Area of the inner quadrilateral?
A hard-level "Area" problem.However lopsided the original quadrilateral, joining the midpoints always gives a parallelogram, and its area is always half. A theorem that makes you want to drag the corners around and test it.
GivenQuadrilateral ABCD has area 48 cm². The midpoints of its four sides are joined
Hints
- A quadrilateral on its own gives you nothing to hold on to. Draw one diagonal, AC, and think in triangles.
- Look at triangle BPQ. P is the midpoint of BA and Q of BC, so triangle BPQ has the same shape as triangle BAC with every side halved.
- Halve every side and the area becomes "half times half" = a quarter. So triangle BPQ is a quarter of triangle BAC.
- Exactly the same holds for triangle DRS on the other side: a quarter of triangle DCA.
- Triangles BAC and DCA together make the whole quadrilateral, so the two blue corners together are a quarter of the whole: 48 ÷ 4 = 12 cm².
- Redraw and do the same with the other diagonal, BD. The two remaining corners (orange) come to a quarter of the whole for exactly the same reason.
- The four corners come to 12 + 12 = 24 cm², exactly half the whole — so the middle is the other half. 48 − 24
Answer 24cm²
More "Area" problems
- Area of the inner triangle?
- Area of triangle DOA?
- Area of quadrilateral ABCD?
- Area of the overlap?
- Total area of the two crescents?
- Area swept by the square?