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Construct a triangle ABC with side BC = 6cm, \angle B = 45\degree,\ \angle A = 105\degree. Then construct another triangle whose side are \frac{3}{4} times the corresponding sides of \DeltaABC.

 

 
 
 
 
 

Answers (1)

Given,

            In \Delta ABC

            BC = 6 cm, \angle B = 45\degree,\ \angle A = 105\degree

            \angle C = 180-(\angle A + \angle B) = 180 - 150

            \angle C = 30\degree

Steps of construction

  1. Draw a line BC = 6 cm
  2. Draw \angle B = 45\degree and \angle C = 30\degree
  3. Let point A be the point where the two rays intersect

\therefore\DeltaABC is the required triangle.

Now we need to make a triangle whose sides are \frac{3}{4} times the corresponding sides of \DeltaABC.

Steps

  1. Draw any ray BX making an acute angle with BC on the opposite side of vertex A.
  2. Locate four points B1, B2, Band B4 in such a way BB_1 = BB_2 =BB_3=BB_4
  3. Join B4C and draw a line from point B3 parallel to B4C so that it cuts BC at C'.
  4. From point C' draw a line parallel to AC and cut AB at A'.
  5. \DeltaA'BC' is the required tringle.

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Safeer PP

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