Hello Basal Originally Posted by
Basal There is an isosceles trapezium with both the oblique sides equal to 8 cm, and the parallel sides being 4, and 9 cm respectively.
It is required to find the diagonals. Now, my answer is 10 cm. Can anyone please verify this ?
Not quite. If the height of the trapezium is $\displaystyle h$ cm, then
$\displaystyle h^2 = 8^2+2.5^2\;*$
and if the diagonal is of length $\displaystyle d$ cm, then
$\displaystyle d^2 = h^2 + 6.5^2$$\displaystyle =8^2+2.5^2 + 6.5^2$
$\displaystyle \Rightarrow d = \sqrt{112.5}$$\displaystyle =10.61$ cm (2 d.p.)
Grandad
* See correction, below!