Question

list the sides in order, underline the side with the shortest length.
24)
25)
26)
list the sides of △abc from the longest to shortest.

1. m∠a = 46°, m∠b = 30° 28) m∠c = 101°, m∠b = 70° 29) m∠a = 59°, m∠c = 61°

find the value of x and list the sides of △abc in order from shortest to longest if the angles have the indicated measures. (hint: find the angle measures first, then decide which sides are the longest)

1. m∠a=(9x + 29)°, m∠b=(93 - 5x)°, and m∠c=(10x + 2)°.
2. m∠a=(9x - 4)°, m∠b=(4x - 16)°, and m∠c=(68 - 2x)°.
3. m∠a=(12x - 9)°, m∠b=(62 - 3x)°, and m∠c=(16x + 2)°.
4. m∠a=(5x + 2)°, m∠b=(6x - 10)°, and m∠c=(x + 20)°.
5. m∠a=(10x)°, m∠b=(5x - 17)°, and m∠c=(7x - 1)°.

answer the following questions.

1. draw △dea with a median (overline{eg}).
2. draw △jkh with an altitude (overline{jp}).
3. find the value of x.

(overline{so}) is an altitude of △sat

Explanation:

Step1: Recall angle - side relationship in a triangle

In a triangle, the longest side is opposite the largest angle and the shortest side is opposite the smallest angle.

Step2: For problem 24

In $\triangle ABC$ with $\angle A = 63^{\circ}$, $\angle B=70^{\circ}$, $\angle C = 47^{\circ}$. Since $47^{\circ}<63^{\circ}<70^{\circ}$, the sides in order from shortest to longest are $\overline{AB},\overline{BC},\overline{AC}$. Underline $\overline{AB}$.

Step3: For problem 25

In $\triangle DEF$ with $\angle D = 125^{\circ}$, $\angle E=30^{\circ}$, $\angle F = 25^{\circ}$. Since $25^{\circ}<30^{\circ}<125^{\circ}$, the sides in order from shortest to longest are $\overline{DE},\overline{DF},\overline{EF}$. Underline $\overline{DE}$.

Step4: For problem 26

In $\triangle ABC$ with $\angle A = 65^{\circ}$, $\angle B$, $\angle C = 40^{\circ}$. First find $\angle B=180^{\circ}-(65^{\circ}+40^{\circ}) = 75^{\circ}$. Since $40^{\circ}<65^{\circ}<75^{\circ}$, the sides in order from shortest to longest are $\overline{AB},\overline{BC},\overline{AC}$. Underline $\overline{AB}$.

Step5: For problem 27

In $\triangle ABC$ with $\angle A = 46^{\circ}$, $\angle B = 30^{\circ}$, then $\angle C=180^{\circ}-(46^{\circ}+30^{\circ}) = 104^{\circ}$. Since $30^{\circ}<46^{\circ}<104^{\circ}$, the sides from longest to shortest are $\overline{AB},\overline{BC},\overline{AC}$.

Step6: For problem 28

In $\triangle ABC$ with $\angle C = 101^{\circ}$, $\angle B = 70^{\circ}$, then $\angle A=180^{\circ}-(101^{\circ}+70^{\circ}) = 9^{\circ}$. Since $9^{\circ}<70^{\circ}<101^{\circ}$, the sides from longest to shortest are $\overline{AB},\overline{AC},\overline{BC}$.

Step7: For problem 29

In $\triangle ABC$ with $\angle A = 59^{\circ}$, $\angle C = 61^{\circ}$, then $\angle B=180^{\circ}-(59^{\circ}+61^{\circ}) = 60^{\circ}$. Since $59^{\circ}<60^{\circ}<61^{\circ}$, the sides from longest to shortest are $\overline{AB},\overline{BC},\overline{AC}$.

Step8: For problem 30

We know that $\angle A+\angle B+\angle C = 180^{\circ}$, so $(9x + 29)+(93 - 5x)+(10x + 2)=180$.
Combining like - terms: $9x-5x + 10x+29 + 93+2=180$.
$14x+124 = 180$.
$14x=180 - 124=56$.
$x = 4$.
$\angle A=(9\times4 + 29)^{\circ}=65^{\circ}$, $\angle B=(93-5\times4)^{\circ}=73^{\circ}$, $\angle C=(10\times4 + 2)^{\circ}=42^{\circ}$.
The sides from shortest to longest are $\overline{AB},\overline{BC},\overline{AC}$.

