Wednesday, October 13, 2021

Geometry Problems of the Day (Geometry Regents, August 2012)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Geometry Regents, August 2012

Part I: Each correct answer will receive 2 credits.

21. In the diagram below, EF is the median of trapezoid ABCD.
If AB = 5x - 9, DC = x + 3, and EF = 2x + 2, what is the value of x?

1) 5
2) 2
3) 7
4) 8

Answer: 1) 5

EF is a midsegment, which means that its length is the average of the two bases. If you think about it, this is no different then the midsegment of a triangle, where the midsegment of a triangle, which has a length equal to the average of the base of the triangle and 0 (which would make it half of the base).

So 5x - 9 + x + 3 = 2(2x + 2)
6x - 6 = 4x + 4
2x = 10
x = 5

This would make DC = 5 + 3 = 8, EF = 2(5) + 2 = 12, and AB = 5(5) - 9 = 16. The average of 8 and 16 is 12, so this checks out.

If you forgot to multiply the midsegment by 2, you would have gotten 2 for x. The other two choices are likely solutions to other incorrect equations that could've been written.

22. In the diagram below of triangle ABC, AB ≅ AC, m∠A = 3x, and m∠B = x + 20.

What is the value of x?

1) 10
2) 28
3) 32
4) 40

Answer: 2) 28

If AB ≅ AC then m∠B = m∠C. And the sum of the three angles is 180.

3x + x + 20 + x + 20 = 180
5x + 40 = 180
5x = 140
x = 28

23. For which polygon does the sum of the measures of the interior angles equal the sum of the measures of the exterior angles?

1) hexagon
2) pentagon
3) quadrilateral
4) triangle

Answer: 3) quadrilateral

The sum of the exterior angles is always 360 degrees. The sum of the interior angles increases with the number of sides of the polygon.

The sun of the interior angles in quadrilaterals is 360 degrees.

24.For a triangle, which two points of concurrence could be located outside the triangle?

1) incenter and centroid
2) centroid and orthocenter
3) incenter and circumcenter
4) circumcenter and orthocenter

Answer: 4) circumcenter and orthocenter

The centroid is ALWAYS inside the triangle. It's the point where three medians meet, and must be 2/3 of the way to the opposite midpoint.

The incenter MUST also be inside the triangle because it is the center of the circle that is inscribed INSIDE the triangle.

The orthocenter will be OUTSIDE of any obtuse triangle because one side must be extended to find the altitude.

The circumcenter may be OUTSIDE of the triangle because it is the center of a circle that has the triangle inscribed inside of it. (And the circle is "circumscribed" about the triangle.) If those three points are all on a minor arc of the circle, then the circumcenter will be outside the triangle.

25. The slope of line L is -1/3. What is an equation of a line that is perpendicular to line L?

1) y + 2 = 1/3 x
2) -2x + 6 = 6y
3) 9x - 3y = 27
4) 3x + y = 0

Answer: 3) 9x - 3y = 27

If the slope of a line is -1/3, then the slope of any line perpendicular to it must be +3.

Choice (1) has a slope of 1/3. Eliminate this choice.

Choice (2) has a slope of -2/6 = -1/3. It is parallel. Eliminate this choice.

Choice (3) has a slope of -(9)/(-3) = 3. This is the correct choice.

Choice (4) has a slope of -3. Eliminate this choice.

I used the formula -A/B for choice (3), but I could have rewritten it in slope-intercept form. Choices (1) and (4) already had 1y, which just needed to be isolated.

More to come. Comments and questions welcome.

More Regents problems.

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