Area of the shaded part?
A hard-level "Area" problem.Spot a pair of parallel lines and try sliding a vertex. The unwanted number drops out and the answer depends on just one square.
GivenA square of side 6 cm and one of side 10 cm stand side by side. A line joins the bottom-left corner A to the top-right corner D.
Hints
- The direct route: the shaded triangle has base 6 + 10 = 16 cm and height 10 cm. That drags 6 into the answer. There is a better way to look at it.
- The base AC and the top edge of the big square are parallel — both are horizontal. That is the way in.
- Parallel lines allow a shear: sliding the apex A along the parallel line does not change the area. So move the apex from A across to B.
- Look at the triangle now: it is exactly half of the big square, cut by a diagonal. Base and height are both 10 cm.
- So 10 × 10 ÷ 2 = 50 cm². The small square's size never mattered. Change 6 cm to anything and the answer stays the same.
Answer 50cm²
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?