Step9: For problem 31

$(9x - 4)+(4x - 16)+(68 - 2x)=180$.
Combining like - terms: $9x+4x-2x-4-16 + 68=180$.
$11x + 48=180$.
$11x=180 - 48 = 132$.
$x = 12$.
$\angle A=(9\times12-4)^{\circ}=104^{\circ}$, $\angle B=(4\times12-16)^{\circ}=32^{\circ}$, $\angle C=(68-2\times12)^{\circ}=44^{\circ}$.
The sides from shortest to longest are $\overline{AB},\overline{AC},\overline{BC}$.

Step10: For problem 32

$(12x - 9)+(62 - 3x)+(16x + 2)=180$.
Combining like - terms: $12x-3x + 16x-9+62 + 2=180$.
$25x+55 = 180$.
$25x=180 - 55=125$.
$x = 5$.
$\angle A=(12\times5-9)^{\circ}=51^{\circ}$, $\angle B=(62-3\times5)^{\circ}=47^{\circ}$, $\angle C=(16\times5 + 2)^{\circ}=82^{\circ}$.
The sides from shortest to longest are $\overline{BC},\overline{AB},\overline{AC}$.

Step11: For problem 33

$(5x + 2)+(6x - 10)+(x + 20)=180$.
Combining like - terms: $5x+6x+x+2-10 + 20=180$.
$12x+12 = 180$.
$12x=180 - 12=168$.
$x = 14$.
$\angle A=(5\times14 + 2)^{\circ}=72^{\circ}$, $\angle B=(6\times14-10)^{\circ}=74^{\circ}$, $\angle C=(14 + 20)^{\circ}=34^{\circ}$.
The sides from shortest to longest are $\overline{AB},\overline{AC},\overline{BC}$.

Step12: For problem 34

$(10x)+(5x - 17)+(7x - 1)=180$.
Combining like - terms: $10x+5x+7x-17-1=180$.
$22x-18 = 180$.
$22x=180 + 18=198$.
$x = 9$.
$\angle A=(10\times9)^{\circ}=90^{\circ}$, $\angle B=(5\times9-17)^{\circ}=28^{\circ}$, $\angle C=(7\times9-1)^{\circ}=62^{\circ}$.
The sides from shortest to longest are $\overline{BC},\overline{AC},\overline{AB}$.

Step13: For problem 35

To draw $\triangle DEA$ with median $\overline{EG}$, first draw $\triangle DEA$. Then find the mid - point $G$ of side $\overline{DA}$ and draw the line segment $\overline{EG}$.

Step14: For problem 36

To draw $\triangle JKH$ with altitude $\overline{JP}$, first draw $\triangle JKH$. Then from vertex $J$, draw a line segment $\overline{JP}$ perpendicular to the opposite side $\overline{KH}$.

Step15: For problem 37

Since $\overline{SO}$ is an altitude of $\triangle SAT$, $\angle SOA = 90^{\circ}$. So $4x-6 = 90$.
$4x=90 + 6=96$.
$x = 24$.

Answer:

1. Sides: $\overline{AB},\overline{BC},\overline{AC}$, Underlined: $\overline{AB}$
2. Sides: $\overline{DE},\overline{DF},\overline{EF}$, Underlined: $\overline{DE}$
3. Sides: $\overline{AB},\overline{BC},\overline{AC}$, Underlined: $\overline{AB}$
4. Sides: $\overline{AB},\overline{BC},\overline{AC}$
5. Sides: $\overline{AB},\overline{AC},\overline{BC}$
6. Sides: $\overline{AB},\overline{BC},\overline{AC}$
7. $x = 4$, Sides: $\overline{AB},\overline{BC},\overline{AC}$
8. $x = 12$, Sides: $\overline{AB},\overline{AC},\overline{BC}$
9. $x = 5$, Sides: $\overline{BC},\overline{AB},\overline{AC}$
10. $x = 14$, Sides: $\overline{AB},\overline{AC},\overline{BC}$
11. $x = 9$, Sides: $\overline{BC},\overline{AC},\overline{AB}$
12. Draw $\triangle DEA$, find mid - point $G$ of $\overline{DA}$ and draw $\overline{EG}$
13. Draw $\triangle JKH$, draw $\overline{JP}\perp\overline{KH}$ from $J$
14. $x = 24$