Lightoj 1178 solution in English

Lightoj 1178 : Problem Link

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In this problem, we are given the lengths of the four sides of a trapezium. We need to calculate its area.

We know the formula for the area of a trapezium is:

$$\text{Area} = \tfrac{1}{2} \times (\text{sum of parallel sides}) \times (\text{distance between them})$$

That is:

$$\text{Area} = \tfrac{1}{2} \times (a + c) \times h$$

Here:

• $a$ and $c$ are the lengths of the parallel sides.

• $h$ is the height (perpendicular distance between $a$ and $c$).

We are given $a$ and $c$, but not $h$. So, we need to calculate $h$.

Step 1: Drop perpendiculars

• From point $C$, drop a perpendicular $CP$ onto $AB$.

• From point $D$, drop a perpendicular $DQ$ onto $AB$.

Thus:

$$CP = DQ = h$$

Also:

$$PQ = c, \quad AP + BQ = a - c$$

Let:

$$AP = x, \quad BQ = a - c - x$$

Step 2: Apply Pythagoras

Now, consider right triangles APC and BDQ.

In triangle APC:

$$AC^2 = AP^2 + h^2$$ $$d^2 = x^2 + h^2 \quad \Rightarrow \quad h^2 = d^2 - x^2 \quad ...(1)$$

In triangle BDQ:

$$BD^2 = BQ^2 + h^2$$ $$b^2 = (a - c - x)^2 + h^2 \quad \Rightarrow \quad h^2 = b^2 - (a - c - x)^2 \quad ...(2)$$

Step 3: Eliminate $h$

From equations (1) and (2):

$$d^2 - x^2 = b^2 - (a - c - x)^2$$

Expanding:

$$d^2 - x^2 = b^2 - (a^2 + c^2 + x^2 - 2ac - 2ax + 2cx)$$ $$d^2 - x^2 = b^2 - a^2 - c^2 - x^2 + 2ac + 2ax - 2cx$$ $$d^2 = b^2 - a^2 - c^2 + 2ac + 2ax - 2cx$$ $$2ax - 2cx = d^2 - b^2 + a^2 + c^2 - 2ac$$ $$x = \frac{d^2 - b^2 + a^2 + c^2 - 2ac}{2a - 2c}$$

So we have found x.

Step 4: Solve for $h$

Now substitute $x$ into equation (1):

$$h^2 = d^2 - x^2$$ $$h = \sqrt{d^2 - \left(\frac{d^2 - b^2 + a^2 + c^2 - 2ac}{2a - 2c}\right)^2}$$

Step 5: Area of the trapezium

Finally, substitute $h$ back into the trapezium formula:

$$\text{Area} = \tfrac{1}{2} \times (a + c) \times h$$

Now we have the formula to calculate the trapezium’s area using the four sides. 

Source Code : SourceCodeLink