The diagonal of the rectangular parallelepiped is 10 cm and makes an angle of 60 degrees with the base plane

The diagonal of the rectangular parallelepiped is 10 cm and makes an angle of 60 degrees with the base plane, the base area is 12 cm2, find the side surface of the parallelepiped.

Consider a right-angled triangle ADD1, in which the angle D is straight, and the angle A = 60, then the angle D1 = 180 – 90 – 60 = 30.

The blood pressure leg lies opposite an angle of 30, which means it is equal to half of the AD1 hypotenuse. AD = AD1 / 2 = 10/2 = 5 cm.

From the same triangle, we define the leg DD1, which is the height of the parallelepiped.

SinA = DD1 / AD1.

DD1 = AD1 * SinA = 10 * √3 / 2 = 5 * √3.

By condition, the base area is 12 cm, AB * BD = 12 cm, then AB = 12 / BD.

Let the length BD = X cm, then AB = 12 / X.

Consider a right-angled triangle ABD, according to the Pythagorean theorem AD ^ 2 = AB ^ 2 + BD ^ 2.

5 ^ 2 = (12 / X) ^ 2 + X ^ 2.

25 = (144 + X4) / X ^ 2.

X4 – 25 * X ^ 2 + 144 = 0.

Let X ^ 2 = Y, then:

Y2 – 25 * Y + 144 = 0.

Let’s solve the quadratic equation.

D = b ^ 2 – 4 * a * c = (-25) ^ 2 – 4 * 1 * 144 = 625 – 576 = 49.

Y1 = (25 – √49) / (2 * 1) = (25 – 7) / 2 = 18/2 = 9.

Y2 = (25 + √49) / (2 * 1) = (25 + 7) / 2 = 32/2 = 16.

Then X1 = √9 = 3, X2 = √16 = 4.

The sides of the base are 3 and 4 cm.

Determine the perimeter of the base P = 2 * (3 + 4) = 14 cm.

Then the lateral surface area is equal to:

Side = R * DD1 = 14 * 5 * √3 = 70 * √3 cm2.

Answer: Side = 70 * √3 cm2.