Mensuration Quiz 1

Area of square = Area of rectangle = 675 cm²

Length of rectangle = 27 cm

∴ Breadth of rectangle = 25 cm

Largest side of the triangle = 25-9 = 16 cm

∵ It is a right angled triangle, therefore it is also hypotenuse.

Let the side of the triangle be x cm.

∴ x² + x² = 16² ⇒ x² = 16²/2

⇒ x = \sqrt{128}\ cm\ = 8\sqrt{2}\ cm\ = 11.31\ cm

Height of house = 63 m

Height of hemispherical portion : Height of lower portion (cuboidal portion) =  1 : 2

⇒ Height of hemispherical portion = \frac{1}{3}\times Total\ height\ = \frac{1}{3}\times 63\ m\ = 21\ m

& Height of cuboidal portion = \frac{2}{3}\times Total\ height\ = \frac{2}{3}\times 63\ m\ = 42\ m

 

Diameter of top = 21 × 2 m = 42 m

⇒ Length and breadth of cuboidal top face = 42 m

∴ Length of cuboidal portion = Breadth of cuboidal portion = 42 m

Hence, Area = 21 × 42 × 42 = 37044 m³

Perimeter of square field = 2(l + b)

= 2 × 2(10 + 4) = 2 × 28 = 56 m

Side of square field = 56/4 = 14 m

Radius of semi-circular field = 14/2 = 7 m

Hence, Circumference of the semi-circular field = (\pi \times 7 + 14)\ m = 36 m

Hence, Length of fence = 36 m

When a 5×5 cube is broken apart into 1 cc cubes, we will get 5×5×5  = 125 1 cc cubes.

All the outer and visible surfaces of the larger cube have at least one of their side painted. The smaller cubes which are not visible to anyone have none of their sides painted.

Therefore, the cubes which are not on the edge of the larger cube and that lie inside the facing sides of the larger cube will have exactly one side painted.

 

In the larger cube, there will be 5×5×5 = 125 cubes. Of these, there will be 3×3×3 = 27 inside the outer layer.

In total, there will be 27 such cubes.

Let the length of the side of his farmhouse be a meter.

∴ It’s area = a² m²

⇒ a² = 24200 ⇒ a=\sqrt{24200} = 155.56\ m

The length of the diagonal, d = \sqrt{2}\ a

⇒ d=\sqrt{2}\times 155.56 = 220\ m

Hence, time taken to cross a distance of 220 m with a speed 6.6 km/hr is given by

Speed, s = \frac{220\ meter}{6.6 kmph} = \frac{220}{6.6\times \frac{5}{18}}

⇒ s = 120 seconds = 2 minutes

Let's take radius of hemisphere and cone be r,

Height of hemisphere h = r.

So the height of the cone = height of the hemisphere = r

Slant height of the cone l² = h² + r²

l = \sqrt{r^2 + r^2} = r^2

(Cone curved surface area) / (Hemisphere curved surface area) =

=\frac{\pi r.r \sqrt{2}}{2\pi r^2} = \frac{\sqrt{2}}{2} = 1 : \sqrt{2}.

Hence, option B is correct.

s = \frac{1}{2}\times(a + b + c) = 18

Therefore area = \sqrt{s(s-a)(s-b)(s-c)}

=\sqrt{18(18-7)(18-14)(18-15)}<br /> <br /> =\sqrt{18(11)(4)(3)}

=\sqrt{18\times 132} = \sqrt{2376} = 48.744 cm²

½ x AC x altitude = 48.744 cm²

Altitude = \frac{48.744\times 2}{AC} = \frac{97.488}{AC}

= \frac{97.488}{14} = 6.9634 approx. = 7cm

Given that, radius of cylinder = radius of cone = r

And, height of cylinder = height of cone = h

\text{ Required ratio } = \frac{\text{ Volume of wasted wood }}{\text{ Volume of utilised wood }}

= \frac{\text{ Volume of cone }}{\text{ Volume of cylinder } - \text{ Volume of cone }}

=\frac{\frac{1}{3}\pi r^2}{\pi r^2h - \frac{1}{3}\pi r^2 h}

To get to know the overall increase in volume of the stadium, we need to apply the net% effect formula twice here.

We know that

x + y + \frac{xy}{100}

Here, x = y = 10%

Applying the formula, we get

= 10 + 10 + \frac{10 \times 10}{100}  = 20 + 1 = 21

Now, let's take x = 21% & y = 10%. Then,

= 21 + 10 + \frac{21 \times 10}{100}

Hence, the volume of the stadium will increased by (31 + 2.1) = 33.1%

Given, ratio of their radii = 5 : 3

Let, R = radius of outer circle and r = radius of inner circle

∴  R = 5x and r = 3x

And, Area enclosed between the circumference of two concentric circles = 16π m

∴  π(R² – r²) = 16π

⇒  (5x)² – (3x)² = 16

⇒  25x² – 9x² = 16

⇒  16x = 16

⇒  x = 1

∴  R = 5x = 5 × 1 = 5 m

∴  Area of outer circle = πR² = π x (5)² = 25π m²