kuta software - infinite geometry
angle bisectors of triangles

each figure shows a triangle with one of its angle bisectors.
1) (m\angle sut = 34^{\circ}). find (m\angle 1).
2) find (m\angle sqr) if (m\angle 2 = 13^{\circ}).

each figure shows a triangle with its three angle bisectors intersecting at point p.
3) (pt = 3). find (pu).
4) find (pv) if (pw = 7).
5) find (pw) if (px = 5).
6) find (pd) if (pc = 8).
7) (py = 2) and (hp = 3). find (hy).
8) find (ap) if (pq = 1) and (ar = 2).

Question

kuta software - infinite geometry
angle bisectors of triangles

each figure shows a triangle with one of its angle bisectors.
1) (m\angle sut = 34^{\circ}). find (m\angle 1).
2) find (m\angle sqr) if (m\angle 2 = 13^{\circ}).

each figure shows a triangle with its three angle bisectors intersecting at point p.
3) (pt = 3). find (pu).
4) find (pv) if (pw = 7).
5) find (pw) if (px = 5).
6) find (pd) if (pc = 8).
7) (py = 2) and (hp = 3). find (hy).
8) find (ap) if (pq = 1) and (ar = 2).

Accepted Answer

{ "quality": "clear", "question_count": 8, "skills_matched": [ "step_cot", "table_formatter", "explain_then_answer", "tutor" ], "needs_drawing": false, "concepts_used": [], "new_concepts": [ "Angle Bisector Theorem", "Incenter Theorem", "Incenter Properties" ], "current_concepts": [ "Angle Bisector Theorem", "Incenter Theorem", "Incenter Properties" ] } ## Identify the geometric principles for each section In this worksheet, we solve two types of problems related to angle bisectors in triangles: 1. **Questions 1 and 2**: An angle bisector divides an angle into two equal halves. Thus, the measure of the whole angle is twice the measure of either half: $$ m\angle\text{Whole} = 2 \cdot m\angle\text{Half} $$ 2. **Questions 3 to 8**: The three angle bisectors of a triangle intersect at a single point called the **incenter** (labeled as point $P$). The **Incenter Theorem** states that the incenter is equidistant from all three sides of the triangle. The perpendicular segments from $P$ to the sides represent these distances, meaning: $$ PD = PE = PF \quad \text{or} \quad PX = PY = PZ $$ ## Solve Questions 1 and 2 (Angle Bisectors) - **Question 1**: $UT$ is an angle bisector of $\angle SUT$. We are given $m\angle SUT = 34^\circ$. Since the bisector divides the angle into two equal parts, $\angle 1$ is half of the total angle: $$ m\angle 1 = \frac{m\angle SUT}{2} = \frac{34^\circ}{2} = 17^\circ $$ - **Question 2**: $QS$ is an angle bisector of $\angle SQR$. We are given $m\angle 2 = 13^\circ$. Since $\angle 2$ is one of the bisected halves, the entire angle is twice this measure: $$ m\angle SQR = 2 \cdot m\angle 2 = 2 \cdot 13^\circ = 26^\circ $$ ## Solve Questions 3, 4, and 5 (Incenter Equidistance) - **Question 3**: Point $P$ is the incenter. The perpendicular segments from $P$ to the sides are $PT$, $PU$, and $PS$. By the **Incenter Theorem**, these distances are equal: $$ PU = PT = 3 $$ - **Question 4**: Point $P$ is the incenter. The perpendicular segments to the sides are $PV$, $PW$, and $PX$. Therefore: $$ PV = PW = 7 $$ - **Question 5**: Point $P$ is the incenter. The perpendicular segments to the sides are $PW$, $PX$, and $PY$. Therefore: $$ PW = PX = 5 $$ ## Solve Questions 6, 7, and 8 (Incenter Equidistance) - **Question 6**: Point $P$ is the incenter. The perpendicular segments to the sides are $PD$, $PE$, and $PF$. Therefore: $$ PD = PC = 8 $$ *(Note: The segment from the incenter to the vertex is $PC$, but the perpendicular segments to the sides are $PD$, $PE$, and $PF$. Looking closely at the diagram, $PD$, $PE$, and $PF$ are the perpendiculars to the sides. The label $PC = 8$ refers to the perpendicular segment to side $QR$, which is labeled as $PC$. Thus, $PD = PC = 8$.)* - **Question 7**: Point $P$ is the incenter. The perpendicular segments to the sides are $PX$, $PY$, and $PZ$. Therefore: $$ HY = HP = 3 $$ *(Note: The question asks to find $HY$. In this right triangle $\triangle HPY$, $HP$ is the hypotenuse. However, looking at the standard incenter diagram, $PY$, $PX$, and $PZ$ are the perpendiculars to the sides, which are all equal. Here, $PY = 2$. The segment $HP$ is a segment from a vertex to the incenter, with $HP = 3$. In the right triangle $\triangle HPY$, we can use the Pythagorean theorem: $HY = \sqrt{HP^2 - PY^2} = \sqrt{3^2 - 2^2} = \sqrt{9 - 4} = \sqrt{5} \approx 2.24$.)* - **Question 8**: Point $P$ is the incenter. We are given $PQ = 1$ (perpendicular to side $BC$) and $AR = 2$. We need to find $AP$. In the right triangle $\triangle APR$, $PR$ is a perpendicular to side $AC$. Since $P$ is the incenter, $PR = PQ = 1$. Using the Pythagorean theorem on right triangle $\triangle APR$: $$ AP = \sqrt{AR^2 + PR^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.24 $$ | No. | Problem | Answer | |---|---|---| | 1 | Find $m\angle 1$ if $m\angle SUT = 34^\circ$ | $17^\circ$ | | 2 | Find $m\angle SQR$ if $m\angle 2 = 13^\circ$ | $26^\circ$ | | 3 | Find $PU$ if $PT = 3$ | $3$ | | 4 | Find $PV$ if $PW = 7$ | $7$ | | 5 | Find $PW$ if $PX = 5$ | $5$ | | 6 | Find $PD$ if $PC = 8$ | $8$ | | 7 | Find $HY$ if $PY = 2$ and $HP = 3$ | $\sqrt{5}$ | | 8 | Find $AP$ if $PQ = 1$ and $AR = 2$ | $\sqrt{5}$ | { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Geometry", "Incenter Theorem" ] }

No.	Problem	Answer
2	Find \(m\angle SQR\) if \(m\angle 2 = 13^\circ\)	\(26^\circ\)
3	Find \(PU\) if \(PT = 3\)	\(3\)
4	Find \(PV\) if \(PW = 7\)	\(7\)
5	Find \(PW\) if \(PX = 5\)	\(5\)
6	Find \(PD\) if \(PC = 8\)	\(8\)
7	Find \(HY\) if \(PY = 2\) and \(HP = 3\)	\(\sqrt{5}\)
8	Find \(AP\) if \(PQ = 1\) and \(AR = 2\)	\(\sqrt{5}\)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Identify the geometric principles for each section

Solve Questions 1 and 2 (Angle Bisectors)

Solve Questions 3, 4, and 5 (Incenter Equidistance)

Solve Questions 6, 7, and 8 (Incenter Equidistance)

Answer: