Question

How to solve the problems​

Answer 1

Answer:

1. 79°

2. 54°

3. 107.5°

4. 44°, 35 cm

5. 76°, 3.5 cm

6. m∠U=36°, m∠M=m∠D=72°,  MD=8.6 cm

7. 78°, 93 cm

8. 81°, 75 cm

Step-by-step explanation:

1. The diagram shows an isosceles triangle because TH = OT. Angles adjacent to the base OH of isosceles triangle are congruent, so

$m\angle H=m\angle O$

The sum of the measures of all interior angles is always 180°, thus

$m\angle H+m\angle O+m\angle T=180^{\circ}\\ \\2m\angle H+22^{\circ}=180^{\circ}\\ \\2m\angle H=180^{\circ}-22^{\circ}\\ \\2m\angle H=158^{\circ}\\ \\m\angle H=79^{\circ}$

2. The diagram shows an isosceles triangle DGO because DG = GO. Angles adjacent to the base DO of isosceles triangle are congruent, so

$m\angle D=m\angle O=63^{\circ}$

The sum of the measures of all interior angles is always 180°, thus

$m\angle D+m\angle O+m\angle G=180^{\circ}\\ \\63^{\circ}+63^{\circ}+m\angle G=180^{\circ}\\ \\m\angle G=180^{\circ}-63^{\circ}-63^{\circ}\\ \\m\angle G=54^{\circ}$

3. The diagram shows an isosceles triangle SLO because LO = SO. Angles adjacent to the base SL of isosceles triangle are congruent, so

$m\angle S=m\angle L$

The sum of the measures of all interior angles is always 180°, thus

$m\angle S+m\angle L+m\angle O=180^{\circ}\\ \\2m\angle L+35^{\circ}=180^{\circ}\\ \\2m\angle L=180^{\circ}-35^{\circ}\\ \\2m\angle L=145^{\circ}\\ \\m\angle L=72.5^{\circ}$

Angles OLE and L (SLO) are supplementary (add up to 180°), so

$m\angle OLE=180^{\circ}-m\angle L\\ \\m\angle OLE=180^{\circ}-72.5^{\circ}\\ \\m\angle OLE=107.5^{\circ}$

4. The diagram shows an isosceles triangle AMR because $m\angle A=m\angle M=68^{\circ}$ (angles adjacent to the side AM are congruent, so triangle AMR is isoseceles).

The sum of the measures of all interior angles is always 180°, thus

$m\angle A+m\angle M+m\angle R=180^{\circ}\\ \\m\angle R+2\cdot 68^{\circ}=180^{\circ}\\ \\m\angle R=180^{\circ}-2\cdot 68^{\circ}\\ \\m\angle R=44^{\circ}$

To legs in isosceles triangle are always congruent, so

$RM=AR=35\ cm$

5. The diagram shows isosceles triangle RYD because YD = RD. Angles adjacent to the base RY are congruent, so

$m\angle R=m\angle Y$

The sum of the measures of all interior angles is always 180°, thus

$m\angle R+m\angle Y+m\angle D=180^{\circ}\\ \\2m\angle Y+28^{\circ}=180^{\circ}\\ \\2m\angle Y=180^{\circ}-28^{\circ}\\ \\2m\angle Y=152^{\circ}\\ \\m\angle Y=76^{\circ}$

To legs in isosceles triangle are always congruent, so

$YD=RD=3.5\ cm$

6. The diagram shows an isosceles triangle UMD because UM = UD. Angles adjacent to the base MD of isosceles triangle are congruent, so

$m\angle D=m\angle M=72^{\circ}$

The sum of the measures of all interior angles is always 180°, thus

$m\angle D+m\angle M+m\angle U=180^{\circ}\\ \\72^{\circ}+72^{\circ}+m\angle U=180^{\circ}\\ \\m\angle U=180^{\circ}-72^{\circ}-72^{\circ}\\ \\m\angle U=36^{\circ}$

To legs in isosceles triangle are always congruent, so

$UM=UD=14\ cm$

The perimeter of isosceles triangle MUD is 36.6 cm, so

$UM+MD+UD=36.6\\ \\MD=36.6-14-14\\ \\MD=8.6\ cm$

7. The sum of the measures of all interior angles is always 180°, thus

$m\angle T+m\angle S+m\angle B=180^{\circ}\\ \\m\angle T+78^{\circ}+24^{\circ}=180^{\circ}\\ \\m\angle T=180^{\circ}-78^{\circ}-24^{\circ}\\ \\m\angle T=78^{\circ}$

Triangle STB is isosceles triangle because $m\angle S=m\angle T=78^{\circ}$ (angles adjacent to the side ST are congruent, so triangle STB is isoseceles).

To legs in isosceles triangle are always congruent, so

$SB=TB\\ \\y+22.5=38.5\\ \\y=38.5-22.5\\ \\y=16$

Hence,

$ST=16\ cm\\ \\TB=SB=38.5\ cm$

and the perimeter of triangle STB is

$P_{STB}=16+38.5+38.5=93\ cm$

8. The diagram shows isosceles triangle CNB because CN = CB. Angles adjacent to the base RY are congruent, so

$m\angle N=m\angle B$

The sum of the measures of all interior angles is always 180°, thus

$m\angle N+m\angle B+m\angle C=180^{\circ}\\ \\2m\angle N+18^{\circ}=180^{\circ}\\ \\2m\angle N=180^{\circ}-18^{\circ}\\ \\2m\angle N=162^{\circ}\\ \\m\angle N=81^{\circ}$

To legs in isosceles triangle are always congruent, so

$CB=CN=2x+90\ m$

The perimeter of the triangle CNB is

$2x+90+2x+90+x=555\\ \\5x+180=555\\ \\x+36=111\\ \\x=111-36\\ \\x=75\ m$

So, $NB=75 \ m$