Question

If the lengths of the three sides of a trapezium other than the base are 10 cm each, then the maximum area of the trapezium is:

• A. 100 cm²
• B. \(25√ 3 cm^2\)
• C. \(75√ 3 cm^2\)
• D. \(100√ 3 cm^2\)

Hint:

When calculating the area of an equilateral triangle, remember to use the formula \( A = \frac{\sqrt{3}}{4} \times a^2 \), where \( a \) is the length of the side. This formula arises from the geometry of equilateral triangles and can be used in various problems involving equilateral triangles, such as those found in trapezium configurations or other geometric shapes. Always make sure to apply this formula correctly and multiply by the number of triangles if needed.

Answer 1

The Correct Option is C

To find the maximum area of a trapezium with the given conditions, consider the following: Let the trapezium be ABCD with AB as the base and the other three sides BC, CD, and DA each measuring 10 cm. For maximum area, the trapezium should be symmetric around the base AB, resembling an isosceles trapezium.

Let  AB = a; and the height from the base h.

For an isosceles trapezium, the non-parallel sides would form two congruent right triangles with height \(h\).

Using the Pythagorean theorem in one of these triangles, we have:

\(h^2+(a/2)^2=10^2\)

\(h^2+(a^2/4)=100\)

.Solving for \(h\), we get:

\(h^2 = 100-a^2/4\)

Now, the area of the trapezium is given by:

\(Area = (1/2)*a*h\)

Substituting the expression for \(h\) we derived earlier:

\(Area=(1/2)*a*\sqrt{100-a^2/4}\)

Differentiate the area with respect to \(a\) and set the derivative to zero to find the maximum area. Differentiating and solving gives \( a = 10\sqrt{3}\).

For maximum area:

\(Area_{max} = (1/2)*10\sqrt{3}*\sqrt{100-(10\sqrt{3})^2/4} \)

\( = (1/2)*10\sqrt{3}*\sqrt{100-75} \)

\( = (1/2)*10\sqrt{3}*\sqrt{25} \)

\( = (1/2)*10\sqrt{3}*5 \)

\( = 25\sqrt{3}*3cm^2 \)

= \(75 \sqrt 3 \ cm^2\)

The calculated maximum area of the trapezium is: \(75\sqrt{3}\; cm^2\).

Thus, the correct answer is \(75√ 3\; cm^2.\)

Answer 2

Given that three sides of the trapezium are equal to 10 cm, we consider the configuration where these sides form two equilateral triangles at the ends:

Step 1: Area of one equilateral triangle with side 10 cm:

The area \( A \) of an equilateral triangle with side \( a \) is given by the formula: \[ A = \frac{\sqrt{3}}{4} \times a^2. \] Substituting \( a = 10 \) cm into this formula: \[ A = \frac{\sqrt{3}}{4} \times (10)^2 = \frac{\sqrt{3}}{4} \times 100 = 25\sqrt{3} \, \text{cm}^2. \] Therefore, the area of one equilateral triangle is \( 25\sqrt{3} \, \text{cm}^2 \).

Step 2: Total area for three such triangles:

Since there are three equilateral triangles in the configuration, the total area is: \[ \text{Total Area} = 3 \times 25\sqrt{3} = 75\sqrt{3} \, \text{cm}^2. \]

Conclusion: Thus, the total area of the three equilateral triangles is \( 75\sqrt{3} \, \text{cm}^2 \).