Ex.11.2 Q6 Constructions Solution - NCERT Maths Class 10

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Question

Let \(ABC\) be a right triangle in which \(AB = 6 \,\rm{cm},\) \(BC = 8 \,\rm{cm}\) and \(\begin{align}\angle \rm{B}=90^{\circ}.\end{align}\)  \(BD\) is the perpendicular to \(AC\). The circle through \(B\), \(C\) and \(D\) is drawn. Construct the tangents from \(A\) to this circle.

Text Solution

 

Steps:

Steps of construction:

(i) Draw \(\begin{align}\rm{BC}=8 \rm{cm}\end{align}\). Draw the perpendicular at \(B\) and cut \(\begin{align}\rm{BA}=6 \rm{cm}\end{align}\)on it. Join \(AC\) right \(\begin{align}\triangle {ABC}\end{align}\)  is obtained.

(ii) Draw \(BD\) perpendicular to \(AC\).

(iii) Since \(\begin{align}\angle \rm{BDC}=90^{\circ}\end{align}\) and the circle has to pass through \(B, C\) and \(D.\) \(BC\) must be a diameter of this circle. So, take \(O\) as the midpoint of \(BC\) and with \(O\) as centre and \(OB\) as radius draw a circle which will pass through \(B\), \(C\) and \(D\).

(iv) To draw tangents from \(A\) to the circle with center \(O\).

(a) Join \(OA\), and draw its perpendicular bisectors to intersect \(OA\) at \(E\).

(b) With \(E\) as center and \(EA\) as radius draw a circle which intersects the  previous circle at \(B\) and \(F\).

(c) Join \(AF\).

\(AB\) and \(AF\) are the required tangents.

Proof:

\(\angle {{ABO}} = \angle {{AFO}} = 90^\circ \)  ( Angle in a semi \(-\) circle)

\(\therefore\) \({AB} \bot {OB}\) and \({AF} \bot {OF}\)

Hence \(AB\) and \(AF\) are the tangents from \(A\) to the circle with centre \(O\).

  
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