Mensuration Exercise 1

In an isoscele right angled triangle,

Area = 1/23.3 × perimeter²

= 1/23.3 × 20² = 17.167 m²

In a triangle,

Area

⇒

∴ Length of perpendicular

= 10 m.

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

=

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

Required number of squares 5²/1² = 25

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

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

Clearly, we have:

ℓ = 9 and ℓ + 2b

= 37 or b = 14.

∴ Area = (ℓ × b)

= (9 × 14) sq. ft.

= 126 sq. ft.

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.

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

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

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

Length of the carpet

=

Area of the room = Area of the carpet

=

∴ Breadth of the room

=

= 7.5 m.

In a rectangle,

⇒

49 = 25 + 2 × area

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

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

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

Area of the inner rectangle

= 15 × 12 = 180 m²

Required area

= (304 – 180) = 124 m²

Area of paper

= 12 × 5 = 60 sq. inch

Area of typing part

= (12 – 2) × (5 – 1)

= (10 × 4) sq. inch

∴ Required fraction

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

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

=

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.

ℓ × b = 100 m²

⇒

Therefore,

⇒ b² – 15b + 50 = 0

⇒ b = 10, 5

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

Therefore, b = 5 m.

Let the length of the room be ℓ m

Then its, breadth = ℓ/2

Therefore,

⇒ ℓ² = 400

⇒ ℓ = 20 m

Also,

⇒ 3ℓh = 270

⇒

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

Side of the inner square

= 55 – 10 = 45

∴ Area of inner square

= 45 × 45 = 2025 sq. m.

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

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