Constructions and Tilings: questions with answers

CA = CB because the crossing arcs have equal radii. OC is common to triangles OAC and OBC. SSS also requires OA = OB, but 8 cm and 5 cm are unequal. The missing equality is OA = OB, so the standard SSS angle-bisection proof does not apply.

D is the intersection beyond chord AB inside the angle. OA = OB, DA = DB and OD common give SSS, so OD bisects the angle into 24 degrees. The other intersection need not lie between O and AB; its position depends on the radii.

A perpendicular bisector passes through a segment's midpoint and meets it at 90 degrees. Line n already passes through the midpoint of AB. It meets AB at 55 degrees, so the missing requirement is a right angle.

Bisection divides the angle into two equal parts. Each part is 120 ÷ 2 = 60 degrees. A third would be 40 degrees.

B lies on the perpendicular bisector because BX = BY = 8 cm. A does not, since AX = 8 cm and AY = 6 cm differ. A line through a point off the bisector cannot be that bisector.

A is equally distant from X and Y, so it lies on their perpendicular bisector. C must also be equally distant from X and Y. Since XC is 4 cm, the arc from Y must have radius 4 cm. The different radii 6 cm and 4 cm make A and C distinct. These two points determine the perpendicular bisector even though they are on the same side of XY.

OX = OY = 4 cm, so O is the midpoint of XY. AX = AY = 5 cm, so A lies on the perpendicular bisector of XY. The two distinct points O and A fix this line. Therefore, OA is perpendicular to XY through O, and 0 further arc crossings are needed.

PX and PY both equal 7 cm, so P lies on the perpendicular bisector of XY. One more point equally distant from X and Y, distinct from P, determines this line together with P. For example, circles centred at X and Y with radius 7 cm intersect at P and a second point Q. The line through P and Q is perpendicular to l.

Two triangles meet at each outer corner. Each contributes 60 degrees, so the corner angle is 60 + 60 = 120 degrees. The apex angles (at the common point) meet at the centre. The base angles meet in pairs at the outer corners.

Each triangle contributes 18 degrees at the common point. The 20 angles total 20 × 18 degrees = 360 degrees, a complete angle. One fewer triangle leaves a gap of 18 degrees. One more overlaps by 18 degrees. The angles at the common point determine the number needed.

For the same nonzero chord, a larger circle radius gives a smaller minor angle; a smaller radius gives a larger one, provided the chord fits. Here the radius changes from 3 to 5 cm, so the copied angle becomes smaller. Equal chord lengths alone do not preserve the angle.

Transfer the straight chord BC by compass. With the vertex radii equal, matching ZY = BC completes three corresponding side equalities. The compass opening measures a straight distance between endpoints, not distance along a curved arc.

The corresponding angle must be 74 degrees. An adjacent angle forming a straight line with it must therefore be 180 − 74 = 106 degrees. Copying 74 into that adjacent position gives the wrong line for these non-right angles.

The corresponding position would require 121 degrees. Reflection across the transversal instead makes that position 59 degrees. Since 121 is not 90 degrees, these differ, so the new line is not parallel to m and the two lines intersect when extended.

Every grid-aligned domino covers one square of each checkerboard colour. Removing this square leaves 7 of each. Pair rows above and below; pair the row-3 cells. Equal colour counts alone are not a general proof that an arbitrary region can be tiled.