Draw a right triangle $ABC$ in which $AB = 6\ cm, BC = 8\ cm$ and $\angle B = 90^o$. Draw $BD$ perpendicular from $B$ on $AC$ and draw a circle passing through the points $B, C$ and $D$. Construct tangents from $A$ to this circle.

Given:

A right triangle $ABC$ in which $AB = 6\ cm, BC = 8\ cm$ and $\angle B = 90^o$.

To do:

We have to draw a right triangle $ABC$ in which $AB = 6\ cm, BC = 8\ cm$ and $\angle B = 90^o$. Draw $BD$ perpendicular from $B$ on $AC$ and draw a circle passing through the points $B, C$ and $D$. Construct tangents from $A$ to this circle.

Solution:

Steps of construction:

(i) Draw a line segment $BC = 8\ cm$.

(ii) From $B$ draw an angle of $90^o$.

(iii) Draw an arc $BA = 6\ cm$ cutting the angle at $A$.

(iv) Join AC. $ABC$ is the required triangle.

(v) Draw perpendicular bisector of $BC$ cutting $BC$ at $M$.

(vi) Taking $M$ as centre and $BM$ as radius, draw a circle.

(vii) Taking $A$ as centre and $AB$ as radius draw an arc cutting the circle at $E$. Join $AE$.

$AB$ and $AE$ are the required tangents.

Justification:

$\angle ABC = 90^o$       (Given)

$OB$ is a radius of the circle.

This implies,

$AB$ is a tangent to the circle.

Also, $AE$ is a tangent to the circle.

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Updated on: 10-Oct-2022

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