Oscar05
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Delmar is constructing an equilateral triangle. He uses his straightedge to draw a segment and labels the endpoints W and X. With the compass opening equal to the length of the segment, he draws a circle centered at point W and then a circle centered at point X. He labels the intersections of the circles as points P and Q.

How should Delmar finish the construction?


Use a straightedge to join points W and P and then points P and X. △WPX△WPX is equilateral.

Use a straightedge to join points P and Q. Label the intersection of PQ¯¯¯¯¯PQ¯ and WX¯¯¯¯¯¯WX¯ as A. Construct the midpoint of AX¯¯¯¯¯AX¯ and label it B. Use a straightedge to join points Q and B. △QAB△QAB is equilateral.

Use a straightedge to join points W and P, P and Q, and then W and Q. △WPQ△WPQ is equilateral.

Use a straightedge to join points P and Q. Label the intersection of PQ¯¯¯¯¯PQ¯ and WX¯¯¯¯¯¯WX¯ as A. Use a straightedge to join points Q and X. △QAX△QAX is equilateral.

Respuesta :

Kalahira
Use a straightedge to join points W and P and then points P and X. â–łWPX is equilateral. Let's see now, Delmar has a line segment WX and has drawn 2 circles whose radius is the length of WX, centered upon W and centered upon X. Sounds to me that all he needs to do is select one of the intersections of those 2 circles and use that at the 3rd point of the equilateral triangle and draw a line from that point to W and another line from that point to X. Doesn't matter which of the two intersections he chooses, just needs to pick one. Looking at the available options, only the 1st one which is "Use a straightedge to join points W and P and then points P and X. â–łWPX is equilateral." matches my description, so that is the correct choice. The other choices tend to do rather bizarre things like create a perpendicular bisector of WX and for some unknown reason, claim that bisector is somehow a side of a desired equilateral triangle.

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