Checkpoint 2: Geometry
Test your understanding of congruence, similarity, geometric reasoning and circle geometry. Covers Lessons 8 through 14.
Multiple Choice
Select the best answer for each question. Each question is worth 1 mark.
1 mark Which congruence test applies when all three sides of one triangle equal the corresponding sides of another?
1 mark Two triangles have angles 40 degrees, 60 degrees, 80 degrees and 40 degrees, 60 degrees, 80 degrees. They are:
1 mark The scale factor between two similar shapes is 3. The ratio of their areas is:
1 mark In a cyclic quadrilateral, opposite angles:
1 mark The angle at the centre of a circle subtended by an arc is 100 degrees. The angle at the circumference on the same arc is:
1 mark A tangent to a circle is perpendicular to:
1 mark Two parallel lines are cut by a transversal. One alternate angle is 55 degrees. A co-interior angle is:
1 mark A chord of length 12 cm is 8 cm from the centre of a circle. The radius is:
1 mark The alternate segment theorem states that the angle between a tangent and a chord equals:
1 mark Which is NOT a valid test for triangle similarity?
Short Answer
Show all working and justify your answers.
Question 11
ABCD is a parallelogram. The diagonals AC and BD intersect at E.
(a) Prove that triangle ABE is congruent to triangle CDE.
(b) Hence prove that the diagonals of a parallelogram bisect each other.
(c) If ABCD is also a rectangle, prove that the diagonals are equal in length.
(d) A student claims that if the diagonals of a quadrilateral bisect each other, it must be a parallelogram. Is this true? Explain.
Question 12
Two circles with centres O1 and O2 touch externally at T. A common tangent at T meets a common external tangent at P. The radii are 5 cm and 12 cm.
(a) Find the distance between the centres O1 and O2.
(b) Find the length of the common external tangent between the points of contact.
(c) A line through T meets the first circle at A and the second circle at B. Prove that the tangents at A and B are parallel.
(d) Evaluate whether the length of the common external tangent depends on the position of P. Explain your reasoning.
Model Answers
(a) In triangles ABE and CDE: AB = CD (opposite sides of parallelogram), angle BAE = angle DCE (alternate angles, AB parallel to CD), angle ABE = angle CDE (alternate angles, AB parallel to CD). Therefore triangle ABE congruent to triangle CDE (AAS).
(b) From congruence, AE = CE and BE = DE. Therefore E is the midpoint of both diagonals, so diagonals bisect each other.
(c) In triangles ABC and BAD: AB = BA (common), BC = AD (opposite sides of rectangle), angle ABC = angle BAD = 90 degrees. Therefore triangle ABC congruent to triangle BAD (SAS), so AC = BD.
(d) True. If diagonals bisect each other, then AE = CE and BE = DE. Vertically opposite angles AEB = CED. Therefore triangle ABE congruent to triangle CDE (SAS). So AB = CD and angle BAE = angle DCE, which means AB is parallel to CD. Similarly, AD is parallel to BC. Hence ABCD is a parallelogram.
Marking guidance: 1 mark for (a), 1 mark for (b), 1 mark for (c), 2 marks for (d).
(a) Distance = $5 + 12 = 17$ cm.
(b) Difference of radii projected = $12 - 5 = 7$ cm. Tangent length = $\sqrt{17^2 - 7^2} = \sqrt{289 - 49} = \sqrt{240} = 4\sqrt{15} \approx 15.5$ cm.
(c) Let angle AO1T = angle between radius O1A and line O1O2. Since triangle AO1T is isosceles (O1A = O1T = radius), angle O1AT = angle O1TA. Similarly for triangle BO2T. The tangent at A is perpendicular to O1A, and the tangent at B is perpendicular to O2B. Since angle AO1T = angle BO2T (corresponding angles in similar configurations), the tangents make equal angles with line AB, hence they are parallel.
(d) The length of the common external tangent depends only on the radii of the two circles, not on the position of P. All common external tangents between two given circles have the same length because they form congruent right triangles with the line of centres.
Marking guidance: 1 mark for (a), 1 mark for (b), 2 marks for (c), 1 mark for (d).