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Triangle Centers and Constructions: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A right-triangle comparison

Triangle ABC is right at C. CA is 28 cm and CB is 45 cm.

O, G, and H are its circumcenter, centroid, and orthocenter. Find CO, CG, and GO exactly. State the order of C, G, and O on their common line, and identify H.

Only the given right angle and leg lengths are shown.

Text description of this figure

ABC is right at C. The two legs CA and CB have lengths 28 and 45 centimeters. No centers or helper lines are given.
1. Hint 1
  Locate the circumcenter and orthocenter before comparing their positions with a median.
2. Hint 2
  The line from C to the hypotenuse midpoint is a median. Keep its whole length separate from its centroid portions.
Answer

$\begin{gathered}CO=\frac{53}{2}\text{ cm}\\CG=\frac{53}{3}\text{ cm}\\GO=\frac{53}{6}\text{ cm}.\end{gathered}$
The order is C, G, O; H is C.

Full solution

Pythagoras gives
$\begin{aligned}AB^2&=28^2+45^2\\AB^2&=2809\\AB&=53\text{ cm}.\end{aligned}$

O is the hypotenuse midpoint.
Its radius to C equals its radius to A, half AB.
Thus
$CO=\frac{53}{2}\text{ cm}.$

CO is a median because O is the midpoint of AB.
The centroid lies between C and O, with the vertex portion longer.
$\begin{aligned}CG&=\frac23\left(\frac{53}{2}\right)\\CG&=\frac{53}{3}\text{ cm}.\end{aligned}$

The remaining portion is
$\begin{aligned}GO&=\frac13\left(\frac{53}{2}\right)\\GO&=\frac{53}{6}\text{ cm}.\end{aligned}$

The order is C, G, O.
The two perpendicular legs are altitude lines meeting at C, so H is C, distinct from both interior G and hypotenuse midpoint O.

The same median connects the three center locations.

Answer

$\begin{gathered}CO=\frac{53}{2}\text{ cm}\\CG=\frac{53}{3}\text{ cm}\\GO=\frac{53}{6}\text{ cm}.\end{gathered}$
The order is C, G, O; H is C.

Key idea

Different center definitions can locate different points on one line.
Problem 2 A line through the off-line point

P is a given point on line l. Q is off l and is not on the perpendicular to l at P.

Using only a compass and unmarked straightedge, construct line $m$ perpendicular to $l$ at $P$ .

Then construct line $n$ through $Q$ perpendicular to $m$ .

Retain all construction arcs. Prove that $n$ is parallel to $l$ , and justify the equal-distance and perpendicular steps.

The two given points and original line are shown.

Text description of this figure

P is on line l. Q is off l and does not lie on the perpendicular to l at P. No answer line is supplied.
1. Hint 1
  Two successive perpendicular directions return to the original parallel direction.
2. Hint 2
  First make P a midpoint using a circle centered there. For the second construction, select two points on the first perpendicular equally far from Q.
3. Hint 3
  For the off-line construction, use a centered circle that cuts the line twice, then use the second intersection of equal circles through Q.
Answer

A valid construction has m perpendicular to l at P and n perpendicular to m through Q; n is parallel to l.

Full solution

Draw a positive-radius circle centered at P.
It cuts l at U and V on opposite sides of P.
Equal radii make P the midpoint of UV.

Draw equal-radius arcs centered at U and V with radius greater than half UV.
Choose an off-line crossing W.
Since W is equally far from U and V, it lies on their perpendicular bisector.
The midpoint P lies there too, so m, the line PW, is perpendicular to l at P.

Q is off m by hypothesis.
Draw a circle centered at Q with radius greater than QP.
P is inside that circle and on m, so the circle cuts m at two distinct points S and T.
The two radii QS and QT are equal.

Draw circles centered at S and T with radius SQ.
Nondegenerate triangle SQT and its equal sides give
$ST<2SQ.$
The circles therefore meet at Q and at a second point Z across m.

Both Q and Z are equally far from S and T.
Their line n is the perpendicular bisector of ST, hence perpendicular to m.
Since l is also perpendicular to m, n is parallel to l and passes through Q.

The two right angles justify the final parallel.

Answer

A valid construction has m perpendicular to l at P and n perpendicular to m through Q; n is parallel to l.

