In a straight triangular prism ABCA1B1C1, angle ACB = 90 degrees, AC = BC = a. Straight line B1C makes

| March 9, 2021 | education

In a straight triangular prism ABCA1B1C1, angle ACB = 90 degrees, AC = BC = a. Straight line B1C makes an angle of 30 degrees with the face plane АА1В1В. Find the area of the lateral surface of the prism.

Since the base is a right-angled and isosceles triangle, its hypotenuse AB is equal to: AB ^ 2 = 2 * a ^ 2.

AB = a * √2 cm.

Let’s draw the height of CH, which is in an isosceles triangle and its median, then AH = BH = a * √2 / 2 cm.

Then CH ^ 2 = a ^ 2 – a ^ 2 * / 2.

CH = a * √2 / 2 cm.

In a right-angled triangle CB1H, the angle CB1H, by condition, is equal to 30, then the leg CH lies opposite it, and the hypotenuse CB1 = 2 * CH = a * √2.

From a right-angled triangle CBB1, BB1 ^ 2 = CB1 ^ 2 – CB ^ 2 = (a * √2) ^ 2 – a ^ 2 = a ^ 2.

BB1 = a cm.

The perimeter of the base of the prism is: Rosn = AB + BC + AC = a * √2 + a + a = a * (2 + √2) cm.

Then Sbok = Rosn * BB1 = a * (2 + √2) * a = a ^ 2 * (2 + √2) cm2.

Answer: The lateral surface area is a2 * (2 + √2) cm2.



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