Let KLMN be a trapezoid (see added picture). From the point M put down the trapezoid height MP, then quadrilateral KLMP is square and KP=MP=10.
A triangle MPN is right and <span>isosceles, because
</span>m∠N=45^{0}, m∠P=90^{0}, so m∠M=180^{0}-45^{0}-90^{0}=45^{0}.Then PN=MP=10.
The ttapezoid side KN consists of two parts KP and PN, each of them is equal to 10, then KN=20 units.
Area of KLMN is egual to
sq. units.
Because a rectangular pyramid's base is square, the cross section would be as well
4.5 / 2 = 2.25cm radius for the smaller circle
3.14 (pi) × 2.25² = 15.89625cm²
10.5 / 2 = 5.25cm radius for the larger circle
3.14 × 5.25² = 86.54625cm²
86.54625 - 15.89625 = <span>70.65cm</span>²
The answer is 70.7cm².
It is a similar triangle so the side length of the smaller triangle (x and x-2) are in the same ratio as the bigger triangle (2x+5 and 2x-1). As the ratio is the same, you can equate them and solve for x.