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# If ED = √8 cm and the area of ∆ABC = trapezoid ACDE, the length of AC is?

Can you help to solve it?

### 4 Answers

- stanschimLv 712 months agoFavorite Answer
AC = 2. See below for explanation.

Notice that there are two similar triangles, ABC and EBD. The area of EBD is twice that of ABC, since the trapezoid area is equal to that of triangle ABC.

But, the square of corresponding sides of similar triangles are proportional to their areas giving:

Area ABC / Area EBD = (AC)^2 / (ED)^2 = (AC)^2 / 8 = 1/2

Therefore AC = 2 units.

- Anonymous12 months ago
Area of ABC is half of the triangle AED, and we can see that they're uniform. The area of a triangle is base times height by two. Therefor the base and the side need to both be divided by a factor of square root of two. Therefor dividing ED by square root of two, we get AC, which is two.

- az_lenderLv 712 months ago
So, the segment AC cuts the area of triangle BDE into two equal halves. Since ACDE is designated as a "trapezoid," we can assume that AC is parallel to DE.

Drop a perpendicular from B onto DE. Let K be the point where that perpendicular hits DE, and let J be the intersection of BK with AC.

Area of triangle ABC = (1/2)*(AC)*(BJ);

Area of triangle ADE = (1/2)*(DE)*(BK).

We are given that the first area is half the 2nd, so

(AC)*(BJ) = (1/2)*(DE)*(BK).

By theorems of similar triangles, we know that

AC/DE = BJ/BK, so

BJ/BK = (1/2)(BK)/(BJ) =>

BK = (BJ)*sqrt(2), and

DE = (AC)*sqrt(2);

therefore AC = sqrt(8)/sqrt(2) = 2 cm.