# Ex.11.2 Q6 Constructions Solution - NCERT Maths Class 10

## 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 \(O\)*.*