Question Detail

What will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of the rectangle as base and a vertex on the opposite side of the rectangle ?

Answer: Option D

Explanation:

As far as questions of Area or Volume and Surface area are concerned, it is all about formulas and very little logic. So its a sincere advice to get all formulas remembered before solving these questions.

Lets solve this,

\begin{aligned}
\text{Area of rectangle =}l*b\\
\text{Area of triangle =}\frac{1}{2}l*b\\
\text{Ratio =}l*b:\frac{1}{2}l*b \\
= 1:\frac{1}{2} \\
= 2:1

\end{aligned}

One little thing which should be taken care in this type of question is, be sure you are calculating ration in the given order of the question.
If it is ratio of triangle and rectangle then we have to write triangle formula first. cheers :)

1. A courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm. The total number of bricks required is :

16000
18000
20000
22000

Answer: Option C

Explanation:

\begin{aligned}
\text{Number of bricks =}\frac{\text{Courtyard area}}{\text{1 brick area}} \\
= \left( \frac{2500 \times 1600}{20 \times 10} \right) \\
= 20000
\end{aligned}

2. What are the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?

Answer: Option B

Explanation:

In this type of questions, first we need to calculate the area of tiles. With we can get by obtaining the length of largest tile.
Length of largest tile can be obtained from HCF of length and breadth.
So lets solve this,

Length of largest tile = HCF of (1517 cm and 902 cm)
= 41 cm

Required number of tiles =
\begin{aligned}
\frac{\text{Area of floor}}{\text{Area of tile}} \\
= \left(\frac{1517\times902}{41 \times 41}\right)\\
= 814

\end{aligned}

3. The height of an equilateral triangle is 10 cm. find its area.

\begin{aligned} \frac{120}{\sqrt{3}} cm^2 \end{aligned}
\begin{aligned} \frac{110}{\sqrt{3}} cm^2 \end{aligned}
\begin{aligned} \frac{100}{\sqrt{3}} cm^2 \end{aligned}
\begin{aligned} \frac{90}{\sqrt{3}} cm^2 \end{aligned}

Answer: Option C

Explanation:

Let each side be a cm, then
\begin{aligned}
\left(\frac{a}{2}\right)^2+{10}^2 = a^2 \\
<=>\left(a^2-\frac{a^2}{4}\right) = 100 \\
<=> \frac{3a^2}{4} = 100 \\
a^2 = \frac{400}{3} \\
Area = \frac{\sqrt{3}}{4}*a^2 \\
= \left(\frac{\sqrt{3}}{4}*\frac{400}{3}\right)cm^2 \\
= \frac{100}{\sqrt{3}}cm^2
\end{aligned}

4. The area of incircle of an equilateral triangle of side 42 cm is :

\begin{aligned} 462 cm^2 \end{aligned}
\begin{aligned} 452 cm^2 \end{aligned}
\begin{aligned} 442 cm^2 \end{aligned}
\begin{aligned} 432 cm^2 \end{aligned}

Answer: Option A

Explanation:

\begin{aligned}
\text{Radius of incircle} = \frac{a}{2\sqrt3} \\
= \frac{42}{2\sqrt3} \\
= 7\sqrt{3} \\
\text{Area of incircle =} \\
\frac{22}{7}*49*3 = 462 cm^2
\end{aligned}

5. A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is.

200 m
150 m
148 m
140 m

Answer: Option A

Explanation:

Let the triangle and parallelogram have common base b,
let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then
\begin{aligned}
\text{Area of triangle =}\frac{1}{2}*b*h1\\
\text{Area of rectangle =}b*h2\\
\text{As per question }\\
\frac{1}{2}*b*h1 = b*h2 \\

\frac{1}{2}*b*h1 = b*100 \\

h1 = 100*2 = 200 m \\

\end{aligned}