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 determine the maximum area of a trapezium under specified conditions, let the trapezium be ABCD, with AB as the base. The sides BC, CD, and DA each measure 10 cm. For maximum area, the trapezium must be symmetric about the base AB, forming an isosceles trapezium.

Let the base AB be denoted by \(a\) and the height by \(h\).

In an isosceles trapezium, the non-parallel sides, along with the height, form two identical right-angled triangles.

Applying the Pythagorean theorem to one of these triangles yields:

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

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

Solving for \(h\):

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

The area of a trapezium is calculated as:

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

Substituting the derived expression for \(h\):

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

To find the maximum area, differentiate the area with respect to \(a\) and set the derivative to zero. This calculation results in \( a = 10\sqrt{3}\).

The maximum area is computed as:

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

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

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

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

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

= \(75 \sqrt 3 \ cm^2\)

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

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