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ABC and AED are triangles with BC parallel to ED. Find the area of BCDE if the area of ABC is 16. 
Triangle_Barrons.png [ 45.21 KiB | Viewed 8136 times ]
(1) BC = 8
(2) ED = 5
Attachment:
Triangle_Barrons.png [ 45.21 KiB | Viewed 8136 times ]
(1) BC = 8
(2) ED = 5
11 Oct 2013, 03:56
Kudos
Bookmarks
ABC and AED are triangles with BC parallel to ED. Find the area of BCDE if the area of ABC is 16.
(1) BC = 8. We don't know where is ED positioned. Not sufficient.
(2) ED = 5. We don't know the other sides. Not sufficient.
(1)+(2) Notice that angles of triangles ABC and AED are equal, which means that the triangles are similar.
Property:
If two similar triangles have sides in the ratio \(\frac{x}{y}\), then their areas are in the ratio \(\frac{x^2}{y^2}\).
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{SIDE^2}{side^2}\).
Since BC/ED=8/5, then \(\frac{AREA_{ABC}}{area_{AED}}=\frac{8^2}{5^2}=\frac{64}{25}\). As given that the area of ABS is 16, then we can find the area of AED and then find the area of BCDE. Sufficient.
Answer: C.
Hope it's clear.
(1) BC = 8. We don't know where is ED positioned. Not sufficient.
(2) ED = 5. We don't know the other sides. Not sufficient.
(1)+(2) Notice that angles of triangles ABC and AED are equal, which means that the triangles are similar.
Property:
If two similar triangles have sides in the ratio \(\frac{x}{y}\), then their areas are in the ratio \(\frac{x^2}{y^2}\).
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{SIDE^2}{side^2}\).
Since BC/ED=8/5, then \(\frac{AREA_{ABC}}{area_{AED}}=\frac{8^2}{5^2}=\frac{64}{25}\). As given that the area of ABS is 16, then we can find the area of AED and then find the area of BCDE. Sufficient.
Answer: C.
Hope it's clear.
14 Oct 2013, 04:41
Kudos
Bookmarks
Bunuel
ABC and AED are triangles with BC parallel to ED. Find the area of BCDE if the area of ABC is 16.
(1) BC = 8. We don't know where is ED positioned. Not sufficient.
(2) ED = 5. We don't know the other sides. Not sufficient.
(1)+(2) Notice that angles of triangles ABC and AED are equal, which means that the triangles are similar.
Property:
If two similar triangles have sides in the ratio \(\frac{x}{y}\), then their areas are in the ratio \(\frac{x^2}{y^2}\).
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{SIDE^2}{side^2}\).
Since BC/ED=8/5, then \(\frac{AREA_{ABC}}{area_{AED}}=\frac{8^2}{5^2}=\frac{64}{25}\). As given that the area of ABS is 16, then we can find the area of AED and then find the area of BCDE. Sufficient.
Answer: C.
Hope it's clear.
(1) BC = 8. We don't know where is ED positioned. Not sufficient.
(2) ED = 5. We don't know the other sides. Not sufficient.
(1)+(2) Notice that angles of triangles ABC and AED are equal, which means that the triangles are similar.
Property:
If two similar triangles have sides in the ratio \(\frac{x}{y}\), then their areas are in the ratio \(\frac{x^2}{y^2}\).
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{SIDE^2}{side^2}\).
Since BC/ED=8/5, then \(\frac{AREA_{ABC}}{area_{AED}}=\frac{8^2}{5^2}=\frac{64}{25}\). As given that the area of ABS is 16, then we can find the area of AED and then find the area of BCDE. Sufficient.
Answer: C.
Hope it's clear.
Wonderfully done - one question please: "the area of ABC is 16" (in the stimulus) - is there anything we can infer from this information? Or does this information help us in any way before looking at S1 and S2?
