The base of the straight prism ABCDA1B1C1D1 is the isosceles trapezoid ABCD, the bases of which are 11 and 21

| September 6, 2021 | education

The base of the straight prism ABCDA1B1C1D1 is the isosceles trapezoid ABCD, the bases of which are 11 and 21, the sin of the angle CAD = 0.6; perimeter of the diagonal section of the prism 76. Find the area of the lateral surface of the prism.

Expression for determining the area of ​​the lateral surface of a prism:
Sside = pосн * h.
To determine the perimeter of the base, consider a trapezoid ABCD, it is isosceles, its bases AD = 21 cm, BC = 11 cm.Let us draw a diagonal AC, it forms an angle <CAB with the base, sin CAB = 0.6. CD = AB, as the lateral sides of an isosceles trapezoid. Let’s draw the heights of the trapezoid СK, BН, and AН = KD, BC = НK, as a projection of a smaller base to a larger one. АН = КD = (АD-ВС) / 2 = 5 cm.
AK = AD-KD = 21-5 = 16 cm.
Consider a triangle AСK, it is rectangular <K = 90˚, AK = 16 cm, sin CAB = 0.6.
(sin CAB) ² = (0.6) ².
(sin CAB) ² = (0.36).
1- (sin CAB) ² = (cos CAB) ².
(cos CAB) ² = 1-0.36 = 0.64.
cos CAB = 0.8.
cos CAB = AK / AC.
AC = AK / cos CAB = 16 / 0.8 = 20 cm.
sin CAB = CK / AC.
CK = AC * sin CAB = 20 * 0.6 = 12 cm.
Consider a triangle CKD <K = 90˚ KD = 5 cm. CK = 12 cm.
By the Pythagorean theorem:
CD = √ (CK² + KD²) = √ (144 + 25) = √169 = 13 cm.
Let’s go back to the trapezoid, find its perimeter:
psc = AD + BC + 2 * CD = 21 + 11 + 2 * 13 = 58 cm.
The perimeter of the diagonal section is:
p sec = AC + A`C` + AA` + CC`.
AC = A`C` as the diagonals of the same trapezoids, AA` = CC` as the height and sides of the prism.
p sec = 2 * AC + 2CC` = 76.
CC` = (76-2АС) / 2 = (76-2 * 20) / 2 = 18 cm.
Side surface area:
S = root * h = 58 * 18 = 1044 cm².
Answer: the area of ​​the lateral surface of the prism is 1044 cm².



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