QUESTION IMAGE
Question
use the diagram and proof below to answer questions 10-11.
statement | reason
--- | ---
\\(\overline{jk} \cong \overline{dk}\\) and \\(\overline{ae}\\) bisects \\(\overline{hc}\\) | 1) given
- \\(\overline{hk} \cong \overline{ck}\\) | 2) ______ (#10.)
- ______ (#11.) | 3) vertical angles are congruent
- \\(\triangle cdk \cong hjk\\) | 4) side-angle-side
***#10.) which reason completes line 2 of the proof above?
(a) angle bisector definition
(b) midpoint definition
(c) segment bisector definition
(d) isosceles triangle definition
#11.) which statement completes line 3 of the proof above?
(a) \\(\angle kjh \cong \angle kdc\\)
(b) \\(\angle khj \cong \angle kcd\\)
(c) \\(\angle jkh \cong \angle dkc\\)
(d) \\(\angle fke \cong \angle bka\\)
**#12.) \\(\triangle mno\\) is an isosceles triangle with base angles \\(\angle n\\) and \\(\angle o\\). given that \\(\overline{mo} = 5y + 7\\) in., \\(\overline{mn} = 4y + 12\\) in., and \\(\overline{on} = 2y + 10\\) in., determine the perimeter of the triangle.
(a) 40 in
(b) 84 in
(c) 18 in.
(d) 21 in.
Question 10
To determine the reason for \(\overline{HK} \cong \overline{CK}\), we analyze the definitions:
- Angle Bisector Definition: Applies to angles, not segments. Eliminate (A).
- Midpoint Definition: A midpoint divides a segment into two congruent parts. But here, \(\overline{AE}\) bisects \(\overline{HC}\), so \(K\) is the midpoint of \(\overline{HC}\), making \(\overline{HK} \cong \overline{CK}\). This matches.
- Segment Bisector Definition: Describes a line/segment bisecting another, not the congruence of sub - segments. Eliminate (C).
- Isosceles Triangle Definition: Relates to triangle sides/angles, not segment bisecting. Eliminate (D).
Vertical angles are formed by intersecting lines and are congruent. We need vertical angles related to the triangles \(\triangle CDK\) and \(\triangle HJK\).
- \(\angle JKH\) and \(\angle DKC\) are vertical angles (formed by intersecting lines \(JK\) - \(DK\) and \(HK\) - \(CK\)).
- Option (A): \(\angle KJH\) and \(\angle KDC\) are not vertical angles.
- Option (B): \(\angle KHJ\) and \(\angle KCD\) are not vertical angles.
- Option (D): \(\angle FKE\) and \(\angle BKA\) are vertical angles but not related to the triangles in the proof.
Step 1: Identify equal sides in isosceles triangle
In \(\triangle MNO\), base angles are \(\angle N\) and \(\angle O\), so the equal sides (legs) are \(\overline{MO}\) and \(\overline{NO}\)? Wait, no—wait, base angles are \(\angle N\) and \(\angle O\), so the legs are \(\overline{MN}\) and \(\overline{MO}\)? Wait, no, in an isosceles triangle, the sides opposite equal angles are equal. Base angles \(\angle N\) and \(\angle O\) are opposite \(\overline{MO}\) and \(\overline{MN}\) respectively? Wait, no, let's correct: In \(\triangle MNO\), base angles are \(\angle N\) and \(\angle O\), so the sides opposite them (\(\overline{MO}\) and \(\overline{MN}\)) are equal. So \(MO = MN\).
Given \(MO=5y + 7\), \(MN = 4y+12\). Set them equal:
\(5y + 7=4y + 12\)
Step 2: Solve for \(y\)
Subtract \(4y\) from both sides: \(y+7 = 12\)
Subtract 7 from both sides: \(y = 5\)
Step 3: Find side lengths
- \(MO=5(5)+7=25 + 7=32\)? Wait, no, that can't be. Wait, maybe I mixed up the sides. Wait, base angles are \(\angle N\) and \(\angle O\), so the equal sides are \(\overline{MN}\) and \(\overline{ON}\)? Wait, no, let's re - examine the triangle label. The triangle is labeled \(M\), \(N\), \(O\) with base \(MN\)? Wait, the diagram shows \(M\) and \(N\) at the base, \(O\) at the top. So base angles are \(\angle M\) and \(\angle N\)? Wait, the problem says "base angles \(\angle N\) and \(\angle O\)". So the legs are \(MO\) and \(NO\)? Wait, no, in a triangle, the base angles are the two angles adjacent to the base. If base angles are \(\angle N\) and \(\angle O\), the base is \(MN\), and the legs are \(MO\) and \(NO\). Wait, but then \(MO = NO\). Wait, \(MO = 5y+7\), \(ON=2y + 10\). Let's try that.
Set \(MO = ON\):
\(5y+7=2y + 10\)
Subtract \(2y\) from both sides: \(3y+7 = 10\)
Subtract 7: \(3y=3\)
\(y = 1\)
Wait, that gives \(MO=5(1)+7 = 12\), \(ON=2(1)+10 = 12\), \(MN=4(1)+12 = 16\). Perimeter would be \(12 + 12+16=40\). Wait, but let's check again. Wait, maybe the equal sides are \(MN\) and \(MO\). Let's try \(MN = MO\):
\(4y + 12=5y+7\)
Subtract \(4y\): \(12=y + 7\)
Subtract 7: \(y = 5\)
Then \(MO=5(5)+7 = 32\), \(MN=4(5)+12 = 32\), \(ON=2(5)+10 = 20\). Perimeter \(32 + 32+20=84\). Wait, but the options include 84 (option B). Wait, maybe I misidentified the equal sides. Let's recall: In an isosceles triangle, the sides opposite the base angles are equal. Base angles are \(\angle N\) and \(\angle O\), so the sides opposite them are \(\overline{MO}\) (opposite \(\angle N\)) and \(\overline{MN}\) (opposite \(\angle O\))? No, \(\angle N\) is at vertex \(N\), opposite side is \(\overline{MO}\); \(\angle O\) is at vertex \(O\), opposite side is \(\overline{MN}\). So if \(\angle N=\angle O\), then \(\overline{MO}=\overline{MN}\). So \(MO = MN\):
\(5y + 7=4y+12\)
\(y = 5\)
Then \(MO = 5(5)+7=32\), \(MN=4(5)+12 = 32\), \(ON=2(5)+10 = 20\). Perimeter \(=32 + 32+20=84\) inches.
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B. Midpoint Definition