Key idea

A tool construction is justified by the equal distances its compass settings create.
Problem 3 An interior point and a circle

P is strictly inside triangle ABC. F, E, and D are its perpendicular feet on side segments AB, AC, and BC respectively. The given vertex distance is
$AP=\sqrt{221}\text{ cm}.$
The side portion is
$AF=10\text{ cm}.$
PE and PD are each 11 cm. Find PF and decide whether P is the incenter. At A, justify the equal-distance-to-bisector implication and its reverse by triangle congruence.

Explain why the circle centered at P with radius PF touches each side segment exactly once.

A schematic records the given perpendicular and lengths.

Text description of this figure

P is inside triangle ABC. The local right-triangle detail shows the perpendicular PF to AB, AF of length 10 centimeters, and AP of length square root of 221 centimeters. The detail is schematic, not to scale.
1. Hint 1
  Identify the hypotenuse in right triangle AFP before comparing the three line distances.
2. Hint 2
  Equal perpendicular distances from an interior point select internal bisectors. Use two vertices for uniqueness.
3. Hint 3
  For a point T of a side line different from its foot, compare PT with the perpendicular by Pythagoras.
Answer

$PF=11\text{ cm}$ . P is the incenter, and the circle has radius 11 cm and touches each side segment once.

Full solution

AP is the hypotenuse of right triangle AFP.
$\begin{aligned}PF^2&=221-10^2\\PF^2&=121\\PF&=11\text{ cm}.\end{aligned}$

At A, right triangles AFP and AEP have common hypotenuse AP and equal perpendicular legs PF and PE.
HL proves them congruent.
Their equal angles at A are the two interior parts because F and E lie on the forward side rays.
Thus AP is the internal bisector at A.

For the reverse implication, suppose AP is known to be the internal bisector.
The two right angles at F and E agree, the two angle parts at A agree, and AP is shared.
AAS then gives congruent triangles AFP and AEP, so their perpendicular legs PF and PE agree.
This supplies the forward equidistance direction without assuming it.

At B, right triangles BFP and BDP share hypotenuse BP and have equal perpendicular legs.
HL and the supplied forward-ray feet make BP the internal bisector at B.
The same HL argument applies at C.
In particular, P is the unique intersection of the two distinct internal bisectors at A and B, so it is the incenter.

If T is on line AB and differs from F, the nonzero side FT gives
$PT^2=PF^2+FT^2.$
Therefore PT is longer than PF, so only F lies on both that line and the circle.
The same argument works for the other two lines.
The supplied feet lie on the side segments, so these are side-segment contacts.

The circle stays in the triangle: to pass across a side line, a radius segment from P would first have to reach that line and then continue beyond its minimum distance of 11 cm.
A point of this circle is only 11 cm from P.
Its three contacts therefore touch the sides without crossing them.

Answer

$PF=11\text{ cm}$ . P is the incenter, and the circle has radius 11 cm and touches each side segment once.

Key idea

An incenter is identified by interior location and three perpendicular side distances.
Problem 4 Four vertices and one circle

ABCD is a nondegenerate parallelogram. Decide exactly when a circle can pass through all four vertices.

In the possible case, specify the center and compass radius. Prove both directions of your condition and prove uniqueness.

The initial diagram asserts only the given parallelogram.

Text description of this figure

ABCD is a nondegenerate parallelogram. No diagonal intersection, equal diagonal condition, center, or circle is supplied.
1. Hint 1
  A possible center must satisfy two endpoint equal-distance conditions, one for each diagonal.
2. Hint 2
  The diagonals of a parallelogram bisect each other. Compare their perpendicular bisectors at their common midpoint.
3. Hint 3
  Use the previously proved rectangle test for congruent diagonals after you locate the only possible center.
Answer

Such a circle exists exactly when the parallelogram is a rectangle. Its center is the diagonal intersection S and its radius is SA; it is unique.

Full solution

Let S be the intersection of the diagonals.
A parallelogram has S as the midpoint of both AC and BD.
Their distinct directions make their perpendicular bisectors nonparallel; both pass through S, so S is their only common point.

Suppose a circle through all four vertices has center O.
Its equal distances to A and C put O on the perpendicular bisector of AC.
Its equal distances to B and D put O on the perpendicular bisector of BD.
Thus O must be S.

