As shown in the figure, △ ABC is an equilateral triangle, which is cut by a rectangle parallel to BC, and ab is cut into three equal parts. Then the area of the shadow part in the figure is () A. 19B. 29C. 13D. 49
∵ AB is cut into three equal parts, ∵ AEH ∵ AFG ∵ ABC, ∵ aeaf = 12, aEAb = 13 ∵ s △ AFG: s △ ABC = 4:9s △ AEH: s △ ABC = 1:9 ∵ s △ AFG = 49S △ ABCs △ AEH = 19S △ ABC ∵ s shadow area = s △ afg-s △ AEH = 49S △ abc-19s △ ABC = 13s △ ABC, so select: C
In △ ABC, BC = 10 cm and the height ad on the edge of BC = 4 cm. A moving point P starts from B and moves along b-c-a
0 < s ≤ 20 B. 0 < s < 20 C. 0 < s < 10 d. 0 < s ≤ 4
b
20 at C
If △ ABC ≌ Δ DEF is known, BC = EF = 6cm, and the area of △ ABC is 18cm2, then the height of EF is ()
A. 6cmB. 7cmC. 8cmD. 9cm
Let the area of △ def be s, the height of edge EF be h, ∵ △ ABC ≌ △ def, BC = EF = 6cm, and the area of △ ABC is 18cm2. The areas of two triangles are equal, that is, s = 18 and S = 12 · EF · H = 18, ∥ H = 6, so a is selected
If △ ABC is all equal to △ def, BC = EF = 6, and the area of △ ABC is 8, then the height of EF is______
It should be very simple
8/3