1. In the figures below, there is a pair of parallel lines and a third line cutting across them. In each figure, the measure of one angle is given and another angle is marked. Find out its measure and write if in the figure itself.
2. In the figure below, AB and CD are parallel. Compute all the angles in the figure.
∠A = 70o, ∠B = 60o
∠E = 180o - (70o + 60o) = 180o - 130o = 50o
(Sum of the angles of a triangle is 180o)
∠ACD = 180 - 70 = 110o
∠CDE = 60o (Corresponding angles)
3. In the figure below, a parallelogram is divided into four triangles by the diagonals. Calculate the angles of all these triangles.
In Δ AOB, ∠ABO = 55o
[180 - (100 + 25)]
In Δ BOC, ∠BOC = 80o (linear pair)
∠OBC = 75o (Alternate angles)
In ΔCOD, ∠COD = 100o
∠OCD = 25o (Alternate angles)
∠CDO = 55o (Alternate angles)
In ΔAOD, ∠AOD = 80o
∠OAD = 25o (Alternate angles)
∠ADO = 75o (Alternate angles)
In the figure above, Ab and DE are parallel. Compute the angles of both triangles.
∠ABC = 60o (Corresponding angles)
∠CDE = 80o (Corresponding angles)
∠C = 40o [180 -(80 + 60)]
In the figure above, AB and CD are parallel. What is the relation between the angles of the small and large triangles?
The alternate angles formed when the line AD intersect the parallel lines AB and CD are ∠BAP and ∠CDP.
∴ They are equal.
The alternate angles formed when the line BC intersect the parallel lines AB and CD are ∠ABP and ∠DCP.
∴ They are equal.
∠APB and ∠CPO are opposite angles and so equal.
The angles of the small and large triangles are same.
6. Draw a line AB and a line CD parallel to it. Draw a line EF cutting across these lines at the points M and N. Measure and write down the pairs of corresponding angles, alternate angles, co-interior angles and co-exterior angles.
∠BME = ∠ DNE; ∠AME = ∠CNE;
∠AMF = ∠CNF; ∠BMF = ∠ DNF
∠AMF = ∠DNE; ∠CNE = ∠BMF
∠BMF + ∠DNE = 180o
∠AMF + ∠CNE = 180o
∠AMF + ∠CNF = 180o
∠BMF + ∠DNF = 180o
7. In a figure, three parallel lines are cut by a fourth line. one angle 30o is marked. Find the remaining angles.
In this figure, ∠1 = 30o
∠3 = 30o, ∠5 = 30o, ∠7 = 30o
∠9 = 30o, ∠11 = 30o
∠2 = 150o, ∠4 = 150o, ∠6 =150o
∠8 = 150o , ∠10 = 150o, ∠12 = 150o
8. In the figure, ∠A = 70o and ∠B = 40o, Find ∠ACD.
Have ∠A, ∠B, and ∠ACD any relation? Examine other triangles of thee types. Is this true for all triangles?
Draw CE parallel to BA.
∠A = ∠ACE = 70o (Alternate angles)
∠B = ∠ECD = 40o (Corresponding angles)
∴ ∠ACD = ∠ACE + ∠ ECD = 70o + 40o = 110o
That is ∠ACD = ∠A + ∠B.
This is true for all triangles.
It is confirmed and ∠A = x. ∠B = y
then, ∠ACD = x + y
Practice in Related Chapters
|Numbers and Algebra|
|Relation of Parts|
|Between the Lines|
|Lines in Unison|
|Area of a Triangle|
|Square and Square Root|