AMC10 2005 A

AMC10 2005 A · Q25

AMC10 2005 A · Q25. It mainly tests Functions basics, Area & perimeter.

In $\triangle ABC$ we have AB = 25, BC = 39, and AC = 42. Points D and E are on AB and AC respectively, with AD = 19 and AE = 14. What is the ratio of the area of triangle ADE to the area of the quadrilateral BCED?

在$ riangle ABC$中，$AB = 25$，$BC = 39$，$AC = 42$。点$D,E$分别在$AB,AC$上，$AD = 19$，$AE = 14$。$ riangle ADE$的面积与四边形$BCED$的面积之比是多少？

(A) $\frac{266}{1521}$ $\frac{266}{1521}$

(B) $\frac{19}{75}$ $\frac{19}{75}$

(D) $\frac{19}{56}$ $\frac{19}{56}$

(E) 1 1

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) We have $\dfrac{\mathrm{Area}(ADE)}{\mathrm{Area}(ABE)}=\dfrac{AD}{AB}=\dfrac{19}{25}$ and $\dfrac{\mathrm{Area}(ABE)}{\mathrm{Area}(ABC)}=\dfrac{AE}{AC}=\dfrac{14}{42}=\dfrac{1}{3},$ so $\dfrac{\mathrm{Area}(ABC)}{\mathrm{Area}(ADE)}=\dfrac{25}{19}\cdot\dfrac{3}{1}=\dfrac{75}{19},$ and $\dfrac{\mathrm{Area}(BCED)}{\mathrm{Area}(ADE)}=\dfrac{\mathrm{Area}(ABC)-\mathrm{Area}(ADE)}{\mathrm{Area}(ADE)}=\dfrac{75}{19}-1=\dfrac{56}{19}.$ Thus $\dfrac{\mathrm{Area}(ADE)}{\mathrm{Area}(BCED)}=\dfrac{19}{56}.$

（D）我们有 $\dfrac{\mathrm{Area}(ADE)}{\mathrm{Area}(ABE)}=\dfrac{AD}{AB}=\dfrac{19}{25}$，并且$\dfrac{\mathrm{Area}(ABE)}{\mathrm{Area}(ABC)}=\dfrac{AE}{AC}=\dfrac{14}{42}=\dfrac{1}{3},$ 所以 $\dfrac{\mathrm{Area}(ABC)}{\mathrm{Area}(ADE)}=\dfrac{25}{19}\cdot\dfrac{3}{1}=\dfrac{75}{19},$ 并且 $\dfrac{\mathrm{Area}(BCED)}{\mathrm{Area}(ADE)}=\dfrac{\mathrm{Area}(ABC)-\mathrm{Area}(ADE)}{\mathrm{Area}(ADE)}=\dfrac{75}{19}-1=\dfrac{56}{19}.$ 因此 $\dfrac{\mathrm{Area}(ADE)}{\mathrm{Area}(BCED)}=\dfrac{19}{56}.$

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