PROBLEM SET 1.4
1. Construct a trapezoid given its four sides.
GSP file Be sure to examine all three pages.
2. Construct a right triangle given its hypotenuse c and a leg a.
3. Construct an Isosceles triangle ABC with AB = AC, given angle A and its perimeter (the sum of the three sides).
4. Construct all points equidistant from three given lines
a. The three lines do not intersect at a single point, no two are parallel.
b. Exactly two lines are parallel
c. The three lines are concurrent.
5. Construct a triangle given a, h(a), and m(a). Click HERE for a GSP file with script tool.
7. Construct an equilateral triangle given its height.
Another approach
Start with segment BD.
Construct a circle with center at D and radius BD.
Construct a circle (arc shown) with center at B, locating two points S and T on the first circle equidistant from B and D. Now BTDS is a rhombus with four sides of length BD and one diagonals BD and 2BD. That is, BTDS is a rhombus divided into two equilateral triangles by diagonal BD.Now the angle bisector of angle SBD Creates angles of 30 degrees as does the angle bisector of angle TBD so extending the angle bisectors to points A and C on the perpendicular to BD at D give an equilateral triangle ABC with the given height.
Angle BDA is the external angle of triangle BDC and therefore the measure of Angle C is 60 degrees and so the measure of angle A also be 60 degrees by the same reasoning.
Another approach to constructing an equilateral triangle given its height.
8. If P is on one side of a given angle A. Find a point X on the other side of the angle such AX + PX = s, the length of a given segment.
9. Trisect a line segment using a theorem about the medians of a triangle.
Given a line AM. Let CB be any line segment with M as its midpoint.
Construct AC and AB.
Construct Median CN to AB.
The intersection of CN and AM is a trisection point on AM. The other trisection point is found by bisection the longer part of AM.
Proof: The point of intersection is the point of concurrencuy of the medians of triangle ABC and by the theorem of the medians, that point divides AM in the ratio of 2:1.
10. Construct a quadrilateral given three of its sides and two diagonals.
Comment: Making a plan (i.e., "Investigation" -- In a quadrilateral, two adjacent sides and a diagonal create a triangle. Thus we could create two triangles, appropriately oriented and the fourth side would be determined by the vertices not on the common base. In what follows, Sunny has created triangles by (a, b, d2) and (c, b, d1).
Is the quadrilateral unique up to congruence?
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11. Construct a triangle given two of its sides and the median to the third side. OPEN GSP file.
12. Construct a triangle given its three medians.
Approach 1: Let P be the point of concurrency for the medians of triangle ABC. The construct the subtriangle APC. Open GSP file. Examine script tool.
Approach 2: Create a triangle from the three medians and use parallel lines to analyze and construct triangle ABC. Open GSP file. Examine script tool.
For those with interest in EMAT 6680, See Assignment 6, Problems 1 and 3
13. a. If AB is a diameter of a semicircle and C is any point on the semicircle, prove that angle ACB is 90 degrees.
Write-up by Stephanie Henderson
b. Use the result from part (a) to construct a right triangle given its hypotenuse and one side (Same problem as Problem 2, but different basis for the construction.)
c. Solve the problem in part (b) without using the theorem in Part (a).
Solutions to Problem 1.14.13c are given in Problem 1.4.2.
14. Construct a right triangle given the hypotenuse c and the sum of the remaining sides a and b.
15. Construct a triangle given a side a, the angle opposite A, and the altitude h to one of the other sides.
17. In triangle ABC, construct point P and Q on sides AB and BC, respectively, such that PQ is parallel to AC and PQ = AP + QC
18. The point of intersection of an altitude with the side of a triangle is called the foot of the altitude. Given the three feet of the altitudes of a triangle, construct the triangle.
Note that in the sketch at the right we would be given the point D. E, and F and asked to construct the triangle. The three points determine a triangle, called the orthic triangle of triangle ABC.
The dashed lines are the perpendiculars from the respective vertices to the opposite sides. However, if we take the three points and construct the orthic triangle, the dashed lines are the angle bisectors of the angles of the orthic triangle. Why?
This gives us a means to construct ABC:
Given three points D, E, and F.
Construct the orthic triangle.
Construct the angle bisectors of each of the angles of triangle DEF
Construct the perpendiculars to each of the angle bisectors at D, E, and F. These will line along the sides of the triangle ABC an the vertices will be determined by the intersection points of these three perpendiculars.
Write-up by Stephanie Britt
Construct the orthic triangle first.
The altitudes of the desired triangle are the angle bisectors of the orthic triangle.Construct the angle bisectors.
From there construct the perpendicular lines through each altitude at the vertices of the orthic triangle.
Where the perpendicular lines intersect are the vertices of the triangle we want.
19. a. Construct Hexagon with each angle 120 degrees, but is not regular.
Take a regular hexagon and construct a segment parallel to one of the sides of the regular hexagon. Now each adjacent side of the hexagon we are constructing will have a side forming a 120 degree angle. Therefore each adjacent side will be parallel to adjacent sides of a regular hexagon. Continue, but vary the lengths of the segments.
b. How many nonsimilar hexagons with all interior angle 120 degrees are there?
c. Prove that in the hexagon in part (a)
20.
a. Prove that if HG and EF are two perpendicular segments connecting arbitrary points on opposite sides of the square, then HG = EF
Consider perpendiculars to the opposite sides from G and from F.
b. Suppose X, Y and Z, W are pairs of points on opposite sides of a square that remain after the sides of the square have been erased.
21. Given Points P and Q on the same side of line l, find (construct) point X on l such that angle PXA is twice the size of angle QXB. (note different label that in textbook)
Analysis:
We know a construction for EQUAL angles. So we look for a point related to Q that would be on a path forming equal angles with P. Consider a circle with center at Q and tangent to l. A tangent line from that circle to the desired point X would have an angle twice the size of angle QXB. In other words it would be equal to the angle PXA .
So construct a perpendicular from P and extend to a point P' on the other side of l such that PP' is bisected by l.
See GSP file
