Question

type the correct answer in the box.

the diagram shows a section of a bridge between the points a and k. the length of line segment \\(\overline{ak}\\) is 640 meters. \\(\triangle abc\\), \\(\triangle cdf\\), and \\(\triangle fjk\\) are similar, and \\(2ac = cf = 2fk\\). the first pillar, \\(\overline{bg}\\), is 20 meters tall.

the area of \\(\triangle cdf\\) is \\(\boxed{}\\) square meters.

Explanation:

Step1: Find the ratio of sides of similar triangles

Given \(2AC = CF = 2FK\), let \(AC = x\), then \(CF = 2x\) and \(FK=x\). So the ratio of \(AC:CF:FK = 1:2:1\). The total length \(AK=AC + CF+FK=x + 2x+x = 4x\). We know \(AK = 640\) meters, so \(4x=640\), solving for \(x\) gives \(x = 160\) meters. Thus, \(AC = 160\) meters, \(CF = 320\) meters, \(FK = 160\) meters.

Since \(\triangle ABC\), \(\triangle CDF\), and \(\triangle FJK\) are similar, the ratio of their corresponding sides is equal to the ratio of \(AC:CF:FK=1:2:1\). Let the height of \(\triangle ABC\) be \(h_1 = BG = 20\) meters. For similar triangles, the ratio of heights is equal to the ratio of corresponding sides. Let the height of \(\triangle CDF\) be \(h_2\). Then \(\frac{h_1}{h_2}=\frac{AC}{CF}=\frac{1}{2}\), so \(h_2 = 2\times h_1=2\times20 = 40\) meters.

Step2: Calculate the area of \(\triangle CDF\)

The base of \(\triangle CDF\) is \(CF = 320\) meters and the height is \(h_2 = 40\) meters. The area of a triangle is given by the formula \(A=\frac{1}{2}\times base\times height\). So the area of \(\triangle CDF\) is \(\frac{1}{2}\times320\times40\).

First, calculate \(\frac{1}{2}\times320 = 160\), then \(160\times40 = 6400\). Wait, no, wait. Wait, the base of \(\triangle ABC\) is \(AC = 160\)? Wait, no, the base of \(\triangle ABC\) is \(AC\)? Wait, no, in \(\triangle ABC\), the base is \(AC\)? Wait, no, the base of \(\triangle ABC\) is \(AC\) (the horizontal segment), and the height is \(BG\). For \(\triangle CDF\), the base is \(CF\) and the height is \(DH\) (the pillar height). Wait, maybe I made a mistake. Wait, the base of \(\triangle ABC\) is \(AC\) (length 160), height \(BG = 20\). The base of \(\triangle CDF\) is \(CF = 320\), and since the ratio of sides is \(1:2\), the height of \(\triangle CDF\) is \(2\times20 = 40\). Then area of \(\triangle CDF\) is \(\frac{1}{2}\times CF\times DH=\frac{1}{2}\times320\times40\)? Wait, no, wait, maybe the base of \(\triangle ABC\) is \(AG + GC\)? Wait, no, in \(\triangle ABC\), the base is \(AC\) (the length from \(A\) to \(C\)), and the height is \(BG\). Since \(\triangle ABC\) is isoceles (because \(BG\) is a pillar, so \(AG = GC\)), so the base of \(\triangle ABC\) is \(AC = 160\), so the base length (the horizontal side) is \(AC = 160\), height \(BG = 20\). Then for \(\triangle CDF\), the base (horizontal side) is \(CF = 320\), and the height (vertical side) is \(DH\). Since the triangles are similar, the ratio of sides is \(1:2\), so the height of \(\triangle CDF\) is \(2\times20 = 40\). Then the area of \(\triangle CDF\) is \(\frac{1}{2}\times320\times40\)? Wait, no, that gives 6400, but that seems wrong. Wait, maybe the base of \(\triangle ABC\) is \(AC\) (the length of the base, which is \(AG + GC\), and since \(BG\) is the pillar, \(AG = GC\), so \(AC = 2\times AG\). Wait, maybe I messed up the base. Let's re - evaluate.

