Question

geometry unit 5 test
grade 11th
name: uanna woods

1. uv, vw, and uw are mid - segments, and ts = 42, uw = 23, and vw = 19. label the triangle and find the perimeter of arst? 4 pts
2. solve for x. 4 points
3. find de. 5 points
4. solve for x. 4 points
5. find m∠klj. 4 points
6. if km is the perpendicular bisector of jl, find x. 3 points
7. find m∠yqr. 4 points
8. the altitudes of a triangle intersect at the:

a. circumcenter b. incenter c. centroid d. orthocenter

1. the angle bisectors of a triangle intersect at the:

a. circumcenter b. incenter c. centroid d. orthocenter

1. classify point c.

a. circumcenter b. incenter c. centroid d. orthocenter

1. classify point c.

a. circumcenter b. incenter c. centroid d. orthocenter

Explanation:

Step1: Analyze problem 1

Since \(UV, VW, UW\) are mid - segments, by the mid - segment theorem, the sides of \(\triangle RST\) are twice the lengths of the mid - segments. \(RS = 2UV\), \(RT=2VW\), \(ST = 2UW\). Given \(UV = 23\), \(VW=19\), \(UW\) is related to \(TS = 42\) (where \(UW=\frac{TS}{2}=21\)). The perimeter \(P=RS + RT+ST=2(23)+2(19)+42\).
\[P = 46+38 + 42=126\]

Step2: Analyze problem 2

In a parallelogram - like figure (assuming parallel sides), we can set up an equation based on the properties of parallel lines and equal - length segments. If the two expressions \(10x - 72\) and \(3x-4\) are equal (assuming they represent equal - length sides), then \(10x-72=3x - 4\).
\[10x-3x=-4 + 72\]
\[7x=68\]
\[x=\frac{68}{7}\]

Step3: Analyze problem 3

If \(DE\) is a mid - segment of \(\triangle ABC\), then \(DE=\frac{1}{2}AC\). First, set up an equation to find \(x\) using the relationship between the segments. If \(13x-37=2(2x + 13)\) (assuming a mid - segment relationship).
\[13x-37=4x+26\]
\[13x-4x=26 + 37\]
\[9x=63\]
\[x = 7\]
Then \(DE=2x+13=2\times7+13=27\)

Step4: Analyze problem 4

Set up an equation based on the angle - relationship. If \((8x-3)^{\circ}+(16x - 33)^{\circ}=90^{\circ}\) (assuming a right - angle relationship).
\[8x+16x=90 + 3+33\]
\[24x=126\]
\[x=\frac{126}{24}=\frac{21}{4}\]

Step5: Analyze problem 5

Set up an equation for the angles. If \((5x-2)^{\circ}+(9x-62)^{\circ}=180^{\circ}\) (assuming a linear - pair relationship).
\[5x+9x=180 + 2+62\]
\[14x=244\]
\[x=\frac{122}{7}\]
\(m\angle KLJ=(9x - 62)^{\circ}=9\times\frac{122}{7}-62=\frac{1098}{7}-\frac{434}{7}=\frac{664}{7}\approx94.86^{\circ}\)

Step6: Analyze problem 6

Since \(KM\) is the perpendicular bisector of \(JL\), then \(JK = KL\). So \(4x + 9=11x-61\).
\[11x-4x=9 + 61\]
\[7x=70\]
\[x = 10\]

Step7: Analyze problem 7

Set up an equation for the angles. If \((5x + 6)^{\circ}+(16x-27)^{\circ}=180^{\circ}\) (assuming a linear - pair relationship).
\[5x+16x=180-6 + 27\]
\[21x=201\]
\[x=\frac{201}{21}=\frac{67}{7}\]
\(m\angle YQR=(5x + 6)^{\circ}=5\times\frac{67}{7}+6=\frac{335}{7}+\frac{42}{7}=\frac{377}{7}\approx53.86^{\circ}\)

Step8:

The altitudes of a triangle intersect at the orthocenter. So the answer is D. Orthocenter.

Step9:

The angle - bisectors of a triangle intersect at the incenter. So the answer is B. Incenter.

Step10:

If the point \(C\) is the intersection of the perpendicular bisectors of the sides of the triangle, it is the circumcenter. So the answer is A. Circumcenter.

Step11:

If the point \(C\) is the intersection of the medians of the triangle, it is the centroid. So the answer is C. Centroid.

Answer:

1. \(P = 126\)
2. \(x=\frac{68}{7}\)
3. \(x = 7\), \(DE=27\)
4. \(x=\frac{21}{4}\)
5. \(x=\frac{122}{7}\), \(m\angle KLJ=\frac{664}{7}\)
6. \(x = 10\)
7. \(x=\frac{67}{7}\), \(m\angle YQR=\frac{377}{7}\)
8. D. Orthocenter
9. B. Incenter
10. A. Circumcenter
11. C. Centroid