Area of the inner square?
A hard-level "Area" problem.A tilted square is "the boxing square minus the four corner triangles". Once you see the four are congruent, it is one calculation and a multiplication.
GivenA 10 cm square. Each side is divided 3 : 7, all going the same way round
Hints
- The inner square is tilted, so we cannot read off its side. Look instead at the four right triangles left around it.
- Take the top-left one. Its two legs are the pieces the side was cut into, 3 cm and 7 cm, so its area is 3 × 7 ÷ 2 = 10.5 cm².
- The other three are the same size, because turning 90° about the centre carries each one exactly onto the next. Dividing all four sides the same way round is what makes this work.
- So the four together are 10.5 × 4 = 42 cm². Work out one and multiply.
- The big square is 10 × 10 = 100 cm². Take the four triangles off and the middle is left: 100 − 42
Answer 58cm²
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?