What follows is a portion of the Common Core Geometry exam. Other parts will be posted on other days. Illustrations will be added at a later time when they become available.
Part II is posted here.
June 2016 Geometry Regents, Part III
32. A barrel of fuel oil is a right circular cylinder where the inside measurements of the barrel are a diameter of 22.5 inches and a height of 33.5 inches. There are 231 cubic inches in a liquid gallon. Determine and state, to the nearest tenth, the gallons of fuel that are in a barrel of fuel oil.
The Volume of a cylinder is pi*r2*h. So V = pi(11.25)2(33.5) = 13319.9
To convert cubic inches into gallons, divide by 231: 13319.9 / 231 = 57.66...
Answer: 57.7 gallons.
33. Given: Parallelogram ABCD, EFG, and diagonal DFB
Prove: Triangle DEF ~ triangle BGF
You could write a two-column proof or a paragraph proof. For this blog, paragraph is a little easier.
Angle DFE is congruent to Angle BFG because they are vertical angles. AD is parallel to BC because opposite sides of a parallelogram are parallel. BD is a transversal. Angle ADB is congruent to CBD because they are alternate interior angles. Therefore Triangle DEF ~ triangle BGF because of AA (Angle-Angle Theorem).
34. In the diagram below, Triangle A'B'C' is the image of Triangle ABC after a transformation.
Describe the transformation that was performed.
Explain why Triangle A'B'C' ~ Triangle ABC.
The transformation was a Dilation of scale factor 2.5 centered on (0, 0). Point A(-2, 4) -> A'(-5, 10). Point B(-2, -4) -> B'(-5, -10). Point C(4, -4) -> C'(-10,-10).
-5/-2 = 2.5. 10/4 = 2.5. -10/-4 = 2.5. 10/4 = 2.5
Dilations preserve shape so the angles are the same size. Therefor the triangles are similar.
June 2016 Geometry Regents, Part IV
35. Given: Quadrilateral ABCD with diagonals AC and BD that bisect each other, and < = <2
Make sure you restate the Given information. Make sure that you restate what you want to prove as your final statement. (In this case, you are proving two things, so one of them will be in the middle, right before you start proving the second half.) Do NOT use what you are trying to prove and a reason why something must be true.
| Statement | Reason |
| 1. AC and BD bisect each other. | Given |
| 2. <1 = <2 | Given |
| 2.5 ABCD is a parallelogram | a quadrilateral with diagonals that bisect each other is a parallelogram
(edited to add missing step) |
| 3. AB || CD | Opposite sides of a parallelogram are parallel. |
| 4. Angle DCA = Angle 1 | Alternate interior angles |
| 5. Angle DCA = Angle 2 | Transitive Property of Congruence |
| 6. Triangle ACD is isosceles | If the base angles of a triangle are congruent, then the triangle is Isosceles |
| 7. AD = CD | The sides opposite congruent base angles of an isosceles triangle are congruent |
| 8. ABCD is a rhombus | A parallelogram with consecutive sides congruent is a rhombus |
| 9. Angle AEB is a right angle. | Diagonals of a rhombus are perpendicular. |
| 10. Triangle AEB is a right triangle. | A triangle with a right angles is a right triangle. |
36.
A water glass can be modeled by a truncated right cone (a cone which is cut parallel to its base)
as shown below.
The diameter of the top of the glass is 3 inches, the diameter at the bottom of the glass is 2 inches,
and the height of the glass is 5 inches.
The base with a diameter of 2 inches must be parallel to the base with a diameter of 3 inches in
order to find the height of the cone. Explain why.
Determine and state, in inches, the height of the larger cone.
Determine and state, to the nearest tenth of a cubic inch, the volume of the water glass.
To find the height, you need to have similar triangles. To have similar triangles, the bases must be parallel so that the base angles are congruent.
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The radius of the large circle is 1.5. The radius of the smaller circle is 1. The height to the smaller circle is x. The height to the larger circle is x + 5.
Set up a proportion: x / 1 = (x + 5) / 1. 5
The Volume of a cone = (1/3)(pi)r2h
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This comment has been removed by a blog administrator.
ReplyDeleteFor 35, the quadrilateral was never proved a parallelogram.
ReplyDeleteYou are absolutely correct. Good catch.
ReplyDeleteEither I misread "quadrilateral" as "parallelogram", or I left out a line when I typed up the proof. Who can remember at this point? Whichever it was, that would be -1 for me!
Edited as Step 2.5