Question:A parallelogram has a base of 64 centimeters and a height of 35 centimeters. A diagonal is drawn, dividing the...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
A parallelogram has a base of \(64\) centimeters and a height of \(35\) centimeters. A diagonal is drawn, dividing the parallelogram into two congruent triangles. What is the area, in square centimeters, of one of the triangles?
1. TRANSLATE the problem information
- Given information:
- Parallelogram base = 64 cm
- Parallelogram height = 35 cm
- A diagonal divides the parallelogram into two congruent triangles
- Find: Area of one triangle
2. INFER the strategic approach
- Key insight: When a diagonal is drawn in a parallelogram, it creates two triangles that are congruent (exactly the same size and shape)
- This means each triangle has exactly half the area of the original parallelogram
- Strategy: Find the parallelogram's area first, then divide by 2
3. Calculate the parallelogram's area
- Area of parallelogram = \(\mathrm{base \times height}\)
\(\mathrm{= 64 \times 35}\)
\(\mathrm{= 2240}\) square centimeters
4. Find the area of one triangle
- Since each triangle has half the parallelogram's area:
- Area of one triangle = \(\mathrm{2240 \div 2}\)
\(\mathrm{= 1120}\) square centimeters
Answer: 1120
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Not recognizing that each triangle has half the parallelogram's area
Students might try to calculate the triangle's area directly using the parallelogram's base and height in the triangle formula: \(\mathrm{(1/2) \times 64 \times 35}\). While this actually gives the correct answer (1120), many students get confused about which measurements to use or whether this approach is valid. Some might incorrectly think they need additional information like the diagonal length, leading to confusion and guessing.
The Bottom Line:
The key insight is recognizing that a diagonal always divides any parallelogram into two congruent triangles. Once you understand this geometric relationship, the problem becomes a straightforward calculation of finding half the parallelogram's area.