Mensuration 1

In an isoscele right angled triangle, the perimeter is 20 metre. Find its area.

In an isoscele right angled triangle,

Area = 1/23.3 × perimeter²

= 1/23.3 × 20² = 17.167 m²

The area of a triangle is 615 m². If one of its sides is 123 metre, find the length of the perpendicular dropped on that side from opposite vertex.

In a triangle,

Area

⇒

∴ Length of perpendicular

= 10 m.

The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. The area of the triangle is

Let ABC be the isosceles triangle and AD be the altitude.

Let AB = AC = x.

Then, BC = (32 – 2x).

Since, in an isosceles triangle, the altitude bisects the base. So,

BD = DC = (16 – x).

In ∆ADC, AC² = AD² + DC²

⇒ x² = (8)² + (16 – x)²

⇒ 32x = 320

⇒ x = 10.

∴ BC = (32 – 2x)

= (32 – 20) cm

= 12 cm.

Hence, required area

=

In a triangle ABC, points P, Q and R are the mid-points of the sides AB, BC and CA respectively. If the area of the triangle ABC is 20 sq. units, find the area of the triangle PQR

Consider for an equilateral triangle. Hence ∆ABC consists of 4 such triangles with end points on mid points AB, BC and CA

⇒ ¼ ar (∆ABC) = ar (∆PQR)

⇒ ar (∆PQR) = 5 sq. units

How many squares are there in a 5 inch by 5 inch square grid, if the grid is made up one inch by one inch squares?

Required number of squares 5²/1² = 25

If the ratio of areas of two squares is 9: 1, the ratio of their perimeter is:

Let the area of two squares be 9x and x respectively.

So, sides of both squares will be

and respectively. [since, side = ]

Now, perimeters of both squares will be

and respectively. [since, perimeter = 4 × side]

Thus, ratio of their perimeters

= 3: 1

The area of a square field is 576 km². How long will it take for a horse to run around at the speed of 12 km/h?

Area of field = 576 km². Then,

each side of field

Distance covered by the horse Perimeter of square field

= 24 × 4 = 96 km

∴

= 96/12= 8 h

A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, then what is the area of the parking space in square feet?

Clearly, we have:

ℓ = 9 and ℓ + 2b

= 37 or b = 14.

∴ Area = (ℓ × b)

= (9 × 14) sq. ft.

= 126 sq. ft.

A farmer wishes to start a 100 square metres rectangular vegetable garden. Since he has only 30 m barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. The dimension of the garden is:

We have: 2b + ℓ = 30

⇒ ℓ = 30 – 2b.

Area = 100 m²

⇒ ℓ × b = 100

⇒ b(30 – 2b) = 100

⇒ b² – 15b + 50 = 0

⇒ (b – 10) (b – 5) = 0

⇒ b = 10 or b = 5.

When b = 10, ℓ = 10 and when b = 5, ℓ = 20.

Since the garden is rectangular,

so its dimension is 20 m × 5 m.

A rectangular plot 15 m ×10 m, has a path of grass outside it. If the area of grassy pathway is 54 m², find the width of the path.

Let the width of the path = W m

then, length of plot with path

= (15 + 2w) m

and breadth of plot with path

= (10 + 2 W) m

Therefore, Area of rectangular plot (without path)

= 15 × 10 = 150 m²

and Area of rectangular plot (with path)

= 150 + 54 = 204 m²

Hence, (15 + 2w) × (10 + 2w) = 204

⇒ 4w² + 50 w – 54 = 0

⇒ 2w² + 25 w – 27 = 0

⇒ (w – 2) (w + 27) = 0

Thus w = 2 or –27

∴ with of the path = 2 m

The area of a rectangular field is 144 m². If the length had been 6 metres more, the area would have been 54 m² more. The original length of the field is

Let the length and breadth of the original rectangular field be x m and y m respectively.

Area of the original field

= x × y = 144 m²

∴ … (i)

If the length had been 6 m more, then area will be

(x + 6) y = 144 + 54

⇒ (x + 6) y = 198 … (ii)

Putting the value of x from eq (i) in eq (ii), we get

⇒ 144 + 6y = 198

⇒ 6y = 54 ⇒ y = 9 m

Putting the value of y in eq (i) we get x = 16 m

The length and breadth of a playground are 36 m and 21 m respectively. Poles are required to be fixed all along the boundary at a distance 3 m apart. The number of poles required will be

Given, playground is rectangular.

Length = 36 m, Breadth = 21 m

Now, perimeter of playground

= 2( 21 + 36) = 114

Now, poles are fixed along the boundary at a distance 3m.

∴ Required no. of poles = ¹¹⁴⁄₃ = 38

The cost of carpeting a room 18 m long with a carpet 75 cm wide at ₹4.50 per metre is ₹810. The breadth of the room is:

Length of the carpet

Area of the room = Area of the carpet

∴ Breadth of the room

= 7.5 m.

If the perimeter and diagonal of a rectangle are 14 and 5 cms respectively, find its area.

In a rectangle,

49 = 25 + 2 × area

A square carpet with an area 169 m² must have 2 metres cut-off one of its edges in order to be a perfect fit for a rectangular room. What is the area of rectangular room?

Side of square carpet

After cutting of one side,

Measure of one side

= 13 – 2 = 11 m

and other side = 13 m (remain same)

∴ Area of rectangular room

= 13 × 11 = 143 m²

A picture 30 inch × 20 inch has a frame 2½ inch wide. The area of the picture is approximately how many times the area of the frame?

