Question

let ABC be a right angle in which a b is equal to 6 CM BC is equals to 8 cm and b is equals to 90° BD is perpendicular from B on C the circle through BCD is thrown constant the tangent from a to the circle​

Answer 1

Answer:

Step-by-step explanation:

Steps of construction:

1. Draw a line segment BC = 8 cm.

2. Make a right angle at the point B i.e., ∠CBX = 90°

3. Draw a are of radius 6 cm as centre B which intersect BX at the point A.

4. Join AC  so, ABC is required right angle triangle.

5. Draw a arc taking centre B which intersect AC at the point K and L respectively taking K and L centre draw two arcs of same radius which intersect at the point M.

6. Join BM so, ∠BDC = 90°

7. Draw perpendicular bisector of BC.

8. Draw a circle taking radius equal to OB and centre O which passes through B, D and C.

9. Draw a arc taking centre A and radius equal to AB intersect the circle at point P so, AP is the tangents

Answer 2

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