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Q. 5) Solve any one sub-question from the following:
1) The area of rhombus is 120cm2. If one of its diagonals is
24cm long. What is the length of the other diagonal and side
of rhombus.



Answer :

Let's solve the problem step-by-step.

### Step 1: Understanding the Properties of a Rhombus

A rhombus is a type of polygon that has the following characteristics:
1. All sides have the same length.
2. The diagonals bisect each other at right angles (90 degrees).

### Step 2: Given Information

- Area of the rhombus ([tex]\(A\)[/tex]) = 120 cm²
- Length of one diagonal ([tex]\(d_1\)[/tex]) = 24 cm

### Step 3: Finding the Length of the Other Diagonal ([tex]\(d_2\)[/tex])

The formula for the area of a rhombus in terms of its diagonals is:
[tex]\[ \text{Area} = \frac{1}{2} \times d_1 \times d_2 \][/tex]

We need to find [tex]\(d_2\)[/tex]. Rearranging the formula to solve for [tex]\(d_2\)[/tex]:
[tex]\[ d_2 = \frac{2 \times \text{Area}}{d_1} \][/tex]

Substitute the given values:
[tex]\[ d_2 = \frac{2 \times 120}{24} \][/tex]
[tex]\[ d_2 = \frac{240}{24} \][/tex]
[tex]\[ d_2 = 10 \, \text{cm} \][/tex]

So, the length of the other diagonal is 10 cm.

### Step 4: Finding the Length of the Side of the Rhombus

Each diagonal splits the rhombus into two right-angled triangles. In one of these triangles:
- The halves of the diagonals are the legs of the right triangle.
- The side of the rhombus is the hypotenuse.

Let's denote the side of the rhombus as [tex]\(s\)[/tex].

Using the Pythagorean theorem in one of these right triangles:
[tex]\[ s = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2} \][/tex]

Substitute the given values:
[tex]\[ s = \sqrt{\left(\frac{24}{2}\right)^2 + \left(\frac{10}{2}\right)^2} \][/tex]
[tex]\[ s = \sqrt{12^2 + 5^2} \][/tex]
[tex]\[ s = \sqrt{144 + 25} \][/tex]
[tex]\[ s = \sqrt{169} \][/tex]
[tex]\[ s = 13 \, \text{cm} \][/tex]

### Final Answer

- The length of the other diagonal ([tex]\(d_2\)[/tex]) is 10 cm.
- The length of a side of the rhombus ([tex]\(s\)[/tex]) is 13 cm.