of tiles)

⇒ Area of floor = 675 × 3000 = 2025000 sq.cm = 202.5 sq.m

⇒ Cost of flooring = 202.5 × 20

⇒ Cost = Rs. 4050

∴ Total cost of flooring is Rs. 4050

Considering Jyoti’s Diagram first,

The pentagon given is divided into two equal trapeziums by Jyoti, whose height, h = 15/2 = 7.5 cm

Also, the parallel sides of each trapezium are of length 15cm and 30 cm.

As we know that,

Since the pentagon is made of two equal trapeziums, so the area of pentagon = twice the area of trapezium

⇒ Area of pentagon = 2 × area of trapezium

⇒ Req. area = 45 × 7.5 = 337.5 sq.cm

∴ area of pentagon as found by Jyoti is 337.5 sq. cm

Now, in case of Rashida,

She divided the pentagon to form a square and a triangle.

Side of square, a = 15 cm

Height of triangle, H = 30 – 15 = 15 cm

Area of square = a² = 15² = 225 sq.cm

⇒ Area of triangle = 112.5 sq.cm

Area of pentagon = Area of triangle + Area of square

⇒ 112.5 + 225

⇒ 337.5 sq.cm

∴ We see that, the area of pentagon found by both ways comes out to be same. Hence, It does not matter whichever way we divide a figure.

The dimensions of rectangular sheet are 36 cm by 25 cm.

The Area of rectangular sheet = 36 × 25 = 900 sq.cm

From this sheet, 56 circular buttons are cut as shown,

Diameter of each circle = 3.5 cm

⇒ radius of each button, r = 1.75 cm

Area of one button = πr²

⇒ Area of each button = 9.625 sq.cm

Total area of buttons cut = area of each button × 56

⇒ Total Area of buttons = 539 sq. cm

Area of remaining sheet (area shaded with blue) = (Area of rectangular sheet) – (total area of buttons)

⇒ Req. area = 900 – 539 = 361 sq.cm

∴ Area of the remaining sheet is 361 sq.cm

Since the circle is inscribed in a square of side 28 cm, this means that the diameter of the circle is equal to the side of square.

Diameter of circle, d = 28cm

⇒ Radius of circle, r = 14 cm

As we know that,

Area of a circle = π × r²

⇒ Area of circle = 616 sq.cm

∴ The area of circle inscribed in the square is 616 sq.cm

The shaded region is formed of four semicircles and as the base of these semicircles lie on the side of a square, therefore the diameter of the semicircle is equal to the side of the square.

⇒ diameter of semicircle = side of square = d

It is shown that, r + d + r = 42

⇒ d = 21 cm

⇒ Radius, r = 10.5 cm

⇒ A = 173.25 sq.cm

Area of shaded region = 4 × (Area of semicircle)

⇒ 4 × A

⇒ 4 × 173.25

⇒ 693 sq.cm

∴ Area of the shaded region is 693 sq.cm

Diameter of bigger semicircle, D = 21 cm

⇒ Radius of bigger semicircle, R = D÷ 2

Diameter of smaller semicircle, d = 21 cm

⇒ Radius of smaller semicircle, r = d÷ 2

⇒ A₁ = 173.25 sq.cm

Similarly,

⇒ A₂ = 43.31 sq.cm

Area of shaded region = 2 × (A₁ – A₂)

⇒ Req. area = 2 × (173.25 – 43.31)

⇒ 2 × 129.94

⇒ 259.87 sq.cm

∴ Area of shaded region is 259.87 sq.cm

As we know that,

From the above figure we see,

Diameter of bigger semi-circle, D = 42 cm

⇒ S1 and S4 are bigger semicircles with Radius, R = 21 cm

Also, S2, S3, S5 and S6 are smaller semi-circles with diameter, d equal to the radius of bigger semi-circle.

