The diagonal of the rectangular parallelepiped is larger than the sides of the base by 3 and 2 cm.

| March 1, 2021 | education

The diagonal of the rectangular parallelepiped is larger than the sides of the base by 3 and 2 cm. Find the total surface area if the height of the parallelepiped is 2√2 cm.

Let AD1 be equal to X cm, then, by hypothesis, BD = (X – 3) cm, AB = (X – 2) cm.

Consider two right-angled triangles: AD1D and ABD, in which AD is a leg in one triangle and a hypotenuse in the other. Let us express AD in both triangles according to the Pythagorean theorem.

AD ^ 2 = AD1 ^ 2 – DD1 ^ 2 = X ^ 2 – (2 * √2) ^ 2 = X ^ 2 – 8.

AD ^ 2 = AB ^ 2 + BD ^ 2 = (X – 2) ^ 2 + (X – 3) = X ^ 2 – 4 * X + 4 + X ^ 2 – 6 * X + 9 = 2 * X ^ 2 – 10 * X + 13.

Let’s equate the right-hand sides of the equations.

X ^ 2 – 8 = 2 * X ^ 2 – 10 * X + 13.

X ^ 2 – 10 * X + 21 = 0.

Let’s solve the quadratic equation.

D = b2 – 4 * a * c = (-10) ^ 2 – 4 * 1 * 21 = 100 – 84 = 16.

X1 = (10 – √16) / (2 * 1) = (10 – 4) / 2 = 6/2 = 3. (Not suitable, because then BD = (X – 3) = 0).

X2 = (10 + √16) / (2 * 1) = (10 + 4) / 2 = 14/2 = 7.

Then, BD = 7 – 3 = 4 cm, AB = 7 – 2 = 5 cm.

Determine the base area.

Sbn = AB * BD = 5 * 4 = 20 cm2.

Let us determine the area of ​​the lateral surface.

Sside = P * h, where P is the perimeter of the base, h is the height.

Side = 2 * (AB + BD) * DD1 = 2 * 9 * 2 * √2 = 36 * √2.

Then S floor = 2 * S main + S side = 2 * 40 + 36 * √2 = 80 + 36 * √2.

Answer: Floor = 80 + 36 * √2 cm2.



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