Length of frame

= 30 + 2.5 × 2 = 35 inch

Breadth of frame

= 20 + 2.5 × 2 = 25 inch

Now, area of picture

= 30 × 20 = 600 sq. inch

Area of frame

= (35 × 2.5) + (25 × 2.5) = 150

The floor of a rectangular room is 15 m long and 12 m wide. The room is surrounded by a verandah of width 2 m on all its sides. The area of the verandah is:

Area of the outer rectangle = 19 × 16 = 304 m²

Area of the inner rectangle

= 15 × 12 = 180 m²

Required area

= (304 – 180) = 124 m²

A typist uses a paper 12 inch by 5 inch length wise and leaves a margin of 1 inch at the top and the bottom and a margin of ½ inch on either side. What fractional part of the paper is available to him for typing?

Area of paper

= 12 × 5 = 60 sq. inch

Area of typing part

= (12 – 2) × (5 – 1)

= (10 × 4) sq. inch

∴ Required fraction

A rectangular lawn 70 m × 30 m has two roads each 5 metres wide, running in the middle of it, one parallel to the length and the other parallel to the breadth. Find the cost of gravelling the road at the rate of ₹4 per square metre.

Total area of road

= Area of road which parallel to length + Area of road which parallel to breadth – overlapped road

= 70 × 5 + 30 × 5 – 5 × 5

= 350 + 150 – 25

= 500 – 25 = 475 m²

∴ Cost of gravelling the road

= 475 × 4 = ₹1900

The length and breadth of the floor of the room are 20 feet and 10 feet respectively. Square tiles of 2 feet length of different colours are to be laid on the floor. Black tiles are laid in the first row on all sides. If white tiles are laid in the one-third of the remaining and blue tiles in the rest, how many blue tiles will be there?

Area left after laying black tiles

= [(20 – 4) × (10 – 4)] sq. ft.

= 96 sq. ft.

Area under white tiles

sq. ft

= 32 sq. ft.

Area under blue tiles

= (96 – 32) sq. ft = 64 sq. ft.

Number of blue tiles

The ratio between the length and the breadth of a rectangular park is 3: 2. If a man cycling along the boundary of the park at the speed of 12km / hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

Perimeter = Distance covered in 8 min.

Let length = 3x metres and breadth = 2x metres.

Then, 2 (3x + 2x) = 1600 or x = 160.

∴ Length = 480 m and Breadth = 320 m.

∴ Area = (480 × 320) m² = 153600 m².

In measuring the side of a square, an error of 5% in excess is made. The error % in the calculated area is:

ℓ × b = 100 m²

Therefore,

⇒ b = 10, 5

If we take b = 10, then garden becomes a square

Therefore, b = 5 m.

The length of a room is double its breadth. The cost of colouring the ceiling at ₹25 per sq. m is ₹5,000 and the cost of painting the four walls at ₹240 per sq. m is ₹64,800. Find the height of the room.

Let the length of the room be ℓ m

Then its, breadth = ℓ/2

⇒ ℓ² = 400

⇒ ℓ = 20 m

Also,

⇒ 3ℓh = 270

A rectangular plate is of 6 m breadth and 12 m length. Two apertures of 2 m diameter each and one apertures of 1 m diameter have been made with the help of a gas cutter. What is the area of the remaining portion of the plate?

Given, Length = 12 m and Breadth = 6 m

∴ Area of rectangular plate

= 12 × 6 = 72 m²

Since, two apertures of 3 m diameter each have been made from this plate.

∴ Area of these two apertures

= π(1)² + π(1)²

= π + π = 2 π

Area of 1 aperture of 1m diameter

∴ Total area of aperture

= = ⁹⁹⁄₁₄

∴ Area of the remaining portion of the plate

= 72 - ⁹⁹⁄₁₄ sq. m

= ⁹⁰⁹⁄₁₄ sq. m ≈ 64.5 sq.m

Four sheets 50 cm × 5 cm are arranged without overlapping to form a square having side 55 cm. What is the area of inner square so formed?

Side of the inner square

= 55 – 10 = 45

∴ Area of inner square

= 45 × 45 = 2025 sq. m.

In the adjoining figure, AC + AB = 5 AD and AC - AD = 8. Then the area of the rectangle ABCD is

AC + AB = 5AD

⇒ AC + a = 5b ...(1)

AC - AD = 8

⇒ AC = b + 8 ...(2)

Using (1) and (2) ,

a + b + 8 = 5b

⇒ a + 8 = 4b ...(3)

Using Pythagorous theorem,

a² + b² = (b + 8)² = b² + 64 + 16b

⇒ a² = 16b + 64 = (4b - 8)² = 16b² + 64 - 64b

⇒ 16b² - 80b = 0

⇒ b = 0 or 5

Putting b = 5 in (3),

a = 4b - 8 = 20 - 8 = 12

Area of rectangle = 12 × 5 = 60

ABCD is a square, F is the mid-point of AB and E is a point on BC such that BE is one-third of BC. If area of ∆FBE = 108 m², then the length of AC is:

Let the side of the square be x, then, BE = x/3 and BF = x/2 Area of ∆FEB Now, ⇒ x² = 108 × 12 = 1296 In ∆ADC, we have AC² = AD² + DC² = x² + x² = 2x² = 2 × 1296 = 2592 ⇒ AC

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