Wait, the key is that for similar triangles, the ratio of areas is the square of the ratio of corresponding sides. The ratio of sides of \(\triangle ABC\) to \(\triangle CDF\) is \(1:2\) (since \(AC:CF = 1:2\)). The area of \(\triangle ABC\): the base of \(\triangle ABC\) is \(AC = 160\) meters (wait, no, \(AC\) is the length from \(A\) to \(C\), but the base of the triangle \(\triangle ABC\) is \(AC\) (the horizontal segment), and the height is \(BG = 20\) meters. So area of \(\triangle ABC\) is \(\frac{1}{2}\times AC\times BG=\frac{1}{2}\times160\times20 = 1600\) square meters.

Since the ratio of sides of \(\triangle ABC\) to \(\triangle CDF\) is \(1:2\), the ratio of their areas is \((\frac{1}{2})^2=\frac{1}{4}\)? Wait, no, wait, the ratio of sides is \(1:2\), so the ratio of areas is \((\frac{1}{2})^2\) if the ratio of corresponding sides is \(1:2\)? Wait, no, if the ratio of corresponding sides is \(a:b\), the ratio of areas is \(a^{2}:b^{2}\). Wait, no, I think I mixed up. Wait, if \(\triangle ABC\) and \(\triangle CDF\) are similar with a scale factor of \(k=\frac{AC}{CF}=\frac{1}{2}\), then the area of \(\triangle ABC\) to the area of \(\triangle CDF\) is \(k^{2}=\frac{1}{4}\)? No, wait, no. Wait, the scale factor from \(\triangle ABC\) to \(\triangle CDF\) is \(2\) (since \(CF = 2\times AC\)), so the ratio of areas is \(2^{2}=4\). So area of \(\triangle CDF\) is \(4\times\) area of \(\triangle ABC\).

Area of \(\triangle ABC\): base \(AC = 160\) meters, height \(BG = 20\) meters. So area \(A_{ABC}=\frac{1}{2}\times160\times20 = 1600\) square meters. Then area of \(\triangle CDF\) is \(4\times1600 = 6400\)? Wait, no, that can't be. Wait, maybe the base of \(\triangle ABC\) is \(AC\) (the length from \(A\) to \(C\)), but actually, in the diagram, \(\triangle ABC\) has base \(AC\) (the horizontal side) and height \(BG\). \(\triangle CDF\) has base \(CF\) and height \(DH\). Since the triangles are similar, \(\frac{AC}{CF}=\frac{BG}{DH}=\frac{1}{2}\), so \(DH = 2\times BG = 40\) meters, and \(CF = 2\times AC\). Let's recalculate the area of \(\triangle CDF\): base \(CF = 320\) meters, height \(DH = 40\) meters. So area \(A=\frac{1}{2}\times320\times40=6400\)? Wait, but let's check the length of \(AC\) again. \(AK = AC + CF+FK\), and \(2AC = CF = 2FK\), so let \(AC = x\), \(CF = 2x\), \(FK=x\), so \(AK=x + 2x+x=4x = 640\), so \(x = 160\), so \(AC = 160\), \(CF = 320\), \(FK = 160\). Then the base of \(\triangle ABC\) is \(AC = 160\), height \(BG = 20\). The base of \(\triangle CDF\) is \(CF = 320\), height \(DH\). Since the triangles are similar, \(\frac{BG}{DH}=\frac{AC}{CF}=\frac{1}{2}\), so \(DH = 40\). Then area of \(\triangle CDF\) is \(\frac{1}{2}\times320\times40 = 6400\). Wait, but maybe I made a mistake in the base. Wait, the base of \(\triangle ABC\) is actually \(AG + GC\), and since \(BG\) is the pillar, \(AG = GC\), so \(AC=AG + GC = 2AG\), so \(AG = 80\) meters. Then the base of \(\triangle ABC\) is \(AC = 160\) meters (which is \(2\times80\)), and height \(BG = 20\) meters. Then for \(\triangle CDF\), the base is \(CF = 320\) meters (which is \(2\times160\)), and height \(DH = 40\) meters. So area is \(\frac{1}{2}\times320\times40 = 6400\). Wait, but let's check with the ratio of areas. The ratio of sides of \(\triangle ABC\) to \(\triangle CDF\) is \(1:2\), so the ratio of areas is \(1:4\). Area of \(\triangle ABC\) is \(\frac{1}{2}\times160\times20 = 1600\), so area of \(\triangle CDF\) is \(4\times1600 = 6400\).

Wait, but maybe I messed up the base. Wait, the length of \(AC\) is \(160\) meters, but the base of the triangle \(\triangle ABC\) is \(AC\) (the horizontal segment), and the height is \(BG\). So the area calculation is correct.

Answer:

\(6400\)