The two radii SA and SB are equal.
Since S bisects both diagonals,
$\begin{gathered}AC=2SA\\BD=2SB.\end{gathered}$
The diagonals are congruent.
The congruent-diagonal test for a parallelogram therefore makes ABCD a rectangle.

Conversely, suppose ABCD is a rectangle.
Its congruent diagonals bisect each other.
Their four halves are equal, so a circle centered at S with radius SA passes through A, B, C, and D.

Every possible center has already been forced to be S.
The distance to A fixes the radius, so there is exactly one such circle.

The solution circle is shown only in the case where it exists.

Answer

Such a circle exists exactly when the parallelogram is a rectangle. Its center is the diagonal intersection S and its radius is SA; it is unique.

Key idea

Two endpoint loci can locate a forced center before a circle is drawn.
Problem 5 Two triangles in the same drawing

In nondegenerate triangle ABC, D, E, and F are the midpoints of BC, CA, and AB. G is the centroid of ABC.

On paper, construct the median from D of triangle DEF with a compass and unmarked straightedge. Then prove that G is also the centroid of DEF, identifying the order of the relevant points.

The original median AD and given centroid G are visible.

Text description of this figure

D, E, and F are the three side midpoints of ABC. G is the given centroid of ABC. The midpoint triangle is drawn, but no midpoint of EF is marked.
1. Hint 1
  Look for the midpoint M of EF on the original median AD.
2. Hint 2
  The midsegment parallels give similar triangles at A. Compare the halfway position on AD with the centroid position.
3. Hint 3
  Treat D as a vertex of DEF. Check the ordered ratio from D to G to M.
Answer

The two triangles have the same centroid G. If M is the midpoint of EF, the order on AD is A, M, G, D and DG is twice GM.

Full solution

Construct the perpendicular bisector of EF using equal-radius circles centered at E and F with radius greater than half EF.
Join their two crossings; this line meets EF at midpoint M.
Draw DM.
Since D is the opposite vertex, DM is the required median of DEF.

The positive midsegments form a nondegenerate triangle DEF: its sides EF and DE are parallel to the distinct side directions BC and AB.
Let line AD meet EF at T.

Since EF is parallel to BC, triangles AFT and ABD are similar by AA.
The midpoint F makes their scale factor one half.
Thus AT is half AD, and FT is half BD.
Similarly, triangles AET and ACD make ET half CD.

BD and CD are equal, so FT and ET are equal.
Therefore T is the midpoint M of EF.
This establishes that DM is a median of DEF.

The original centroid and the new midpoint satisfy
$\begin{gathered}AG=\frac23 AD\\AM=\frac12 AD.\end{gathered}$
Consequently the order is A, M, G, D.

The two portions from D toward M are
$\begin{gathered}DG=\frac13 AD\\GM=\frac16 AD.\end{gathered}$
Hence DG is twice GM.

The centroid of DEF is the unique interior point two thirds of its median DM from D.
G has exactly that ordered position.
It is therefore the centroid of DEF as well as ABC.

Read the new median from D toward M.

Answer

The two triangles have the same centroid G. If M is the midpoint of EF, the order on AD is A, M, G, D and DG is twice GM.

Key idea

Midsegment similarity can put an old centroid in a new median division.
Problem 6 A foot beyond the side

Triangle ABC is obtuse at B. AB is 47 cm and BC is 65 cm. The perpendicular from C to line AB has foot D beyond B, with BD of length 16 cm.

Construct this altitude on paper with a compass and unmarked straightedge, retaining the arcs. Then find CD and CA exactly, and justify which two right triangles you use.

The construction must reach the side extension.

Text description of this figure

ABC is obtuse at B. AB is 47 centimeters and BC is 65 centimeters. The supporting line is extended beyond B, but no perpendicular or foot is supplied.
1. Hint 1
  Extend the full side line before constructing the perpendicular from C.
2. Hint 2
  To find the altitude length, use triangle CBD with BC as its hypotenuse.
3. Hint 3
  The order A, B, D means the other right triangle uses the sum of AB and BD.
Answer

$\begin{gathered}CD=63\text{ cm}\\CA=63\sqrt2\text{ cm}.\end{gathered}$
A valid construction makes CD perpendicular to line AB beyond B.

