Hexaware Mensuration Quiz 1

Let breadth = x. then, length = 3x. then,

(3x - 15) (x + 15) – 3x × x = 150

⇒ 45x - 15x – 225 = 150
⇔ 30x = 150 + 225
⇒x = 12.5
So, length of the rectangle = 3x

Therefore, 3 × 12.5 cm ⇒ 37.5 cm.

One of the parallel sides of the trapezium is a = 30 m, let another parallel side be b, height h = 25 m.

The given area of trapezium = 720 m^{2}[\latex]

We know, by formula;

Area of a trapezium = ½ h (a+b)

720 = ½ (25) (30+b)

30+ b = \frac{720×2}{25}[\latex]

b = 57.6 - 30 = 27.6 m

Area of the carpet = Area of the room = (15 × 15) m2
⇒ 225 m2.

Length of the carpet = Area/ Width

⇒\frac{225}{25} m ⇒ 9 m[\latex]

So, Cost of carpeting = Rs. (9 × 15)

⇒ Rs. 135

Here we will use formula :-

Area = \sqrt s (s-a)(s-b)(s-c) [\latex]

Where , a = 17cm, b = 18 cm and c = 19 cm

s = \frac {1}{2}(a+b+c)[\latex]
= s = 27
\text{ Area } = \sqrt 27 (27-17)(27-18)(27-19) [\latex]
Area = 139.43 cm^{2}[\latex].

To solve this question, we can apply the net% effect formula

Net % effect = x + y + \frac{xy}{100}% [\latex]

Here, x = y = 75%

Applying the net% effect, we get

=75 + 75 + \frac {75 × 75}{100} %[\latex]

= 150 + 56.25 = 206.25%

Let the diagonals of the squares be 3x and 7x respectively.

So, ratio of their areas =

\frac{1}{2} \times (3x)^{2} : \frac {1}{2} \times (7x)^{2}[\latex]
=9x^{2} : 49x ^{2}[\latex]
= 9 : 29

Let the garden be ABCD.

Area of pavement = Area of A₂B₂C₂D₂ – Area of A₁B₁C₁D₁
=[(100 + 8) (50 + 8)] – [(100 – 8)(50 – 8)]
=[(108 × 58) – (92 × 42)]
=2400 sq m

∴ Cost of construction = 2400× 10 = 24000/-

Circumference of circular shape = π × diameter

= 22/ 7 × 140 = 440cm

Length of wire = 440cm

∴Perimeter of rectangle = 2(length + breadth) ⇒ 2(l + b) = 440

⇒ (l + b) = 440/ 2 = 220

Given ratio of sides = 3 : 5. So, Smallest side = 3 / 8 × 220 = 82.5 cm.

According to the question,



2πr₁ = 154 ⇒ r₁ = 24.5 cm

and, 2πr₂ = 66 ⇒ r₂ = 10.5 cm

∴ Area between two concentric circles

= πr₁² – πr₂²

= π(r₁² – r₂²)

= π(r₁ + r₂)(r₁ – r₂)

= \frac {22}{7} × 24.5 × 10.5 = 808.5 cm^{2}[\latex]

Let the breadth of the rectangular field be 'x' m. Then, length of the field will be

x +\frac {x\times 20}{100} = \frac{6x}{5}[\latex]

Now, x\times \frac{6x}{5} = 600[\latex]

or, 6x^{2} = 600 \times 5[\latex]

or, x^{2} = 500[\latex]

or, x = 22.36 m