12 multiple-choice questions, progressively harder.
Points AAA and BBB lie on a circle with center OOO. Their distances from OOO are (3x+1)(3x+1)(3x+1) cm and (5x−11)(5x-11)(5x−11) cm. Find the diameter's length.
Solution
Correct answer: B
Both distances are radii, so
3x+1=5x−11.3x+1=5x-11.3x+1=5x−11.
Subtract 3x3x3x and add 11 to obtain
12=2x.12=2x.12=2x.
Dividing by 2 gives x=6x=6x=6. The radius in centimeters is
3(6)+1=19.3(6)+1=19.3(6)+1=19.
Double the radius to find the diameter's length.
2×19=382\times19=382×19=38
The diameter is 383838 cm.
Diameter ABABAB contains center OOO, with AO=(2y+4)AO=(2y+4)AO=(2y+4) cm and OB=(5y−8)OB=(5y-8)OB=(5y−8) cm. Find ABABAB.
Correct answer: D
The two radii are equal, so
2y+4=5y−8.2y+4=5y-8.2y+4=5y−8.
Subtract 2y2y2y and add 8 to obtain
12=3y.12=3y.12=3y.
Dividing by 3 gives y=4y=4y=4. Each radius is 2(4)+4=122(4)+4=122(4)+4=12 cm. Since OOO is between AAA and BBB, segment addition gives
AB=12+12=24AB=12+12=24AB=12+12=24
The diameter is 242424 cm long.
The equilateral-triangle procedure is attempted with A=BA=BA=B. Which statement identifies the missing hypothesis and its consequence?
Correct answer: A
Coincident endpoints give AB=0AB=0AB=0.
This course's circle has a positive radius, and the construction begins with distinct endpoints.
The proposed starting data supply neither a nonzero base nor the required positive compass setting, so they do not produce a triangle.
An equilateral triangle made by the two-circle construction has perimeter 515151 cm. What compass radius was used?
Correct answer: C
Each of the three sides equals the construction radius. Divide the perimeter into three equal side lengths.
51÷3=1751\div3=1751÷3=17
The compass radius was 171717 cm.
For an equilateral construction on AB=9AB=9AB=9 cm, which candidate has the required two circle distances?
The circle centered at AAA requires AC=9AC=9AC=9 cm.
The circle centered at BBB separately requires BC=9BC=9BC=9 cm.
The chosen intersection must satisfy both conditions.
Two disks in the same plane share a center and have radii 555 cm and 888 cm. Where is every point on the smaller disk's boundary circle?
Every point on the smaller boundary circle lies in the common plane and is exactly 555 cm from the common center.
That equals the smaller radius and is less than the larger radius, so the point is on the smaller boundary and inside the larger disk.
An angle copy uses source circle radius 888 cm and source chord length 666 cm. The target circle at PPP has already been drawn with radius 888 cm and meets the baseline at DDD. Which circle locates the next mark?
The radius setting has already been transferred to PPP.
The next step transfers the chord: set the compass to AB=6AB=6AB=6 cm and center it at DDD to locate the target intersection.
Two source rays bound a reflex opening of 260∘260^\circ260∘. Their smaller opening is copied using the angle-copy procedure. What reflex opening do the target rays then bound?
The smaller source opening is 360∘−260∘=100∘360^\circ-260^\circ=100^\circ360∘−260∘=100∘. The procedure copies that 100∘100^\circ100∘ opening. The other target opening is
360∘−100∘=260∘.360^\circ-100^\circ=260^\circ.360∘−100∘=260∘.
A circle has center OOO and positive radius rrr, and ABABAB is a chord with distinct endpoints. Which additional fact permits the equation AB=AO+OB=2rAB=AO+OB=2rAB=AO+OB=2r?
The positive radius gives AO=BO=r>0AO=BO=r>0AO=BO=r>0, so OOO is neither endpoint.
If OOO lies on segment ABABAB, it is therefore between AAA and BBB.
Segment addition then gives
AB=AO+OB=2r.AB=AO+OB=2r.AB=AO+OB=2r.
The other facts do not supply the required betweenness.
A circle centered at PPP is drawn with positive compass opening UVUVUV, meeting ray PXPXPX at QQQ. Without changing the opening, a second circle is drawn centered at QQQ, and RRR is chosen on it. Which reason establishes QR=UVQR=UVQR=UV?
The first circle has center PPP and radius UVUVUV, giving PQ=UVPQ=UVPQ=UV. The unchanged opening is therefore also PQPQPQ.
The second circle has center QQQ and that same opening as its radius. Since RRR lies on it, QR=PQQR=PQQR=PQ, and hence QR=UVQR=UVQR=UV.
An equilateral triangle is constructed on AB=6AB=6AB=6 cm. Point MMM is the midpoint of ABABAB. What is AM+BCAM+BCAM+BC?
The midpoint gives AM=6÷2=3AM=6\div2=3AM=6÷2=3 cm. The construction gives BC=6BC=6BC=6 cm.
AM+BC=3+6=9AM+BC=3+6=9AM+BC=3+6=9
The sum is 999 cm.
Which statement matches the justifications established in this lesson?
The distance condition of a circle establishes the segment copy and the three equal sides.
The full angle-copy argument is deferred to lesson 4.5, where triangle congruence is available; a measured drawing supplies evidence with limited precision.
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