At the base of the rectangular parallelepiped there is a square, and the height is 1 m greater than the side of the base.

| August 10, 2021 | education

At the base of the rectangular parallelepiped there is a square, and the height is 1 m greater than the side of the base. Find the surface area of a rectangular parallelepiped if its diagonal is 11 m.

Let the side of the square at the base of the prism be X cm, then the side edge of the prism will be (X + 1) cm.

Consider a right-angled triangle ABD. By the Pythagorean theorem, the hypotenuse BD will be equal to: BD ^ 2 = X ^ 2 + X ^ 2 = 2 * X ^ 2.

Consider a right-angled triangle B1BD, in which the hypotenuse B1D, according to the Pythagorean theorem will be equal to: B1D ^ 2 = BD ^ 2 + B1B ^ 2.

112 = 2 * X ^ 2 + (X + 1) ^ 2.

121 = 2 * X ^ 2 + X ^ 2 + 2 * X + 1.

3 * X ^ 2 + 2 * X – 120 = 0.

Let’s solve the quadratic equation.

D = b ^ 2 – 4 * a * c = 2 ^ 2 – 4 * 3 * (-120) = 4 + 1440 = 1444.

x1 = (-2 – √1444) / (2 * 3) = (-2 – 38) / 6 = -40 / 6 = -20/3. (Doesn’t fit because <0).

X2 = (-2 + √1444) / (2 * 3) = (-2 + 38) / 6 = 36/6 = 6.

AB = BC = CD = AC = 6 cm.

AA1 = BB1 = CC1 + DD1 = 6 + 1 = 7 cm.

Determine the surface area of ​​the parallelepiped.

S = 2 * S main + S side.

S main = AB * BC = 6 * 6 = 36 cm2.

Side = 4 * AB * AA1 = 4 * 6 * 7 = 168 cm2.

S = 2 * 36 + 168 = 240 cm2.

Answer: The surface area is 240 cm2.



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