Saturday, September 11, 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 III: Each correct answer will receive 4 credits. Partial credit is available.


35. Given: Triangle ABC, BD bisects angle ABC, BD ⊥ AC

Prove: AB ≅ CB


Answer:


Prove that the triangles are congruent, which means that the corresponding parts are congruent. Since BD is perpendicular to AC, the two smaller triangles are both right triangles.

Make you state the Given and that your last line is what you want to prove.

Statement:Reason:
1. BD bisects angle ABC, BD ⊥ ACGiven.
2. Angles ADB and CDB are right anglesDefinition of Perpendicular
3. Angle ABD ≅ Angle CDBAll right angles are congruent.
4. Angle CBD ≅ angle ABDDefinition of angle bisector.
5. BD ≅ BDReflexive Property.
6. Triangle ABD ≅ triangle CBD.ASA
7. AB ≅ CBCPCTC (Corresponding Parts of Congruent Triangles are Congruent.





36. Quadrilateral MATH has coordinates M(-6,-3), A(-1,-3), T(-2,-1), and H(-4,-1). The image of quadrilateral MATH after the composition rx-axis ° T7,5 is quadrilateral M"A"T"H". State and label the coordinates of M"A"T"H". [The use of the set of axes below is optional.]

Answer:


The composition of transformations should be read as the Reflection OF THE Translation. In other words, you must do the Translation first (getting M'A'T'H') before you can reflect. You will lose half credit if you do it the other way around.

M(-6,-3) --> M'(-6 + 7, -3 + 5), M'(1, 2)
A(-1,-3) --> A'(-1 + 7, -3 + 5), A'(6, 2)
T(-2,-1) --> T'(-2 + 7, -1 + 5), T'(5, 4)
H(-4,-1) --> H'(-4 + 7, -1 + 5), H'(3, 4)

Reflecting over the x-axis will flip the sign of the y-coordinate and leave the x-coodinate alone.

M'(1, 2) --> M"(1, -2)
A'(6, 2) --> A"(6, -2)
T'(5, 4) -->T"(5, -4)
H'(3, 4) --> H"(3, -4)





37. Trapezoid TRAP, with median MQ, is shown in the diagram below. Solve algebraically for x and y.


Answer:


The median is average of the two bases, so it is half the size of the sum of the bases.

Angles T and P are supplementary. Angle A is superfluous, it doesn't help with anything. It's only there to throw you off.

2(16y + 1) = 12y + 1 + 18y + 6
32y + 2 = 30y + 7
2y = 5
y = 2.5

12x - 4 + 7x + 13 = 180
19x + 9 = 180
19x = 171
x = 9




More to come. Comments and questions welcome.

More Regents problems.

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