Question

name
5 - 2 lesson quiz
envision geometry
savvasrealize.com
bisectors in triangles
use the diagram shown for items 1 and 2.

1. select all the conditions that would be enough to prove

that ( p ) is the incenter of ( \triangle h j k ).
( square ) a. ( l, m ), and ( n ) are the midpoints of ( overline{h k}, overline{h j} ) and ( overline{k j} ).
( square ) b. ( overline{p l} cong overline{p m} cong overline{p n} )
( square ) c. ( overline{p k} cong overline{h p} cong overline{p j} )
( square ) d. ( \triangle h j k ) is an acute triangle.
( square ) e. ( overline{k p}, overline{h p} ), and ( overline{j p} ) are angle bisectors of the triangle

1. assume ( p ) is the incenter of ( \triangle h j k ). if ( l p = 4 x + 10 ) and ( m p = 8 x - 2 ),

what is the radius of the inscribed circle of ( \triangle h j k )?
(a) 3
(b) 12
(c) 18
(d) 22

1. what is the radius of the circumscribed circle

of ( \triangle a b c )?
(a) 1.3
(b) 4.0
(c) 5.6
(d) 5.8

1. which point is the center of a circle that contains

( r, s ), and ( t )?
(a) ( l )
(c) ( n )
(b) ( m )
(d) ( p )

1. when constructing the incenter of a triangle, why

do you first construct the angle bisectors of two
angles? select the true statement.
(a) because you need the angle bisectors of all the angles.
(b) angle bisectors are perpendicular to each other.
(c) angle bisectors are equidistant from the vertices of the triangle.
(d) the intersections of angle bisectors are equidistant from the sides
of the triangle.

Explanation:

Step1: Use the property of incenter

The incenter of a triangle is equidistant from the sides of the triangle. If \(P\) is the incenter, then \(PL = PM=PN\) (where \(PL\), \(PM\), \(PN\) are the distances from \(P\) to the sides of \(\triangle HJK\)). Also, the incenter is the intersection of the angle - bisectors of the triangle. So, if \(\overline{KP}\), \(\overline{HP}\), and \(\overline{JP}\) are angle bisectors of the triangle, \(P\) is the incenter.

• Option A: Mid - points are related to the centroid (using the mid - point theorem for the centroid, not the incenter).
• Option B: Since \(PL\), \(PM\), \(PN\) are the distances from \(P\) to the sides of \(\triangle HJK\), if \(PL\cong PM\cong PN\), \(P\) is equidistant from the sides, so \(P\) is the incenter.
• Option C: \(PK\), \(HP\), \(PJ\) are segments from \(P\) to the vertices. The incenter is equidistant from the sides, not necessarily from the vertices.
• Option D: The type of triangle (acute) does not determine the incenter.
• Option E: If \(\overline{KP}\), \(\overline{HP}\), and \(\overline{JP}\) are angle bisectors, by the definition of the incenter (the intersection of angle bisectors), \(P\) is the incenter.

Step2: Solve for \(x\) in problem 2

Since \(P\) is the incenter, \(PL = PM\). Given \(PL = 4x + 10\) and \(PM=8x - 2\).
Set up the equation \(4x + 10=8x - 2\).
Subtract \(4x\) from both sides: \(10 = 4x-2\).
Add 2 to both sides: \(12 = 4x\).
Divide both sides by 4: \(x = 3\).
Then \(PL=4x + 10=4\times3+10=12 + 10=22\) (Since \(PL\) is the radius of the inscribed circle, and \(PL = PM\) when \(P\) is the incenter)

Step3: Analyze problem 5

The incenter of a triangle is the point of intersection of the angle bisectors of the triangle. The incenter is equidistant from the sides of the triangle.
When constructing the incenter, we construct the angle bisectors of two angles (the third angle bisector will pass through the intersection of the first two by the property that the angle bisectors of a triangle are concurrent). The reason is that the intersection of angle bisectors (by the Angle - Bisector Theorem) is equidistant from the sides of the triangle.

Answer:

1. B. \(\overline{PL}\cong\overline{PM}\cong\overline{PN}\), E. \(\overline{KP}\), \(\overline{HP}\), and \(\overline{JP}\) are angle bisectors of the triangle
2. D. 22
3. D. The intersections of angle bisectors are equidistant from the sides of the triangle.