Q.If length of three sides of a trapezium other than base are equal to $10$ cm, then find the area of the trapezium when it is maximum.

The key idea is to express the area of the isosceles trapezium as a function of $x$, then maximise it using differentiation.

Step 1: Express the height.

From the right triangle at each end, the height $h = \sqrt{10^2 - x^2} = \sqrt{100 - x^2}$.

Step 2: Write the area function.

The parallel sides are $AB = 10 + 2x$ (base) and $DC = 10$ (top).

Area $A = \frac{1}{2}(\text{sum of parallel sides}) \times h = \frac{1}{2}(10 + 2x + 10)\sqrt{100 - x^2} = (10 + x)\sqrt{100 - x^2}$.

Step 3: Differentiate and set to zero.

$A'(x) = \sqrt{100 - x^2} + (10 + x) \cdot \frac{-x}{\sqrt{100 - x^2}} = 0$

Multiply through by $\sqrt{100 - x^2}$:

$(100 - x^2) - x(10 + x) = 0 \implies 100 - x^2 - 10x - x^2 = 0 \implies 2x^2 + 10x - 100 = 0$ …

The problem gives us a trapezium where three sides — the top and both slant sides — are each 10 cm. Only the longer base is free to vary. The figure shows an isosceles trapezium, so the two slant sides are equal and the feet of the perpendiculars from the top corners split the base into three segments: two equal overhangs of length $x$ on each side, and a middle segment of 10 cm (same as the top). The height $h$ comes from the right triangle formed by the slant side, the overhang $x$, and the height itself.

The key insight: as $x$ increases, the base gets longer (good for area) but the height shrinks (bad for area). Somewhere in between, the product of base and height reaches a maximum. This is a classic optimisation problem — express area as a function of $x$, differentiate, set to zero, and verify it's a maximum.

Let's work through it.

1. Set up the geometry. Let the two equal overhangs be $AP = QB = x$ cm. The top $DC = 10$ cm, so the bottom base $AB = 10 + 2x$ cm. In right triangle $ADP$, the slant side $AD = 10$ cm and the horizontal leg $AP = x$ cm. By Pythagoras:

The height is defined only when $0 < x < 10$ (if $x = 0$, the trapezium becomes a rectangle; if $x = 10$, the height becomes zero and the figure collapses).

2. Write the area function. Area of a trapezium = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$. Here:

Simplify:

3. Differentiate to find critical points. Use the product rule. Let $u = 10 + x$ and $v = (100 - x^2)^{1/2}$. $u' = 1$, $v' = \frac{1}{2}(100 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{100 - x^2}}$. So:

Put over a common denominator:

4. Set $A'(x) = 0$. The denominator is positive for $0 < x < 10$, so we only need the numerator to be zero:

Divide by 2:

Solve:

$x = \frac{-5 \pm \sqrt{25 + 200}}{2} = \frac{-5 \pm 15}{2}.$ …

Method: Area-Optimization Problems Requiring an Auxiliary Right Triangle

For shapes built from a mix of fixed and variable side lengths, this method turns the geometry into a single-variable area function.

Steps

Step 1: Identify the varying quantity and draw the auxiliary right triangle

Where a slanted fixed-length side meets the base, dropping a perpendicular creates a right triangle whose legs are the varying "overhang" and the shape's height — connected by Pythagoras since the slant length is fixed.

Step 2: Express the height in terms of the variable

Step 3: Write the area formula and substitute

Use the standard area formula for the shape (for a trapezium, $\frac12(\text{sum of parallel sides})\times\text{height}$), writing the parallel sides in terms of $x$ too, then substitute the height expression from Step 2.

Step 4: Differentiate (product rule) and solve …

Common Mistakes

Mistake 1: Forgetting the top side is also part of the sum of parallel sides

Why it's wrong: since the top $DC=10$ cm is one of the fixed 10 cm sides too, the base works out to $AB = 10+2x$, so the sum of the parallel sides is $10 + (10+2x)$, not just $2x$ or $10+2x$ alone. Missing this changes the area formula incorrectly. Correct approach: always re-derive which sides are parallel from the figure and add both actual lengths, even when one of them coincides with a given fixed length.

Mistake 2: Skipping the check that the critical value actually gives a maximum …