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. The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is :

22 cm
20 cm
18 cm
10 cm

Answer: Option C

Explanation:

\begin{aligned}
\text{Area of triangle, A1 = }\frac{1}{2}*base*height \\
= \frac{1}{2}*15*12 = 90 cm^2 \\
\text{Area of second triangle =} 2*A1 \\
= 180 cm^2 \\
\frac{1}{2}*20*height = 180 \\
=> height = 18 cm

\end{aligned}

2. The area of a square is 69696 cm square. What will be its diagonal ?

373.196 cm
373.110 cm
373.290 cm
373.296 cm

Answer: Option D

Explanation:

If area is given then we can easily find side of a square as,
\begin{aligned}Side = \sqrt{69696} \\
= 264 cm \\
\text{we know diagonal =}\sqrt{2}\times side \\
= \sqrt{2}\times 264 \\
= 1.414 \times 264 \\
= 373.296 cm \end{aligned}

3. A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered.If the area of the field is 680 sq.ft, how many feet of fencing will be required ?

88 feet
86 feet
84 feet
82 feet

Answer: Option A

Explanation:

We are given with length and area, so we can find the breadth.
as Length * Breadth = Area
=> 20 * Breadth = 680
=> Breadth = 34 feet

Area to be fenced = 2B + L = 2*34 + 20
= 88 feet

4. The perimeters of 5 squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square equal in area to the sum of the area of these square is:

124 cm
120 cm
64 cm
56 cm

Answer: Option A

Explanation:

Clearly first we need to find the areas of the given squares, for that we need its side.

Side of sqaure = Perimeter/4

So sides are,

\begin{aligned}
\left(\frac{24}{4}\right),\left(\frac{32}{4}\right),\left(\frac{40}{4}\right),\left(\frac{76}{4}\right),\left(\frac{80}{4}\right) \\
= 6,8,10,19,20 \\
\text{Area of new square will be }\\
= [6^2+8^2+10^2+19^2+20^2] \\
= 36+64+100+361+400 \\
= 961 cm^2 \\
\text{Side of new Sqaure =}\sqrt{961} \\
= 31 cm \\
\text{Required perimeter =}(4\times31) \\
= 124 cm

\end{aligned}

5. 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}