Geometry Problems of the Day (Geometry Regents, January 2014)

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, January 2014

Part III: Each correct answer will receive 4 credits. Partial credit is possible.


35. In the diagram of triangle BCD shown below, BA is drawn from vertex B to point A on DC, such that BC ≅ BA.

In triangle DAB, m∠D = x, m∠DAB = 5x - 30, and m∠DBA = 3x - 60. In triangle ABC, AB = 6y - 8 and BC = 4y - 2. [Only algebraic solutions can receive full credit.]

Find m∠D.

Find m∠BAC.

Find the length of BC.

Find the length of DC.

Answer:

You need to solve these problems algebraically. Unpack everything that you are given and label the diagram.

You are given algebraic expressions for each angle in triangle DAB. The sum of those angles is 180, so you have enough information to solve for x.

x + 5x - 30 + 3x - 60 = 180
9x - 90 = 180
9x = 270
x = 30

Since m∠D = x then m∠D = 30 degrees.

Angle BAC is supplementary to DAB, and m∠DAB = 5x - 30 and x = 30. Therefore, m∠DAB = 5(30) - 30 = 150 - 30 = 120.
m∠BAC + 120 = 180, so m∠BAC = 60

BC ≅ BA so solve the following equation:

6y - 8 = 4y - 2
2y - 8 = -2
2y = 6
y = 3
So BC = 4y - 2 = 4(3) - 2 = 10

If you got a negative answer for the length, then you know you made a mistake.

BC has a length of 3.

Since BA = BC, then BAC is an isoscles triangle. Since m∠BAC = 60, then m∠C = 60 because they are base angles. Since the two base angles are 60, then the vertex angle must also be 60 and BAC is an equilateral triangle. That means that angle DBC is 30 + 60 = 90 degrees, and it is a right angle.

Also, triangle DBC is a 30-60-90 degree right triangle.

You could write a trigonometric statement here, or you can note that the hypotenuse in a 30-60-90 triangle is twice the length of the side opposite the 30 degree angle. DC is 2 * BC = 2 * 10 = 20

36. The coordinates of the vertices of triangle ABC are A(-6,5), B(-4,8), and C(1,6). State and label the coordinates of the vertices of triangle A"B"C", the image of ABC after the composition of transformations T_4,-5 ° r_y-axis.

[The use of the set of axes below is optional.]

Answer:

The most important thing to remember here is that you need to do a tranlations OR THE reflection! The reflection must be done first before you translate it.

If you do it in the incorrect order, you will lose half credit, assuming the rest is correct.

You don't need to use the grid, but if you do, make sure that the intermediate graph is shown along with the final image. The final image needs to be labelled, including the coordinates.

To reflect over the y-axis (x = 0), change the sign of the x-coordinate.

A(-6,5), B(-4,8), and C(1,6) --->>> A'(6,5), B'(4,8), and C'(-1,6)

Next translate each point by adding 4 to each x-coordinate and subtracting 5 from each y-coordinate

A'(6,5), B'(4,8), and C'(-1,6) --->>> A"(10,0), B"(8,3), and C"(3,1)

37. In right triangle ABC below, CD is the altitude to hypotenuse AB. If CD = 6 and the ratio of AD to AB is 1:5, determine and state the length of BD.

[Only an algebraic solution can receive full credit.]

Answer:

If the ratio of AD to AB is 1:5, then label AD = x and AB = 5x. That makes DB = 4x.

According to the Right Triangle Altitude Thereom, the following is true:

AD / CD = CD / DB

(AD)(DB) = (CD)²

(x)(4x) = (6)²
4x² = 36
x² = 9
x = 3

If x = 3, then BD = 4x = 4(3) = 12.

Don't get thrown off by the ratio 1:5. Also, don't forget that DB is 4x, not x.

More to come. Comments and questions welcome.

More Regents problems.

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