Full solution

Draw the full line AB.
Draw a circle centered at C with a radius greater than CA, so A is strictly inside it and the line cuts it at two distinct points U and V.
Their equal distances from C put C on the perpendicular bisector of UV.

Construct that perpendicular bisector with two equal-radius compass crossings.
It contains C and is perpendicular to AB.
Its crossing with AB is D, on the supplied extension beyond B.

Triangle CBD is right at D, with hypotenuse BC.
$\begin{aligned}CD^2&=65^2-16^2\\CD^2&=3969\\CD&=63\text{ cm}.\end{aligned}$

The given order makes
$\begin{aligned}AD&=AB+BD\\AD&=47+16\\AD&=63\text{ cm}.\end{aligned}$

Triangle CAD is right at D, with legs CD and AD.
$\begin{aligned}CA^2&=63^2+63^2\\CA^2&=7938\\CA&=63\sqrt2\text{ cm}.\end{aligned}$

The extension is necessary: the angle at B is obtuse, so a right triangle with its foot on the forward ray BA would incorrectly contain both a right angle and that obtuse angle.

The order A, B, D determines the sum along the line.

Answer

$\begin{gathered}CD=63\text{ cm}\\CA=63\sqrt2\text{ cm}.\end{gathered}$
A valid construction makes CD perpendicular to line AB beyond B.

Key idea

An exterior foot changes segment addition before the right-triangle calculation.
Problem 7 A chain of images

One given nonidentity rotation maps A to B and B to C. The three points are distinct and noncollinear.

From those three points, construct the rotation center and a circle through all three with a compass and unmarked straightedge. Justify why the center you recover belongs to the given rotation, and why the circle is unique.

Include a congruence justification of both directions of the endpoint equal-distance relationship used by the construction. Handle the midpoint itself directly.

Both chords belong to the same given rotation.

Text description of this figure

A maps to B and B maps to C under one given nonidentity rotation. The three points are distinct and noncollinear. Only chords AB and BC are drawn.
1. Hint 1
  Compare the center-to-point distances along the chain of images.
2. Hint 2
  Construct two endpoint loci. The two chords cannot be parallel because the three points are noncollinear.
3. Hint 3
  Once the center is fixed, one vertex distance fixes a circle.
Answer

The perpendicular bisectors of AB and BC meet at the given rotation center O. The circle centered at O with radius OA is the unique circle through A, B, and C.

Full solution

Construct the perpendicular bisector of AB with equal-radius compass arcs whose radius exceeds half AB.
Construct the perpendicular bisector of BC in the same way.
Their original chords are nonparallel, so the two bisectors meet at a unique point O.

Here is the endpoint-locus justification for AB; the argument for BC is identical.
Let M be the midpoint of AB.
If X is on the perpendicular through M and differs from M, triangles XMA and XMB share XM, have equal halves MA and MB, and have equal included right angles.
SAS makes them congruent, so XA and XB agree.

If X is M itself, its two distances are MA and MB.
The midpoint definition gives their equality directly; the connector triangles have collapsed and need no congruence test.

Conversely, let Y have equal distances to A and B.
If Y is off line AB, triangles YMA and YMB have the matching endpoint distances, equal halves MA and MB, and shared YM.
SSS gives equal angles at M.
Those angles form a linear pair, so each is right; Y lies on the perpendicular bisector.

If Y is on line AB, segment addition excludes points at or beyond either endpoint from having equal endpoint distances.
Between A and B, equality selects exactly their midpoint M, which belongs to the perpendicular bisector.
Thus both directions include every possible point.

Let K be the center of the given rotation.
Preservation of distance from that center gives
$\begin{gathered}KA=KB\\KB=KC.\end{gathered}$
The converse places K on both constructed bisectors, so K must be their unique intersection O.

The forward direction now makes O equally far from A and B and from B and C.
Set the compass at O with radius OA and draw the circle.
It passes through all three points, so O is both the rotation center and the circumcenter of ABC.

Any circle through these three noncollinear points must have its center on the same two perpendicular bisectors.
The center is therefore O, and the distance OA fixes the radius.
The shared-rotation hypothesis links the two mappings before this construction; arbitrary image pairs do not automatically share one directed turn.

The recovered center need not be inside the triangle.

