12 multiple-choice questions, progressively harder.
Two vertical objects are measured simultaneously, but their shadows lie on surfaces with different unknown slopes. Is the usual height-over-shadow proportion guaranteed?
Solution
Correct answer: A
The usual AA argument needs matching base and ray-to-ground angles. Different unknown slopes remove that justification. A special setup might still give the same numerical result, so the precise conclusion is not guaranteed.
In a valid mirror setup, EF is eye height, TB object height, FM and MB are ground legs, and EM is a slanted sight line. Which proportion is supported?
Correct answer: C
AA pairs TB with EF and MB with FM. The sight line EM is not interchangeable with a ground leg, and FB is the whole ground separation.
In the crossed survey, A,B,C and T,B,D are collinear in order, while AT and CD are perpendicular to AC. Which ordered similarity states the correspondence correctly?
Correct answer: B
The inaccessible target T matches D, the right-angle vertex A matches C, and the crossing vertex B matches itself. Thus the order is TAB to DCB.
In a valid crossed survey, the ratio of bank segments AB:BC is 7:3 and accessible perpendicular CD is 4.20 m. Find inaccessible perpendicular AT.
Correct answer: D
AA matches AT with CD and AB with CB. The ratio of inaccessible length AT to accessible length CD is seven thirds.
AT=4.20⋅73,=9.80 m.\begin{aligned}AT&=4.20\cdot\frac 73,\\&=9.80\text{ m}.\end{aligned}AT=4.20⋅37,=9.80 m.
A print changes a 4.0 cm horizontal length to 5.0 cm and a 9.0 cm vertical length to 10.8 cm. Can one new similarity scale describe the whole print?
Compare corresponding lengths in both directions.
A single similarity requires one common factor.
Two similar triangular frames have original and image perimeters 72 and 108 cm. Corresponding sides are 2x+4 and 5x-1 cm. Find x.
The known similarity makes the perimeter ratio the common length factor.
The corresponding side lengths are 11 and 16.5 cm, both positive.
A measurement calculation gives 17.8625 m. The task asks for an estimate to the nearest tenth of a meter. Which report is appropriate?
Round the hundredths digit 6 upward into the tenths place. The result is approximately 17.9 m; a measured model does not gain exactness from a long decimal.
A 5 cm model edge corresponds to a 12 m real edge. What is the factor mapping real lengths to model lengths in common units?
Express the real edge in centimeters before forming model divided by real.
k=51200,=1240.\begin{aligned}k&=\frac 5{1200},\\&=\frac 1{240}.\end{aligned}k=12005,=2401.
A mirror is horizontal and the modeled triangles have right angles, but no reflection-angle relation is supplied. What further premise does the usual AA argument need?
The right angles give only one pair. The second pair comes from the equal-angle reflection model, expressed with the mirror perpendicular or its complementary ground angles.
An original triangular boundary has sides a,b,c. A similar image has corresponding sides ka,kb,kc for positive k. Which perimeter equation is justified?
Add the three image sides. Factoring the same multiplier from ka+kb+kc gives k(a+b+c).
An original 1:6000 map is reprinted uniformly. The new print has scale 1:4000. What is new drawing length divided by old drawing length?
The real length stays fixed. Compare the new drawing-to-real ratio with the old one.
LnewLold=1/40001/6000,=1.5.\begin{aligned}\frac{L_{\rm new}}{L_{\rm old}}&=\frac{1/4000}{1/6000},\\&=1.5.\end{aligned}LoldLnew=1/60001/4000,=1.5.
A simultaneous shadow check gives height/shadow ratios 2.4/1.8 and 4.0/3.0 for two calibration objects. What does their agreement show?
Both ratios agree.
2.41.8=4.03.0,=43.\begin{aligned}\frac{2.4}{1.8}&=\frac{4.0}{3.0},\\&=\frac 43.\end{aligned}1.82.4=3.04.0,=34.
That is a useful consistency check, while verticality, ground conditions and simultaneous rays remain model premises.
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