If the diagonal of a square is increased by 4 cm, its area increases by 56 cm 2. Find the ratio of the new area of the square to the initial area of the square.
16 : 9
Understanding the Square Geometry Problem
The problem asks us to find the ratio of the new area of a square to its initial area, given that an increase in the square's diagonal by 4 cm results in an area increase of 56 cm2.
To solve this, we need to use the relationships between the side, diagonal, and area of a square.
Key Formulas for a Square
- If the side of a square is $s$, its diagonal $d$ is $s\sqrt{2}$. This means $s = \frac{d}{\sqrt{2}}$.
- The area $A$ of a square with side $s$ is $A = s^2$.
- Substituting the expression for $s$ in terms of $d$ into the area formula, we get the area in terms of the diagonal: $A = \left(\frac{d}{\sqrt{2}}\right)^2 = \frac{d^2}{2}$.
Setting Up the Equations
Let the initial diagonal of the square be $d_1$ cm. The initial area, $A_1$, is given by:
$$A_1 = \frac{d_1^2}{2}$$
According to the problem, the diagonal is increased by 4 cm. So, the new diagonal, $d_2$, is:
$$d_2 = d_1 + 4$$
The area increases by 56 cm2. So, the new area, $A_2$, is:
$$A_2 = A_1 + 56$$
We can also express the new area $A_2$ in terms of the new diagonal $d_2$:
$$A_2 = \frac{d_2^2}{2} = \frac{(d_1 + 4)^2}{2}$$
Solving for the Initial Diagonal and Area
Now we have two expressions for $A_2$. We can equate them:
$$A_1 + 56 = \frac{(d_1 + 4)^2}{2}$$
Substitute the expression for $A_1$ into this equation:
$$\frac{d_1^2}{2} + 56 = \frac{(d_1 + 4)^2}{2}$$
Multiply the entire equation by 2 to eliminate the denominators:
$$d_1^2 + 112 = (d_1 + 4)^2$$
Expand the right side of the equation:
$$d_1^2 + 112 = d_1^2 + 2(d_1)(4) + 4^2$$
$$d_1^2 + 112 = d_1^2 + 8d_1 + 16$$
Subtract $d_1^2$ from both sides:
$$112 = 8d_1 + 16$$
Subtract 16 from both sides:
$$112 - 16 = 8d_1$$
$$96 = 8d_1$$
Divide by 8 to find $d_1$:
$$d_1 = \frac{96}{8}$$
$$d_1 = 12 \text{ cm}$$
Now we can find the initial area $A_1$:
$$A_1 = \frac{d_1^2}{2} = \frac{12^2}{2} = \frac{144}{2} = 72 \text{ cm}^2$$
Calculating the New Area
The new area $A_2$ is $A_1 + 56$:
$$A_2 = 72 + 56 = 128 \text{ cm}^2$$
Alternatively, the new diagonal $d_2$ is $d_1 + 4 = 12 + 4 = 16$ cm. The new area $A_2$ is:
$$A_2 = \frac{d_2^2}{2} = \frac{16^2}{2} = \frac{256}{2} = 128 \text{ cm}^2$$
Both methods give the same new area, 128 cm2.
Finding the Ratio of New Area to Initial Area
The problem asks for the ratio of the new area to the initial area, which is $A_2 : A_1$.
$$\text{Ratio} = \frac{A_2}{A_1} = \frac{128}{72}$$
To simplify the ratio, we find the greatest common divisor (GCD) of 128 and 72. Both numbers are divisible by 8.
$$\frac{128 \div 8}{72 \div 8} = \frac{16}{9}$$
So, the ratio of the new area to the initial area is 16:9.
Summary of Areas and Ratio
| Parameter | Value |
|---|---|
| Initial Diagonal ($d_1$) | 12 cm |
| Initial Area ($A_1$) | 72 cm2 |
| New Diagonal ($d_2$) | 16 cm |
| New Area ($A_2$) | 128 cm2 |
| Ratio $A_2 : A_1$ | 128 : 72 = 16 : 9 |
The final ratio of the new area of the square to the initial area of the square is 16:9.
Revision Table: Square Area and Diagonal
| Concept | Formula | Notes |
|---|---|---|
| Side ($s$) to Diagonal ($d$) | $d = s\sqrt{2}$ | Diagonal is $\sqrt{2}$ times the side |
| Diagonal ($d$) to Side ($s$) | $s = \frac{d}{\sqrt{2}}$ | Side is diagonal divided by $\sqrt{2}$ |
| Area ($A$) from Side ($s$) | $A = s^2$ | Standard area formula |
| Area ($A$) from Diagonal ($d$) | $A = \frac{d^2}{2}$ | Derived from $A = s^2$ and $s = \frac{d}{\sqrt{2}}$ |
Additional Information: Scaling of Area with Linear Dimensions
For any two similar figures, the ratio of their areas is the square of the ratio of their corresponding linear dimensions (like side, diagonal, perimeter, etc.).
In this problem, the initial square and the new square are similar figures (all squares are similar). The diagonal is a linear dimension. Let $d_1$ be the initial diagonal and $d_2$ be the new diagonal. The ratio of the diagonals is $\frac{d_2}{d_1}$.
The ratio of the areas should be the square of the ratio of the diagonals:
$$\frac{A_2}{A_1} = \left(\frac{d_2}{d_1}\right)^2$$
We found $d_1 = 12$ cm and $d_2 = 16$ cm. Let's check this relationship:
$$\frac{A_2}{A_1} = \frac{128}{72} = \frac{16}{9}$$
And the ratio of the diagonals squared is:
$$\left(\frac{d_2}{d_1}\right)^2 = \left(\frac{16}{12}\right)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$$
This confirms that our calculated ratio of areas is consistent with the scaling property of similar figures.
Important Questions from Quant Based Puzzle
A number is subtracted from 4 times of it and then, the number obtained is added to its (the resultant’s) next number. If this gives the answer as 91, what was the original number?
When twice of a number added to 3 is multiplied by 5 and added to the number itself, it gives 158. What is the square of that number?
In a class of 72 students, the number of boys is twice the number of girls. Find the number of boys.
Two years ago, T was twice as old as P. P is thrice as old as R. In five years, P will be 29. What is the present age of T?
When a number is added to its multiple of 5 and its square, the sum of these three numbers is 91. Find the number.
