Question Detail
If the ratio of the areas of two squares is 225:256, then the ratio of their perimeters is :
- 15:12
- 15:14
- 15:16
- 15:22
Answer: Option C
Explanation:
\begin{aligned}
\frac{a^2}{b^2} = \frac{225}{256} \\
\frac{15}{16} \\
<=> \frac{4a}{4b} = \frac{4*15}{4*16} \\
= \frac{15}{16} = 15:16
\end{aligned}
1. 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}
2. The area of rhombus is 150 cm square. The length of one of the its diagonals is 10 cm. The length of the other diagonal is:
- 15 cm
- 20 cm
- 25 cm
- 30 cm
Answer: Option D
Explanation:
We know the product of diagonals is 1/2*(product of diagonals)
Let one diagonal be d1 and d2
So as per question
\begin{aligned}
\frac{1}{2}*d1*d2 = 150 \\
\frac{1}{2}*10*d2 = 150 \\
d2 = \frac{150}{5} = 30 \\
\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. If the circumference of a circle increases from 4pi to 8 pi, what change occurs in the area ?
- Area is quadrupled
- Area is tripled
- Area is doubles
- Area become half
Answer: Option A
Explanation:
\begin{aligned}
2\pi R1 = 4 \pi \\
=> R1 = 2 \\
2\pi R2 = 8 \pi \\
=> R2 = 4 \\
\text{Original Area =} 4\pi * 2^2 \\
= 16 \pi \\
\text{New Area =} 4\pi * 4^2 \\
= 64 \pi
\end{aligned}
So the area quadruples.
5. 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 ?
- 1:1
- 1:2
- 2:3
- 2:1
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 :)