Answer

The perpendicular bisectors of AB and BC meet at the given rotation center O. The circle centered at O with radius OA is the unique circle through A, B, and C.

Key idea

A chain of one rotation can turn two point-image conditions into a circumcircle construction.
Problem 8 A supplied radius

A nondegenerate triangle ABC and a nonzero segment RS are given on paper. Let t be the length of RS.

Describe how to decide, with compass and straightedge, whether a circle of radius t can have its disk contained in ABC and touch all three side segments. Give the construction when possible and prove your condition is necessary and sufficient.

Justify the angle-bisector compass settings, the uniqueness of the possible center, and the choice of radius.

The triangle has no preconstructed center or circle.

Text description of this figure

An illustrative nondegenerate triangle ABC is shown. The nonzero segment RS is supplied separately and has length t. No internal bisectors, center, or circle are shown.
1. Hint 1
  Find the geometric point that any circle satisfying all three side conditions would have to use.
2. Hint 2
  At an angle vertex, equal first radii and equal later radii create two side pairs for SSS.
3. Hint 3
  Compare the supplied segment with the perpendicular from the constructed center to a side; do not choose the center to compensate for a different radius.
Answer

Construct the incenter I and a perpendicular IF to a side. The required circle exists exactly when RS and IF have equal lengths; then its center is I and radius IF, and it is unique.

Full solution

At A, draw an arc meeting rays AB and AC at X and Y.
Keep that positive radius for the circles centered at X and Y, and take their second intersection Z strictly inside angle A.
Equal radii give AX equal to AY and XZ equal to YZ.
With shared AZ, SSS proves that ray AZ internally bisects the angle.

Repeat at B and label the interior intersection of the two internal bisectors I.
Construct the perpendicular from I to AB with foot F.
Compare IF with RS by copying the supplied length with the compass.

If those lengths agree, draw the circle centered at I with radius IF.
The internal-bisector theorem links the three perpendicular distances; the interior converse includes the third bisector.
Each foot lies on its side segment because the internal bisectors at both endpoints place it on both forward rays.

For each side line, any point other than the foot is farther from I by Pythagoras.
The radius therefore reaches that line exactly at its foot.
A point across a side line would be farther than the perpendicular distance, so the disk stays inside ABC.
This proves sufficiency.

For necessity, suppose a positive-radius disk fits inside and its circle touches every side.
Its center P is strictly inside the triangle.
To reach a side, the radius cannot be less than the perpendicular distance to its line; to keep the disk inside, it cannot be greater.
Hence each of the three perpendicular distances is t.

The interior-point converse at A and B places P on the same two distinct internal bisectors, whose only intersection is I.
Thus the center cannot be changed to fit another radius.
The radius must be IF, so a circle exists only when RS and IF agree.
The center and radius are then both fixed, proving uniqueness.

Compare the constructed perpendicular radius with the supplied segment.

Answer

Construct the incenter I and a perpendicular IF to a side. The required circle exists exactly when RS and IF have equal lengths; then its center is I and radius IF, and it is unique.

Key idea

A prescribed radius must match the unique perpendicular distance selected by the triangle.
Problem 9 A circle from an altitude intersection

Through A, B, and C of nondegenerate triangle ABC, draw the parallels to BC, CA, and AB respectively. They form triangle PQR, with A between Q and R, B between R and P, and C between P and Q.

Let H be the intersection of the altitude lines from A and B of ABC. Explain how to draw a circle through P, Q, and R using H and one compass setting, without separately constructing the side bisectors of PQR.

Prove why the procedure works. Establish all three auxiliary midpoint roles from the parallelograms, account for the third original altitude, and justify uniqueness of the circle.

The midpoint conclusions are left for the proof.

Text description of this figure

The parallels through A, B, and C form PQR. The supplied between-point orders are shown, but there are no equality ticks or constructed altitude lines.
1. Hint 1
  The parallel pairs make quadrilaterals whose opposite sides identify the two halves of each auxiliary side.
2. Hint 2
  Compare the original altitude through a vertex with the auxiliary side passing through that vertex.
3. Hint 3
  After proving both the midpoint and perpendicular conditions, apply the circle-center theorem to PQR.
Answer

Use center H and radius HP. The three original altitude lines are the auxiliary side perpendicular bisectors, so H is also on the third altitude and is the unique circumcenter of PQR.

