Question

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

Answer 1

Hint: We start solving the problem by first drawing the line segment AB and then drawing an arc of length 8 cm taking B as centre. We then construct angle $\angle ABC={{90}^{\circ }}$ intersecting the arc at C. We then join point A and C to complete the construction of the triangle. We then draw an arc taking B as centre intersecting the side AC at E and F. We then draw arcs taking E and F as centre which intersects at O. We then Join the points O and B which intersects side AC at D which is the altitude of the triangle ABC. We then draw arcs taking B and C as centres which intersect at points R and T. We then join points R and T which intersects side BC at S which is the circumcentre. We then draw a circle taking S as centre and SB as the radius. We then draw arcs taking points A and C as centre which intersects at U and V. We then join points U and V which intersects the line segment AS at W. We then draw arcs taking W as centre which intersects the circumcircle at B and X. We then join the points A and X, A and B to complete the construction of tangents AX and AB.

Complete step by step answer:
According to the problem, we are asked to construct a triangle with measurements \[AB=6cm\], $BC=8cm$ and $\angle B={{90}^{\circ }}$. Also, we need to construct tangents from point A to the circle passing through points B, C, and D, where BD is perpendicular from B to AC.
The following are the steps to construct the given figures.
Step1: Let us first draw the line segment AB of length 6 cm.

Step 2: Now, we draw an arc of radius 8 cm taking B as a centre.

Step 3: Now, let us construct angle $\angle ABC={{90}^{\circ }}$, such that it intersects the arc at C.

Step 4: Now, let us join the points A and C to complete the construction of triangle ABC.

Step 5: Now, let us draw an arc of any radius less than the length of side AC taking B as centre intersecting the side AC at two points E and F.

Step 6. Now, let us draw an arc of the radius of 1 cm taking F as centre outside the triangle.

Step 7: Now, let us draw an arc of the radius of 1 cm taking E as centre outside the triangle, which intersects the other arc at point O.

Step 8: Now, let us join the points O and B, which intersects side AC at D.

Step 9: Now, let us draw an arc of the radius of 5 cm taking B as centre.

Step 9: Now, let us draw an arc of radius of 5 cm taking C as centre intersecting the previous arc at R and T.

Step 10: Now, let us join the points R and T which intersects side BC at S, which is the circumcentre of the triangle BDC.

Step 11: Let us draw a circle passing through points D, C and B taking S as centre.

Step 12: Now, let us joint the points A and S.

Step 13: Now, let us draw arcs of a radius of 5 cm taking A as centre as shown below:

Step 14: Now, let us draw arcs of the radius of 5 cm taking S as centre as shown below insecting the previous arcs at U and V:

Step 15: Now, let us join the points R and T which intersects AS at W.

Step 16: Now, let us draw arcs of length AW taking W as centre intersecting the circle at points X and B.

Step 17: Now, let us join points A and X, A and B to construct the required tangents.

Note:
We should perform each step carefully to avoid confusion while constructing the triangles and tangents. We should know that the circumcentre lies on the mid-point of the hypotenuse in a right-angle triangle. We should keep in mind that the BD is the altitude of the triangle, not the perpendicular bisector as perpendicular bisectors need to pass through the vertex in a triangle. Similarly, we can expect problems to construct the incircle of the triangle BDC.