Area of quadrilateral ABCD?
A hard-level "Area" problem.Shearing is the tool for "change the shape, keep the area". Quadrilateral to triangle, pentagon to quadrilateral — you can shed one corner at a time.
GivenBC = 10 cm, CE = 6 cm, A is 9 cm high. D lies on the line through C parallel to AE
Hints
- There is no formula for this quadrilateral — it is neither a trapezoid nor a parallelogram. So rebuild it as a triangle of the same area.
- First draw the diagonal AC, splitting the quadrilateral into two triangles. Leave triangle ABC as it is and think about moving only triangle ACD.
- Take AC as the base of triangle ACD; its apex is D. So draw the line through D parallel to AC and let it meet the extension of BC at E.
- On a parallel line you can slide the apex anywhere: base AC and the height both stay put. So triangle ACD = triangle ACE — moving D to E changes the area by nothing at all.
- Clear the working and look at what is left: quadrilateral ABCD has turned into a single triangle ABE with base BE and height 9 cm. The base is 10 + 6 = 16 cm.
- Now just the triangle formula: 16 × 9 ÷ 2
Answer 72cm²
More "Area" problems
- Area of the inner triangle?
- Area of triangle DOA?
- Area of the overlap?
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- Area swept by the square?
- What area of the table is covered by paper?