Question

∆ABC and ∆PQR are congruent to each other by the correspondence

∆ABC ∆PRQ. Write the parts of ∆PQR that correspond to:

a. ∠C b. side AC c. ∠B

Answer 1

Step-by-step explanation:

Question 1:

In the given figure, ∠∠ACD is an exterior angle of Δ∆ABC. ∠∠B = 40°, ∠∠A = 70°.



Find the measure of ∠∠ACD.

 

ANSWER:

In Δ∆ABC,

∠ACD = ∠A + ∠B   (Exterior angle property)

= 70∘ + 40∘          

= 110∘

Hence, the measure of  ∠∠ACD is 110∘.

Question 2:

In Δ∆PQR, ∠∠P = 70°, ∠∠Q = 65 ° then find  ∠∠R.

ANSWER:

In Δ∆PQR,

∠P + ∠Q + ∠R = 180∘   (Angle sum property)

⇒ 70∘ + 65∘ + ∠R = 180∘         

⇒ 135∘ + ∠R = 180∘  

⇒ ∠R = 180∘ − 135∘  

= 45∘

Hence, the measure of  ∠∠R is 45∘.

Question 3:

The measures of angles of a triangle are x°, ( x−-20)°, (x−-40)°. Find the measure of each angle.

ANSWER:

Let us suppose the angles ∠P, ∠Q, ∠Rof a Δ∆PQR be x°, (x −- 20)°, (x −- 40)° respectively.

∠P + ∠Q + ∠R = 180∘   (Angle sum property)

⇒ x∘ + (x −- 20)° + (x −- 40)° = 180∘         

⇒ 3x −- 60 = 180  

⇒ 3x = 240

⇒ x = 80

Therefore,

∠P = 80∘

∠R = (80 −- 20)°

= 60∘

∠R = (80 −- 40)°

= 40∘

Hence, the measure of each angle is 80∘, 60∘ and 40∘respectively.

Question 4:

The measure of one of the angles of a triangle is twice   the measure of its smallest angle and the measure of the other is thrice the measure of the smallest angle. Find the measures of the three angles.

ANSWER:

Let us suppose the angles of a Δ∆PQR such that ∠P < ∠Q < ∠R.

A.T.Q, 

∠Q = 2∠P

∠R = 2∠P

Now, ∠P + ∠Q + ∠R = 180∘   (Angle sum property)

⇒ ∠P + 2∠P + 3∠P = 180∘         

⇒ 6∠P = 180∘  

⇒ ∠P = 30∘

Therefore,

∠P = 30∘

∠R = 60∘

∠R = 90∘

Hence, the measure of each angle is 30∘, 60∘ and 90∘respectively.

Question 5:

In the given figure, measures of some angles are given. Using the measures find the values of x, y, z. 



ANSWER:

∠NEM + ∠NET =  180∘      (Linear angle property)

⇒ y + 100∘ = 180∘       

⇒ y = 80∘

Also, ∠NME + ∠EMR =  180∘      (Linear angle property)

⇒ z + 140∘ = 180∘       

⇒ z = 40∘

Now, In △NEM

∠N + ∠E + ∠M = 180∘   (Angle sum property)

⇒ x + y + z = 180∘         

⇒ x + 80∘ + 40∘ = 180∘  

⇒ x + 120∘ = 180∘ 

⇒ x = 60∘  

Hence, the values of x, y and z are 60∘, 80∘ and 40∘respectively.

Question 6:

In the given figure, line AB || line DE. Find the measures of ∠∠DRE and ∠∠ARE using given measures of some angles.



ANSWER:

AB || DE and AD is a transversal line.

∠BAR = ∠RDE = 70∘      (Alternate angles)

In △DER

∠D + ∠E + ∠DRE = 180∘   (Angle sum property)

⇒ 70∘ + 40∘ + ∠DRE = 180∘         

⇒ 110∘ + ∠DRE = 180∘ 

⇒ ∠DRE = 70∘

Now, ∠ARE = ∠DRE + ∠RDE   (Exterior angle property)

= 70∘ + 40∘          

= 110∘

Hence, the measures of ∠DRE and ∠ARE are 70∘ and 110∘respectively.

Question 7:

In Δ∆ABC, bisectors of ∠∠A and ∠∠B intersect at point O. If ∠∠C = 70° . Find measure of ∠∠AOB.

ANSWER:

If the bisectors of ∠X and ∠Y of a △XYZ intersect at point O, then ∠XOY=90°+12∠XZY∠XOY=90°+12∠XZY

∴∠AOB=90°+12∠ACB=90°+12(70°)∴∠AOB=90°+12∠ACB=90°+1270°

= 90∘ + 35∘ 

= 125∘  

Hence, the measure of ∠AOB is 125∘.

Question 8:

In the given Figure, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ∠∠BPQ and ∠∠PQD respectively.

Prove that m ∠∠PTQ = 90 °.



ANSWER:

AB || CD and PQ is a transversal line.

∠BPQ + ∠DQP = 180∘      (Angles on the same side of a transversal line are supplementary angles)

⇒12∠BPQ+12∠DQP=180°2⇒12∠BPQ+12∠DQP=180°2

⇒ ∠QPT + ∠PQT = 90∘

In △PQT

∠QPT + ∠PQT + ∠PTQ = 180∘   (Angle sum property)

⇒ 90∘ + ∠PTQ = 180∘         

⇒ ∠PTQ = 90∘ 

Hence proved.

Question 9:

Using the information shown in figure, find the measures of ∠∠a, ∠∠b and ∠∠c.



ANSWER:

c + 100∘ = 180∘                (Linear angle property)      

⇒ c = 80∘

Now, b = 70∘     (Vertically opposite angles)      

a + b + c = 180∘       (Angle sum property)   

⇒ a + 70∘ + 80∘ = 180∘  

⇒ a + 150∘ = 180∘ 

⇒ a = 30∘  

Hence, the values of a, b and c are 30∘, 70∘ and 80∘respectively.

Question 10:

In the given figure, line DE ||   line GF ray EG and ray FG are bisectors of ∠∠DEF and ∠∠DFM respectively. Prove that,

(i)  ∠∠DEG = 12 ∠12 ∠EDF    (ii)  EF = FG.   



ANSWER:

(i) Given: DE || GF

Now, ∠∠DEF = ∠∠GFM              (Corresponding angles as DM is a transversal line)

⇒ 2∠∠DEG = ∠∠DFG         (Ray EG and ray FG are bisectors of ∠∠DEF and ∠∠DFM)

⇒ 2∠∠DEG = ∠∠EDF          (∵ ∠∠EDF = ∠∠DFG, alternate angles as DF is a transversal line)

⇒ ∠∠DEG = 12 ∠12 ∠EDF

(ii) Given: DE || GF

 ∠∠DEG = ∠∠EGF              (Alternate angles as EG is a transversal line)

∴ ∠∠GEF = ∠∠EGF            (∵ ∠∠DEG = ∠∠GEF)

∴ EF = FG                         (Sides opposite to equal angles)