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  5. 6.1 Perpendicular And Angle Bisector

6.1 Perpendicular and Angle Bisector

Authored by Anne Madridano

Mathematics

9th - 11th Grade

CCSS covered

Used 97+ times

6.1 Perpendicular and Angle Bisector
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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Point C is in the interior of ∠DEF. If ∠DEC and ∠CEF are congruent, then ray EC is the _________________________ of ∠DEF. 

angle bisector

perpendicular bisector

midpoint

Tags

CCSS.HSG.CO.C.9

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Media Image

Find GH.

4.6

3.6

9.2

7.2

Tags

CCSS.HSG.CO.C.9

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Media Image

Find QR.

4.7

1.3

2.6

Tags

CCSS.HSG.CO.C.9

4.

FILL IN THE BLANKS QUESTION

3 mins • 1 pt

Media Image

Find AB.

(a)  

Tags

CCSS.HSG.CO.C.9

5.

FILL IN THE BLANKS QUESTION

3 mins • 1 pt

Media Image

Find UW.

(a)  

Tags

CCSS.HSG.CO.C.11

6.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

Tell whether the information in the diagram allows you to conclude that point P lies on the perpendicular bisector of LM. Explain your reasoning.

No. You would need to know that LK ‾ ⊥ MK‾\overline{LK\ }\ \perp\ \overline{MK} .

No. You would need to know that KP‾ ⊥ LM‾\overline{KP}\ \perp\ \overline{LM} .

Yes. LN‾\overline{LN} ≅ MN‾\overline{MN} and NK→\overrightarrow{NK} ⊥\perp LM‾\overline{LM} , so NK→\overrightarrow{NK} is the perpendicular bisector of LM‾\overline{LM} and P is on NK→\overrightarrow{NK} .

Yes. NK→\overrightarrow{NK} ⊥ LM‾\overline{LM} , so NK→\overrightarrow{NK} is the perpendicular bisector of LM‾\overline{LM} and P is on NK→\overrightarrow{NK}

Tags

CCSS.HSG.CO.C.9

7.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Media Image

Tell whether the information in the diagram allows you to conclude that point P lies on the perpendicular bisector of LM. Explain your reasoning.

No. You would need to know that m∠MNP = 90º.

No. You would need to know that either  LN‾ ≅ MN‾\overline{LN}\ \cong\ \overline{MN}  or  LP‾ ≅ MP‾\overline{LP}\ \cong\ \overline{MP}  

Yes.  LN‾ ≅ NP‾\overline{LN}\ \cong\ \overline{NP}  and  NP→ ⊥ LM‾\overrightarrow{NP}\ \perp\ \overline{LM}  so  NP‾\overline{NP}  is the perpendicular bisector of  LM‾\overline{LM}  .

Tags

CCSS.HSG.CO.C.9

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