Question

The parallel sides of a trapezium shaped field are \(25\) m and \(10\) m and non-parallel sides are \(14\) m and \(13\) m. What is the area (in \(m^2\)) of the field?

• A. \(204\)
• B. \(196\)
• C. \(156\)
• D. \(144\)

Hint:

For trapezium area with non-parallel sides known, use right triangles from dropping perpendiculars to find the height first, then apply the area formula.

Answer 1

The Correct Option is B

Step 1: Let the trapezium be \(ABCD\).
\(AB\) and \(CD\) are parallel sides, \(AB = 25\) m, \(CD = 10\) m, \(AD = 14\) m, \(BC = 13\) m.
Step 2: Find the difference of parallel sides and divide into segments.
Difference \(AB - CD = 15\) m. Drop perpendiculars from \(C\) and \(D\) to \(AB\), meeting at points \(E\) and \(F\) respectively, with \(CE \perp AB\) and \(DF \perp AB\).
Thus: \[ AE = \text{part on left side with } AD = \sqrt{14^2 - h^2}, \] \[ BF = \text{part on right side with } BC = \sqrt{13^2 - h^2}. \] Also: \[ AE + CD + BF = AB \quad \Rightarrow \quad \sqrt{14^2 - h^2} + 10 + \sqrt{13^2 - h^2} = 25. \] Simplify: \[ \sqrt{14^2 - h^2} + \sqrt{13^2 - h^2} = 15. \] Step 3: Solve for \(h\).
Let \(x = \sqrt{14^2 - h^2}\) and \(y = \sqrt{13^2 - h^2}\). We have: \[ x + y = 15 \quad \text{and} \quad x^2 - y^2 = 14^2 - 13^2 = 27. \] From \(x^2 - y^2 = (x - y)(x + y)\), \[ (x - y) \cdot 15 = 27 \quad \Rightarrow \quad x - y = \frac{27}{15} = 1.8. \] Solving: \[ x = \frac{15 + 1.8}{2} = 8.4, \quad y = 15 - 8.4 = 6.6. \] Then: \[ h = \sqrt{14^2 - 8.4^2} = \sqrt{196 - 70.56} = \sqrt{125.44} = 11.2 \ \text{m}. \] Step 4: Area of trapezium.
Area = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times h\) \[ = \frac{1}{2} \times (25 + 10) \times 11.2 = 17.5 \times 11.2 = 196 \ m^2. \] \[ \boxed{196 \ m^2} \]