The area of the trapezoid is 30. The height is 5 One of the diagonals is 6. Find the second diagonal.

Let us denote the original trapezoid by ABCD, and its diagonals d1 = 6 and d2, which must be found. Let us draw a straight line parallel to the diagonal d2 = BD through the vertex C. Let us denote the “new” line – e. We have obtained an equal-sized triangle ACK, the area of ​​which is equal to the area of ​​the trapezoid. Note that DK = BC.
Since we have drawn a straight line through vertex C, which is on the same level with vertex B, the height of the new triangle ACK is equal to the height of the original trapezoid ABCD. Let us denote the base by the letter d and find it through the area ACK using the formula. We get that d = 12.
Let’s designate the angle between the diagonal d1 and the base of AH2 as “alpha”. Note that the ACH2 triangle is rectangular, therefore, we can find the sine of the “alpha” angle. It will be equal to the ratio of the opposite leg to the hypotenuse: 5/6. Let us express in terms of cosine, we get the root of (11) / 6. We apply the cosine theorem to express e (^ 2): we get 180 – 24 roots of 11, so e = root from (180 – 24 roots of 11) or approximately 10, 02. Since e = d2, therefore d2 = 10, 02.