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

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