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Question

In a rectangle, diagonal is 4 times of its breadth. The ratio of length to breadth is:

The correct answer is \(\sqrt{15}:1\)

Calculating the Ratio of Length to Breadth in a Rectangle

The question asks us to find the ratio of the length to the breadth of a rectangle given a relationship between its diagonal and breadth.

Let's define the dimensions of the rectangle:

  • Length = $l$
  • Breadth = $b$
  • Diagonal = $d$

In a rectangle, the length, breadth, and diagonal form a right-angled triangle. According to the Pythagorean theorem, the square of the diagonal is equal to the sum of the squares of the length and the breadth.

The Pythagorean theorem for a rectangle can be written as:

$$l^2 + b^2 = d^2$$

We are given a condition relating the diagonal and the breadth:

The diagonal is 4 times its breadth.

This can be written as:

$$d = 4b$$

Now, we can substitute the expression for $d$ from the given condition into the Pythagorean theorem equation:

$$l^2 + b^2 = (4b)^2$$

Let's simplify the right side of the equation:

$$l^2 + b^2 = 16b^2$$

We want to find the ratio of length to breadth, which is $\frac{l}{b}$. To do this, we need to isolate $l^2$ on one side of the equation:

$$l^2 = 16b^2 - b^2$$

$$l^2 = 15b^2$$

Now, take the square root of both sides of the equation to find $l$ in terms of $b$. Since length and breadth are physical dimensions, they must be positive values.

$$\sqrt{l^2} = \sqrt{15b^2}$$

$$l = \sqrt{15} \cdot \sqrt{b^2}$$

$$l = \sqrt{15}b$$

Finally, we can find the ratio of length to breadth, $\frac{l}{b}$:

$$\frac{l}{b} = \frac{\sqrt{15}b}{b}$$

Cancel out $b$ from the numerator and denominator (assuming $b \neq 0$, which must be true for a rectangle):

$$\frac{l}{b} = \sqrt{15}$$

This ratio $\frac{l}{b} = \sqrt{15}$ can be expressed as $l:b = \sqrt{15}:1$.

Comparing this result with the given options, we find that it matches option 1.

Dimension Representation
Length $l$
Breadth $b$
Diagonal $d$

Step Description Equation
1 Pythagorean Theorem $l^2 + b^2 = d^2$
2 Given condition relating diagonal and breadth $d = 4b$
3 Substitute (2) into (1) $l^2 + b^2 = (4b)^2$
4 Simplify $l^2 + b^2 = 16b^2$
5 Isolate $l^2$ $l^2 = 15b^2$
6 Take square root of both sides $l = \sqrt{15}b$
7 Find the ratio $l:b$ $l:b = \sqrt{15}:1$

Revision Table: Rectangle Dimensions and Ratios

Let's quickly summarize the key relationships used:

  • A rectangle's corners are all $90^\circ$.
  • The diagonal divides the rectangle into two right-angled triangles.
  • The Pythagorean theorem ($a^2 + b^2 = c^2$) is fundamental for right triangles.
  • For a rectangle, Length$^2$ + Breadth$^2$ = Diagonal$^2$.

Additional Information: Properties of Rectangles and Pythagorean Theorem

A rectangle is a quadrilateral with four right angles. Opposite sides are equal in length and parallel. The diagonals of a rectangle are equal in length and bisect each other.

The Pythagorean theorem is a cornerstone of geometry. It states that in a right-angled triangle, the area of the square on the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares of the other two sides (legs). In our case, the diagonal is the hypotenuse, and the length and breadth are the legs of the right triangle formed within the rectangle.

Understanding how to apply the Pythagorean theorem to geometric shapes like rectangles is crucial for solving problems involving lengths, breadths, and diagonals.

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Important Questions from Plane Figures

  1. If length of a rectangle is increased to its three times and breadth is decreased to its half, then the ratio of the area of given rectangle to the area of new rectangle is:

  2. The width of the path around a square field is 4.5 m and its area is 105.75 m 2. Find the cost of fencing the field at the rate of Rs. 100 per meter.

  3. What is the area of the square (in cm 2) whose vertices lie on a circle of radius 5 cm?

  4. The circumcentre of an equilateral triangle is at a distance of 3.2 cm from the base of the triangle. What is the length (in cm) of each of its altitudes?

  5. The perimeter of a circular lawn is 1232 m. There is 7 m wide path around the lawn. The area (in m 2) of the path is:

    Take \(\left(\pi=\frac{22}{7}\right)\)

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