Constructions
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Ex 11.2, 6 - Let ABC be a right triangle AB = 6 cm, BC = 8 cm, B = 90

Ex 11.2, 6 - Chapter 11 Class 10 Constructions - Part 2
Ex 11.2, 6 - Chapter 11 Class 10 Constructions - Part 3
Ex 11.2, 6 - Chapter 11 Class 10 Constructions - Part 4
Ex 11.2, 6 - Chapter 11 Class 10 Constructions - Part 5 Ex 11.2, 6 - Chapter 11 Class 10 Constructions - Part 6 Ex 11.2, 6 - Chapter 11 Class 10 Constructions - Part 7

 

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Transcript

Question 6 Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠ B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle. First we draw a rough sketch Δ ABC Now, it is given that circle is drawn through point B, C, D Let’s draw a circle through points B, C, D Now, we see that ∠ BDC = 90° We know that Angle in a semicircle is a right angle So, BC must be the diameter Now, We first construct Δ ABC No need to mark point D on the triangle We find center of circle – which will be mid-point of BC. So, we draw perpendicular bisector of BC, and find center of circle. Through center of circle, we draw circle Then, we draw tangents from point A to the circle Let’s draw Δ ABC first Steps to draw Δ ABC Draw base BC of side 8 cm Draw ∠ B = 90° 3. Taking B as center, 6 cm as radius, we draw an arc Let the point where arc intersects the ray be point A 4. Join AC ∴ Δ ABC is the required triangle Now, let’s draw a circle and construct tangents Steps to draw circle and tangents Draw perpendicular bisector of line BC Let the line intersect BC at point E. Now, E is the mid-point of BC Taking E as center, and BE as radius, draw a circle We need to construct tangents from point A to the circle Join point A to center of Circle E. Make perpendicular bisector of AE Let M be the midpoint of AE 4. Taking M as center and MA as radius, draw a circle. 5. Let blue circle intersect the other circle at B and Q Join AQ Thus, AB and AQ are the required tangents Justification We need to prove that AB and AQ are the tangents to the circle. Join EQ. Now, ∠AQE is an angle in the semi-circle of the blue circle And we know that, Angle in a semi-circle is a right angle. ∴ ∠AQE = 90° ⇒ EQ ⊥ AQ Since EQ is the radius of the circle, AQ has to be a tangent of the circle. Similarly, AB is a tangent of the circle.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo