Thursday, September 09, 2021

Geometry Problems of the Day (Geometry Regents, August 2013)



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 2013

Part I: Each correct answer will receive 2 credits. Partial credit is available.


29.Triangle ABC has vertices A(6,6), B(9,0), and C(3, -3). State and label the coordinates of triangle A'B'C', the image of triangle ABC after a dilation of D1/3.

Answer:


A dilation of 1/3 centered on the origin -- no center of dilation was stated, so the default is the origin -- means that all points in the image will have coordinates that are 1/3 of the pre-image.

Make sure you write something. Don't do it all in your head!

A(6,6) * 1/3 --> A'(2,2)

B(9,0) * 1/3 --> B'(3,0)

C(3,-3) * 1/3 --> C'(1,-1)





30. Using a compass and straightedge, construct the bisector of ∠MJH. [Leave all construction marks.]


Answer:


Make an arc about point J. Mark off the points on JM and JH. Next, make an arc about the point on JM. Make an arc of the SAME size about the point on JH. Draw a line from J through the point where the two new arcs intersect. This is your angle bisector.





31. Find, in simplest radical form, the length of the line segment with endpoints whose coordinates are (-1,4) and (3,-2).

Answer:


Use the distance formula, or make a sketch on graph paper and use the Pythagorean Theorem.

d = SQRT( (-1 - 3)2 + (4 - (-2))2) )
= SQRT ( (4)2 + (6)2 )
= SQRT ( 16 + 36 )
= SQRT (52)
= SQRT (2 * 2 * 13)
= 2 * SQRT (13)







More to come. Comments and questions welcome.

More Regents problems.

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