As shown in the figure, in △ ABC, ∠ a = 150 °, ab = 20cm, AC = 30cm, then the area of △ ABC is () A. 330cm2 B. 450cm2 C. 150cm2 D. 300cm2

Pass point B as BD ⊥ AC,
∵∠A=150°，
∴∠BAD=30°，
∴BD=1
2AB，
∵AB=20cm，
∴BD=10cm，
∵S△ABC=1
2AC•BD=1
2×30×10=150cm2，
Therefore, C

Given that the area of the triangle ABC is 12 square centimeters, calculate the area of the shadow part?

Let the right sides of the isosceles right triangle ABC be a, 12a2 = 12, A2 = 24; the area of sector abd: 18 π A2 = 3 π = 9.42 (square centimeter), the area of blank part BCD: 12-9.42 = 2.58 (square centimeter), the area of semicircle: 12 π (a △ 2) 2 = 12 × 3.14 × 14a2 = 18 × 3.14 × 24 = 9.42 (square)

As shown in the figure, triangle ABC is a right triangle. The area of shadow ① is 23c m2smaller than that of shadow ②. What is the length of BC?

Let BC be x cm in length, because △ ABC area semicircle area = 23 square centimeter, so 12 × 20x-12 × 3 × 102 = 23 10x-150 = 23 & nb

As shown in the figure, in the right triangle ABC, the angle c = 90 degrees, AC = 3 cm, BC = 4 cm, and the radius of the circle O is 4 / 3 cm. Calculate the area of the shadow part The first volume of sixth grade mathematics matching exercises 43 pages

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Known triangle ABC is a right triangle, AC = 4cm, BC = 2cm

3.14 × (4 △ 2) 2 △ 2 + 3.14 × (2 △ 2) 2 △ 2-a-b-c-b = 4 × 2 △ 2-b,
14 × 2 + 3.14 △ 2-a-b-c = 4,
A + B + C = 3.14 × 2 + 3.14 △ 2-4,
=6.28+1.57-4
=7.85-4，
=85 (square centimeter)
A: the area of 85.3 cm

Known triangle ABC is a right triangle, AC = 4cm, BC = 2cm

As shown in the figure, the triangle ABC is an isosceles right triangle, D is the midpoint of the circle, BC is the diameter of the semicircle, ab = BC = 10 cm, calculate the area of the shadow part

Connecting BD, OD and OA, because do ⊥ BC, ab ⊥ BC, so do ∥ AB, then s ⊥ AOD = s ⊥ BOD, while the shadow area = s △ AOB + s sector bod-s △ AOD, = s △ AOB + s sector bod-s △ BOD, = 12 × 10 × 10 ﹥ 2 + 14 × π × (102) 2-12 × 102 × 102, = 25 + 19.625-12.5, = 32.125

As shown in the figure, the three sides of RT △ ABC are known to be 6, 8 and 10 respectively. Take its three sides as the diameter, make three semicircles upward, and calculate the area of the shadow part in the graph

The area of the semicircle with the diameter of AC is π× (6 ￣ 2) 2 × 12 = 92 π = 4.5 π, the area of semicircle with BC diameter is π× (8 ￣ 2) 2 × 12 = 8 π, the area of semicircle with ab diameter is π× (10 ￣ 2) 2 × 12 = 252, π = 12.5 π, the area of triangle ABC is 6 × 8 × 12 = 24, the surface of shadow part is

As shown in the figure, in △ ABC, ∠ a = 150 °, ab = 20cm, AC = 30cm, then the area of △ ABC is () A. 330cm2 B. 450cm2 C. 150cm2 D. 300cm2