12 multiple-choice questions, progressively harder.
P lies on an internal bisector. Its perpendicular distances to the two side lines are (4x−3)(4x-3)(4x−3) cm and (2x+9)(2x+9)(2x+9) cm. Find their common length.
Solution
Correct answer: C
Set the perpendicular lengths equal.
Substitute into either expression.
I is the incenter and IF is perpendicular to side AB. AI is 17 cm and AF is 15 cm. What is the inradius?
Correct answer: A
AI is the hypotenuse of right triangle AFI.
The positive perpendicular length IF is the inradius.
I is the incenter of triangle ABC, with F the perpendicular foot on AB. Why does using radius IA fail to draw the incircle?
Correct answer: B
In right triangle AFI, IA is the hypotenuse and IF is a leg. Radius IA would reach beyond side line AB rather than just reach its foot.
P is inside angle UOV. Its perpendicular feet are X,Y and PX equals PY. Which congruence test applies directly once both feet are known to be on the forward rays?
Correct answer: D
The right triangles share hypotenuse OP and have equal perpendicular legs PX and PY. These are exactly the HL hypotheses.
Which construction is sufficient to locate the incenter of a nondegenerate triangle?
Concurrency puts the intersection of two internal bisectors on the third one. One bisector leaves many possible points; perpendicular bisectors locate a different center.
I is inside triangle ABC and lies on the internal bisectors at A and B. Its perpendicular distance to AB is 4.6 cm. What is its perpendicular distance to BC?
The bisector at B gives equal distances to lines BA and BC. The given distance to AB is therefore also the distance to BC.
The first compass arc at O has radius 5 cm. The later arcs centered at U and V both have radius 7 cm and meet at X strictly inside the angle. Is OX an internal bisector?
SSS uses three matching pairs.
The first pair and second pair do not need to have the same length as each other.
Triangle ABC has ∠A=52∘\angle A=52^\circ∠A=52∘ and ∠B=68∘\angle B=68^\circ∠B=68∘. I is its incenter. What is ∠ACI\angle ACI∠ACI?
First use the triangle angle sum.
CI halves that angle.
In triangle ABC, I is the incenter, ∠A=40∘\angle A=40^\circ∠A=40∘ and ∠B=80∘\angle B=80^\circ∠B=80∘. What is ∠AIB\angle AIB∠AIB?
Triangle AIB has angles 20 degrees at A and 40 degrees at B, because AI and BI are internal bisectors.
The first arc gives OU=OV. Later arcs use unequal radii and meet inside the angle at X. Which SSS equality is no longer guaranteed?
The unequal later settings need not give UX=VX. The original equal radii and the shared side still remain, but two matching side pairs do not establish SSS.
For the incenter I, the perpendicular foot F to line AB lies on ray AB and also on ray BA. Where must F lie?
The overlap of ray AB with ray BA is segment AB. The acute half-angles at the vertices rule out endpoints, so the foot is strictly between A and B.
Which statement completes the concurrency proof? I is interior, its distances to BC and CA are equal, so ...
The two equal perpendicular distances concern the side lines of angle C, and I lies inside that angle. Its interior-point converse identifies the third internal bisector.
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