12 multiple-choice questions, progressively harder.
On median AD, G is the centroid and AD is 42 cm. How much longer is AG than GD?
Solution
Correct answer: C
The whole median has three equal ratio parts.
The vertex portion has two such parts, so the difference between it and the shorter portion is one part, or 14 cm.
On median AD, G is the centroid. The whole length is AD=4x+9AD=4x+9AD=4x+9 cm.
The shorter portion is GD=x+6GD=x+6GD=x+6 cm.
What is x?
Correct answer: A
The whole is three times the shorter portion.
At this value, the shorter portion is 15 cm and the whole is 45 cm, both positive.
On median AD, G is the centroid. AG exceeds GD by 11 cm. What is AD?
Correct answer: D
The two portions contain two parts and one part, so their difference is one ratio part. Therefore GD is 11 cm.
AD and BE are medians meeting at centroid G. AD is 24 cm and BE is 33 cm. What is the sum of AG and GE?
Correct answer: B
Take the vertex portion on AD but the midpoint portion on BE.
Their sum is 27 cm.
D and E are the midpoints of BC and CA. Medians AD and BE meet at G. In the similarity used in the proof, which vertex corresponds to A?
The parallel sides are AB and DE. The corresponding triangles can be written in order as
Thus A corresponds to D, G to G, and B to E.
D and E are the midpoints of BC and CA. Segments AD and BE meet at G. If AB is 30 cm and GD is 4 cm, what is AG?
The midsegment DE is half of AB, so the corresponding large-to-small triangle factor is 2. Similarity gives
Hence AG is 8 cm.
A proof gives AG as two thirds of AD and AH as two thirds of AD. Both G and H lie inside segment AD. What completes the concurrence argument?
Both G and H lie the same distance along the same segment from A. They must be the same point. This identifies the intersections made by two different median pairs.
A segment from A meets BC at D and is perpendicular to BC. Which extra fact would establish that AD is also a median?
Perpendicularity alone does not establish a midpoint. Equal parts BD and DC, together with D on BC, establish the opposite midpoint required for a median.
D is the midpoint of BC in a nondegenerate triangle ABC. P is the midpoint of AD. Can P be the centroid?
At the midpoint P, AP and PD are equal and positive. A centroid on this median must instead satisfy
Equal positive portions cannot also have that ratio.
You have constructed the medians from A and B and located their intersection G. How can you locate the centroid without constructing the third median?
The proved concurrence theorem says all three medians pass through the intersection of any two. Thus G is already located.
In the proof with D the midpoint of BC and E the midpoint of CA, what does the midsegment theorem supply about DE?
DE connects midpoints on sides BC and CA. The remaining side is AB, so DE is parallel to AB and has half its length.
A triangle and its medians are enlarged by a dilation of factor 1.5. A centroid-to-midpoint portion was 8 cm. What happens to that portion and the median division ratio?
A dilation multiplies every length by the same factor.
Multiplying both portions by the same positive factor preserves their two-to-one ratio.
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