ANGLE BISECTORS OF A TRIANGLE WORKSHEET

Problems 1-6 : Each figure shows a triangle with one of its angle bisectors (in blue color).

Problem 1 :

In ΔABC below, if m∠ABC = 34°, find m∠ABD.

anglebisectorofatriangle1.png

Problem 2 :

In ΔPQR below, if m∠SQR = 13°, find m∠PQR.

anglebisectorofatriangle2.png

Problem 3 :

In ΔWXY below, solve for x, if

m∠WXZ = (4x + 5)°

m∠ZXY = (5x - 2)°

anglebisectorofatriangle3.png

Problem 4 :

In ΔABC below, solve for x, if

m∠BAD = (1 + 28x)°

m∠BAC = (59x - 1)°

anglebisectorofatriangle4.png

Problem 5 :

In ΔMNO below, solve for x, if

m∠MNP = (7x + 7)°

m∠MNO = (16x + 4)°

anglebisectorofatriangle5.png

Problem 6 :

In ΔABC below, find m∠ABC, if

m∠ABD = (7x + 5)°

m∠DBC = (9x - 5)°

anglebisectorofatriangle6.png

Problems 7-12 : Each figure shows a triangle with one of its angle bisectors (in blue color).

Problem 7 :

In ΔABC below, if MD = 3, find ME.

anglebisectorofatriangle11.png

Problem 8 :

In ΔJKL below, if MQ = 7, find RM.

anglebisectorofatriangle12.png

Problem 9 :

In ΔABC below, if RM = 5, find QM.

anglebisectorofatriangle7.png

Problem 10 :

In ΔDEF below, if CD = 2 and DM = 3, find MB.

anglebisectorofatriangle8.png

Problem 11 :

In ΔWXY below, if MQ = 3 and XQ = 4, find XM.

anglebisectorofatriangle9.png

Problem 12 :

In ΔSTU below, if MK = 1 and SL = 2, find SM.

anglebisectorofatriangle10.png
tutoring.png

Answers

1. Answer :

anglebisectorofatriangle1.png

In ΔABC above, since BD is the angle bisector of ∠ABC,

m∠ABD = 17°

2. Answer :

anglebisectorofatriangle2.png

In ΔPQR above, since QS is the angle bisector of ∠PQR,

m∠PQS = m∠SQR

m∠PQS = 13°

m∠PQR = m∠PQS + m∠SQR

m∠PQR = 13° + 13°

m∠PQR = 26°

3. Answer :

anglebisectorofatriangle3.png

In ΔWXY above, since XZ is the angle bisector of ∠WXY,

m∠WXZ = m∠ZXY

(4x + 5)° = (5x - 2)°

4x + 5 = 5x - 2

-x = -7

x = 7

4. Answer :

anglebisectorofatriangle4.png

In ΔABC above, since AD is the angle bisector of ∠BAC,

m∠BAD = m∠DAC

m∠BAD = (1 + 28x)°

m∠BAC = m∠BAD + m∠DAC

(59x - 1)° = (1 + 28x)° + (1 + 28x)°

59x - 1 = 1 + 28x + 1 + 28x

59x - 1 = 56x + 2

3x = 3

x = 1

5. Answer :

anglebisectorofatriangle5.png

In ΔMNO above, since NP is the angle bisector of ∠MNO,

m∠PNO = m∠MNP

m∠PNO = (7x + 7)°

m∠MNO = m∠MNP + m∠PNO

(16x + 4)° = (7x + 7)° + (7x + 7)°

16x + 4 = 7x + 7 + 7x + 7

16x + 4 = 14x + 14

2x = 10

x = 5

6. Answer :

anglebisectorofatriangle6.png

In ΔABC above, since BD is the angle bisector of ∠ABC,

m∠ABD = m∠DBC

(7x + 5)° = (9x - 5)°

7x + 5 = 9x - 5

-2x = -10

x = 5

x = 7

m∠ABC = m∠ABD + m∠DBC

m∠ABC = (7x + 5)° + (9x - 5)°

Substitute x = 5.

m∠ABC = (7 ⋅ 5 + 5)° + (9 ⋅ 5 - 5)°

m∠ABC = (35 + 5)° + (45 - 5)°

m∠ABC = 40° + 40°

m∠ABC = 80°

Key concept for problems 7-12 :

The perpendicular distance from the incenter (point of intersection of the three angle bisectors) of a triangle to each side of the triangle is same. 

7. Answer :

anglebisectorofatriangle11.png

In ΔABC above, since AM, BM and CM are angle bisectors,

MD = ME = MF

Since MD = 3,

ME = 3

8. Answer :

anglebisectorofatriangle12.png

In ΔJKL above, since JM, KM and LM are angle bisectors,

MP = MQ = MR

Since MQ = 7,

MR = 7

or

RM = 7 

9. Answer :

anglebisectorofatriangle7.png

In ΔABC above, since AM, BM and CM are angle bisectors,

PM = QM = RM

Since RM = 5,

QM = 5

10. Answer :

anglebisectorofatriangle8.png

In ΔDEF above, since DM, EM and FM are angle bisectors,

MA = MC = MB

In right triangle DMC, using Pythagorean Theorem,

MC2 + CD2 = DM2

MC2 + 22 = 32

MC2 + 4 = 9

MC2 = 5

MC = √5

Since MA = MC = MB,

MB = √5

11. Answer :

anglebisectorofatriangle9.png

In right triangle MQX, using Pythagorean Theorem,

XM2 XQ2 + MQ2

XM2 = 32 + 42

XM2 = 9 + 16

XM2 = 25

XM2 = 52

XM = 5

12. Answer :

anglebisectorofatriangle10.png

In ΔSTU above, since SM, TM and UM are angle bisectors,

MJ = MK = ML

Since MK = 1,

ML = 1

In right triangle SML, using Pythagorean Theorem,

SM2 = SL2 ML2

SM2 = 22 + 12

SM2 = 4 + 1

SM2 = 5

SM = √5

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