Question

to indirectly measure the distance across a river, madeline stands on one side of the river and uses sight-lines to a landmark on the opposite bank. madeline draws the diagram below to show the lengths and angles that she measured. find pr, the distance across the river. round your answer to the nearest foot. (diagram is not to scale.)

Explanation:

Step1: Identify Similar Triangles

Triangles \( \triangle PRO \) and \( \triangle PRE \)? Wait, no, \( \triangle PRE \) and \( \triangle POC \)? Wait, actually, \( \triangle PRE \) and \( \triangle POC \) are similar? Wait, \( \angle PRE = \angle POC = 90^\circ \), and they share \( \angle P \), so by AA similarity, \( \triangle PRE \sim \triangle POC \). So the ratios of corresponding sides are equal. So \( \frac{PR}{PO} = \frac{RE}{OC} \).

Wait, \( PO = PR + RO \), but \( RO = 165 \) ft? Wait, no, \( RO \) is 165 ft? Wait, the diagram: \( R \) and \( O \) are right angles, \( RE = 210 \) ft, \( OC = 330 \) ft, \( RO = 165 \) ft. Wait, actually, \( \triangle PRE \) and \( \triangle POC \) are similar. So \( \frac{PR}{PR + 165} = \frac{210}{330} \). Wait, no, \( PO = PR + RO \), but \( RO = 165 \), \( RE = 210 \), \( OC = 330 \). So the ratio of \( RE \) to \( OC \) is \( \frac{210}{330} = \frac{7}{11} \). And since the triangles are similar, \( \frac{PR}{PO} = \frac{RE}{OC} \), where \( PO = PR + 165 \). So:

\( \frac{PR}{PR + 165} = \frac{210}{330} \)

Simplify \( \frac{210}{330} = \frac{7}{11} \). So:

\( 11 \cdot PR = 7 \cdot (PR + 165) \)

Step2: Solve for PR

Expand the right side: \( 11PR = 7PR + 1155 \)

Subtract \( 7PR \) from both sides: \( 4PR = 1155 \)

Divide both sides by 4: \( PR = \frac{1155}{4} = 288.75 \), wait, that can't be right. Wait, maybe I mixed up the triangles. Wait, maybe \( \triangle PRE \) and \( \triangle POC \) have \( RE \) and \( OC \) as horizontal sides, and \( PR \) and \( PO \) as vertical sides. Wait, no, maybe \( \triangle PRE \) and \( \triangle POC \) are similar, so \( \frac{RE}{OC} = \frac{PR}{PO} \), but \( PO = PR + RO \), but \( RO = 165 \). Wait, maybe I got the sides wrong. Wait, let's re-examine.

Wait, \( RE \) is 210, \( OC \) is 330, \( RO \) is 165. So the horizontal segments are \( RE = 210 \) and \( OC = 330 \), vertical segments are \( PR \) (distance across river) and \( PO = PR + 165 \) (total vertical from P to O). So the ratio of horizontal sides is \( 210/330 = 7/11 \), so the ratio of vertical sides should be the same. So \( PR / (PR + 165) = 7/11 \). Then:

\( 11PR = 7(PR + 165) \)

\( 11PR = 7PR + 1155 \)

\( 4PR = 1155 \)

\( PR = 1155 / 4 = 288.75 \), which is 289 when rounded. But wait, maybe the triangles are \( \triangle PRE \) and \( \triangle POC \) with \( RE \) and \( OC \) as the bases, and \( PR \) and \( PO \) as the heights. But maybe I made a mistake in the similar triangles. Wait, another approach: the two triangles are similar, so \( \frac{PR}{165} = \frac{210}{330 - 210} \)? No, that doesn't make sense. Wait, maybe the correct ratio is \( \frac{PR}{165} = \frac{210}{330 - 210} \)? No, 330 - 210 is 120, that doesn't help. Wait, maybe the triangles are \( \triangle REP \) and \( \triangle OCP \). Wait, \( \angle R = \angle O = 90^\circ \), \( \angle EPR = \angle CPO \) (common angle), so AA similarity. So \( \frac{PR}{PO} = \frac{RE}{OC} \). \( PO = PR + RO = PR + 165 \), \( RE = 210 \), \( OC = 330 \). So:

\( \frac{PR}{PR + 165} = \frac{210}{330} \)

Simplify \( 210/330 = 7/11 \), so:

\( 11PR = 7(PR + 165) \)

\( 11PR = 7PR + 1155 \)

\( 4PR = 1155 \)

\( PR = 1155 / 4 = 288.75 \approx 289 \). Wait, but let's check again. Wait, maybe the vertical side is \( RO = 165 \), and the horizontal sides are \( RE = 210 \) and \( OC = 330 \). So the ratio of horizontal sides is 210:330 = 7:11, so the ratio of vertical sides (PR:PO) should be 7:11. So PO is PR + 165, so PR/(PR + 165) = 7/11. Solving gives PR = 288.75, which is 289 when rounded. Wait, but maybe I messed up the similar triangles. Alternatively, maybe the triangles are \( \triangle PRE \) and \( \triangle POC \), with \( RE \) and \( OC \) as the horizontal legs, and \( PR \) and \( PO \) as the vertical legs. So the ratio of horizontal legs is 210/330 = 7/11, so the ratio of vertical legs is also 7/11. So PR / PO = 7/11, and PO = PR + 165. So:

PR = (7/11) * PO

But PO = PR + 165, so:

PR = (7/11)(PR + 165)

Multiply both sides by 11:

11PR = 7PR + 1155

4PR = 1155

PR = 1155 / 4 = 288.75 ≈ 289. Wait, but let's check with another approach. Let's use proportions. The difference in horizontal sides is 330 - 210 = 120, but no. Wait, maybe the triangles are \( \triangle REO \) and \( \triangle... \) No, the key is AA similarity. Since both triangles have a right angle and share angle P, they are similar. So the ratio of corresponding sides is equal. So RE/OC = PR/PO. RE is 210, OC is 330, PO is PR + 165. So 210/330 = PR/(PR + 165). Solving gives PR = 288.75, which is 289 when rounded to the nearest foot. Wait, but let's check the calculation again. 210*165 = 34650? No, wait, 1155 divided by 4 is 288.75, which is 289 when rounded. So the answer is 289.

Answer:

289