WeCP Mensuration Quiz 1

Let a = 13, b = 14 and c = 15. then ,

s = \frac{1}{2}(a + b +c) = 21.

So, (s – a) = 8, (s – b) = 7 and (s – c) = 6.

So, \text{ Area }= \sqrt{s(s – a)(s – b)(s – c)}

\Rightarrow \sqrt{21\times 8\times 7\times 6 \Rightarrow 84 cm^2.

Hence, option C is correct.

Let x and y be the length and breadth of the rectangle respectively.

Then, x - 4 = y + 3 or x - y = 7.

Area of the rectangle = xy;   area of the square = (x - 4) (y + 3)

So, (x - 4) (y + 3) = xy   ⇔  3x – 4y = 12

Solving (i) and (ii), we get x = 16 and y = 9.

So, the perimeter of the rectangle = 2(x + y)  ⇒  [2 (16 + 9)] cm = 50 cm.

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

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

Here, x = y = 25%

Applying the net% effect, we get

= 25 + 25 + \frac{25 \times 25}{100}\%

= 50 + 6.25 = 56.25%

Hence, the total percentage change in area will be 56.25%

Let the diagonals of the squares be 2x and 5x respectively.

So, ratio of their areas =

\frac{1}{2}\times (2x)^2 : \frac{1}{2}\times (5x)^2

⇒4x² : 25x²  ⇒  4 : 25.

Hence, option B is correct.

Let the garden be ABCD.

Area of pavement = Area of A₂B₂C₂D₂ - Area of A₁B₁C₁D₁

=[(75 + 4) (47 + 4)] – [(75 – 4)(47 – 4)]

=[(79 × 51) – (71 × 43)] =  4029 – 3053

=976 sq m

∴  Cost of construction = 976 x 15 = $15,616/-

Let the edge of the cube be a. Then,

\therefore \sqrt{3a} = 6\sqrt{3} \Rightarrow a = 6.

So, Volume = a³ = (6 × 6 × 6) cm³  ⇒  216 cm³.

Surface area = 6a²  ⇒  (6 × 6 × 6) cm²  ⇒ 216 cm².

Here, r = 7m and h = 24m.

So, l = \sqrt{r^2 + h^2} \Rightarrow \sqrt{7^2 + (24)^2} \Rightarrow \sqrt{625} ⇒ 25 m.

Area of canvas = πrl

\Rightarrow \left(\frac{22}{7}\times 7\times 25 \right)m^2

\Rightarrow 550m^2

∴ Length of canvas =

\left( \frac{\text{ Area }}{\text{ Width }} \right) = \left( \frac{550}{1.25} \right)m

⇒  440 m.

Hence, option C is correct.

Let the depth of the tank be h meters. then,

π × (7)² × h = 1848

\Rightarrow h = \left( 1848\times \frac{7}{22} \times \frac{1}{7\times 7} \right)

= 12m.

Hence, option D is correct.

Let the radii of the cylinder and cone be 3r and 4r and their heights be 2h and 3h respectively.

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

\frac{\pi\times (3r)^2\times 2h}{(1/3)\pi \times (4r)^2\times 3h} = \frac{9}{8} = 9:8[/latex]

Hence, option A is correct.

Volume required in the tank (200 × 150 × 2) = 60000m³

Length of the fuel column flown in 1 min.

= \frac{\left( \frac{20\times 1000}{60} \right)}m  \Rightarrow \frac{1000}{3}m

Volume flows per minute

= \left( \frac{1.5\times 1.25\times 1000}{3} \right)m^3 ⇒  625 m³.<br /> <br /> ∴ Reqd. time =<br /> <br /> \frac{60000}{625} min. ⇒ 96 min

Hence, option B is correct.