Similarity of Triangles Trivia Quiz

Answer: True

To define similarity simply, it is a case in which all the corresponding angles of two or more triangles are equal, but their sides are not necessarily equal.

Congruent triangles are triangles which are identical in every way.

As such, all congruent triangles are similar, but not all similar triangles are congruent.

Answer: 3:5

Triangle ADE was similar to triangle ABC.

A striking property of similar triangles is that the ratio of all their corresponding sides are equal. For example, if triangle ABC is similar to triangle PQR, then:-

AB:PQ = AC:PR = BC:QR

Answer: ~

ABC ~ PQR means that triangle ABC is similar to triangle PQR.

Answer: 2:1

In order to evaluate this answer, you had to use the Basic Proportionality Theorem, which states, "If a line is drawn parallel to a side of a triangle, it divides the other two sides in proportion."

For example, in the above-mentioned case, XP:PY = XQ:QZ. Therefore, since the ratio XP:XY was given, you were required to find out the ratio of XP:PY.
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XP:XY = 2:3
Let XP = 2x and XY be 3x.
PY = XY - XP = 3x - 2x = x.
Therefore, XP:PY = 2x:x = 2:1
But XP:PY = XQ:QZ
Therefore, XQ:QZ = 2:1
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This was why I advised you to draw diagrams.

Answer: True

What I just stated is the converse of the Basic Proportionality Theorem.
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In ABC, DE is a line which cuts AB at D and AC at E.
AD:DB = AE:EC.
Therefore, DE is parallel to the third side of ABC, namely BC.
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Answer: "If a line passes through the midpoint of one side of a triangle and is parallel to a second side, the given line bisects the third side."

Since I have given you the theorem and its converse, use it to solve any subsequent numerical based on them.

Now, I know this is a quiz on Similarity, but I've included the Midpoint and Basic Proportionality Theorems because:

1) The Midpoint Theorem is a special case of similarity. If a line DE passes through the midpoints of the sides AB and AC of a triangle ABC, then triangle ADE automatically become similar to triangle ABC.

2) The Basic Proportionality Theorem is a consequence of similarity. It cannot be proven without similarity.

Answer: All of these can be used to evaluate the ratio of the areas of the triangles

Let me explain a bit more illustratively.
[Note: 'Ao' means 'Area of' and '^2' means 'squared'.)
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In triangles MNO and DEF:-

MN and DE are corresponding sides, MP and DG are corresponding altitudes, and MQ and DH are corresponding medians of triangles MNO and DEF respectively.

Ao MNO: Ao DEF = (MN:DE)^2 = (MP:DG)^2 = (MQ:DH)^2
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I'd give you the proof which shows you how to arrive to this conclusion, but it's really long and tedious.

Answer: True

Each angle of a an equilateral triangle is sixty degrees. Hence, the whole lot of them are similar.

Answer: BDC ~ CEB

It becomes much simpler if you draw a diagram like I told you to.
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In triangles BDC and CEB:
Angle CEB = Angle BDC = 90 degrees
Angle DCB = Angle EBC [As base angles of isosceles triangle ABC are equal]
The third pair of angles are also equal, due to the hypothesis that the sum of all the angles of a triangle is equal to 180 degrees.

Answer: True

This is known as the S.S.S test of similarity.
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AB:PQ = AC:PR = BC:QR

Therefore, ABC ~ PQR.
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