Full solution

The parallels make ABCQ, ACBR, and ABPC parallelograms.
Opposite sides of the first two give the two lengths at A:
$AQ=BC.$
Also,
$AR=BC.$
With A between Q and R, A is their midpoint.

The parallelograms ACBR and ABPC give
$RB=AC.$
Also,
$BP=AC.$
With B between R and P, B is the midpoint of RP.

Parallelograms ABPC and ABCQ give
$PC=AB.$
Also,
$CQ=AB.$
With C between P and Q, C is the midpoint of PQ.

The altitude from A is perpendicular to BC, hence to its parallel QR.
Since it passes through midpoint A, it is the perpendicular bisector of QR.
Likewise the altitude from B is the perpendicular bisector of RP, and the altitude from C is the perpendicular bisector of PQ.

The auxiliary triangle is nondegenerate because its sides follow the three distinct original side directions.
Its perpendicular-bisector concurrence theorem applies.
The first two lines meet at H, so the third passes through H too; this also proves the third original altitude contains H.

The circumcenter property gives
$\begin{gathered}HP=HQ\\HQ=HR.\end{gathered}$
Draw the circle centered at H with compass opening HP.
It passes through all three auxiliary vertices.

Any other circle through P, Q, and R would have its center on the same two intersecting perpendicular bisectors.
Thus its center would be H and its radius HP, proving uniqueness.
The midpoint facts belong to the auxiliary sides, not the original altitude feet.

The original orthocenter is the auxiliary circumcenter.

Answer

Use center H and radius HP. The three original altitude lines are the auxiliary side perpendicular bisectors, so H is also on the third altitude and is the unique circumcenter of PQR.

Key idea

The same lines can locate an altitude intersection and a different triangle’s circle center.
Problem 10 Two circles for one triangle

ABC is equilateral, with each side $22\sqrt3$ cm. D is the midpoint of BC.

Identify a point K that is all four triangle centers and prove why the defining constructions coincide. Find the circumradius and inradius exactly, and state the two compass settings for their circles.

Only the equilateral triangle and given side length are shown.

Text description of this figure

ABC is equilateral and each side is 22 square root of 3 centimeters. No center, midpoint, circle, or altitude is marked.
1. Hint 1
  Compare triangles ABD and ACD before assigning all four roles to AD.
2. Hint 2
  The right triangle formed by half a side gives the whole median length.
3. Hint 3
  The common center is also the centroid. Read the longer and shorter portions in vertex-to-midpoint order.
Answer

K is the common intersection of the medians. The circumcircle uses center K and radius KA of 22 cm; the incircle uses center K and perpendicular radius KD of 11 cm.

Full solution

AD is a median by the midpoint definition.
Triangles ABD and ACD have equal sides AB and AC, equal halves BD and DC, and common AD.
SSS makes them congruent.
Their equal angles at A make AD an internal angle bisector.

The angles at D are equal and form a straight angle, so they are right angles.
Thus AD is an altitude, and its full line is the perpendicular bisector of BC.
Repeat from the other vertices.
The four defining line families are the same, so their unique intersection K is all four centers.

Half of BC is $11\sqrt3$ cm.
The two squares in right triangle ABD are
$\begin{gathered}AB^2=1452\text{ cm}^2\\BD^2=363\text{ cm}^2.\end{gathered}$
Pythagoras gives
$\begin{aligned}AD^2&=1452-363\\AD^2&=1089\\AD&=33\text{ cm}.\end{aligned}$

K is the centroid, so
$\begin{aligned}AK&=\frac23(33)\\AK&=22\text{ cm}.\end{aligned}$

The shorter perpendicular portion is
$\begin{aligned}KD&=\frac13(33)\\KD&=11\text{ cm}.\end{aligned}$

K is also the circumcenter, so center K and opening KA draw the circle through the vertices.
As the incenter, K has perpendicular side distance KD, so center K and opening KD draw the incircle.
Sharing a center does not make these two radii equal.

The vertex distance is twice the perpendicular side distance.

Answer

K is the common intersection of the medians. The circumcircle uses center K and radius KA of 22 cm; the incircle uses center K and perpendicular radius KD of 11 cm.

Key idea

Coinciding constructions identify the center; the ordered median division distinguishes its two circle radii.

Multiple choice

Core practice