⇒ d = R

⇒ A_B = 693 sq.cm

Similarly,

⇒ A_S = 173.25 sq.cm

Area of shaded region = Area(S1) + Area(S4) + Area(S3) + Area(S6) – Area (S2) – Area(S5)

⇒ Area of shaded region = A_B + A_B + A_S + A_S – A_S - A_S

⇒ Area = 2 × A_B

⇒ 2 × 693 = 1386 sq.cm

∴ Area of shaded region is 1386 sq.cm

Let the area of a semicircle be A_S and the area of a quadrant be A_Q

Since there are four semi-circles, out of which two are shaded and two are unshaded, therefore the area of shaded region is equal to:

Since the area of shaded region is equal on both sides of line AOB,

So, if we calculate area of upper portion, then we can simply twice that area in order to get required area.

In figure above, S1, S2, S3,and S4 represents semi-circles and Q1 and Q2 represents quadrants.

⇒ Area of shaded region = 2 × (area of upper shaded portion on line AOB)

⇒ Area of shaded region = 2 × [area(S1) + area(Q1) – area(S2)]

As S1 and S2 are equal because of same radius, therefore their area will also be equal i.e area(S1) = area(S2)

⇒ Area of shaded region = 2 × area(Q1)

⇒ Area (Q1) = 154 sq.cm

⇒ Required area = 2 × area(Q1)

⇒ 2 × 154 = 308 sq.cm

∴ Area of shaded region is 308 sq.cm

Side of square, a = 7cm

Area of square, A_S = a²

⇒ A_S = 7² = 49 sq.cm

⇒ R = 3.5 cm

There is a quadrant of each of four circles that is present inside the square,

⇒ A_Q = 9.625 sq.cm

Shaded area = Area(square) – [4 × area(quadrant)]

⇒ Shaded area = A_S – [4 × A_Q]

⇒ 49 – (4 × 9.625)

⇒ 49 – 38.5

⇒ 10.5 sq.cm

∴ Area of shaded region is 10.5 sq.cm

Let the side of the triangle be ‘a’

As we know that,

⇒ a = 14 cm

Radius of each circle, r = a ÷ 2

⇒ r = 7 cm

Here, θ = 60°

⇒ A_S = 25.67 sq.cm

Required area = Area(Δ ABC) – 3× areaof sector

⇒ Req. area = 49√3 – [3× 25.67]

⇒ 84.87 – 77.01

⇒ 7.86 sq.cm

∴ Area of the portion in the triangle not included in the circles is 7.86 sq.cm

Radius of each circle, r = a

We need to find the shaded area,

If we connect the four centers we will get a square of side 2a.

Area of square,A_S = (2a)² = 4a²

Area of shaded region = A_S – [4 × A_Q]

Since each side of square measures 14 cm, so the radius of each circle is half of the side.

Therefore, radius of circle, R = 7 cm

We need to find the area of the shaded region,

There is a quadrant of each of four circles that is present inside the square,

⇒ A_Q = 38.5 sq.cm

Also, area of square, A_S = 14² = 196 sq.cm

Area of shaded region = A_S – [4 × A_Q]

⇒ 196 – [4 × 38.5]

⇒ 196 – 154

⇒ 42 sq.cm

∴ area of shaded region is 42 sq. cm

AB = BC = 3.5 cm and DE = 2cm

⇒ DC = DE + EC = 2 + 3.5 = 5.5 cm

As we know that,

Height of trapezium here, BC = 3.5 cm

⇒ A_T = 15.75 sq.cm

Now, area of quarter circle = A_Q

⇒ A_Q = 9.625 sq.cm

Required area = A_T – A_Q

⇒ Req. area = 15.75 – 9.625 = 6.125 sq.cm

∴ Area of remaining piece of cardboard is 6.125 sq.cm

The length of rope to which horse is tied will be equal to the radius of the quarter circle that the horse grazes.

⇒ Radius, r = 21m

As we know that,

⇒ Area = 346.5 sq.m

∴ The area that horse can graze is 346.5 m²