Question

Draw the figures below according to the specifications given.

Answer 1

(1)

The steps of construction to construct the required figure are as follows:

1) Draw a circle with centre O and radius OA = 2 cm.

2) Extend line segment OA to a point P such that OP = 6 cm (= 4 cm + 2 cm).

3) Mark a point B on line segment OP such that PB = BO = = 3 cm.

4) Draw a circle with B as the centre and BO as the radius. Name the points where it intersects the previous circle at S and T.

5) Join PS and PT.

(2)

The given figure is:

We know that any tangent to circle is perpendicular to the radius to the point of contact.

PS is the tangent to the given circle and SO is the radius of the circle.

∴ ∠OSP = 90°

Applying Pythagoras theorem in ΔOSP:

OP² = PS² + SO²

Now, we can construct the required figure. The steps of construction are as follows:

1) Draw a circle with centre O and radius OA = 2.5 cm.

2) Extend line segment OA to a point P such that OP = 6.5 cm.

3) Mark a point B on line segment OP such that PB = BO = = 3.25 cm.

4) Draw a circle with B as the centre and BO as the radius. Name the points where it intersects the previous circle at S and T.

5) Join PS, PT and OS.

(3)

We have joined the centre of the circle given in the figure with the point where the tangent intersects the circle.

The given figure is:

We know that any tangent to circle is perpendicular to the radius to the point of contact.

PS is the tangent to the given circle and SO is the radius of the circle.

∴ ∠OSP = 90°

Applying Pythagoras theorem in ΔOSP:

OP² = PS² + SO²

⇒ SO² = OP² − PS²

Now, we can construct the required figure. The steps of construction are as follows:

1) Draw a circle with centre O and radius OA = 3.32 cm.

2) Extend line segment OA to a point P such that OP = 5.2 cm.

3) Mark a point B on line segment OP such that PB = BO = = 2.6 cm.

4) Draw a circle with B as the centre and BO as the radius. Name the points where it intersects the previous circle at S and T.

5) Join PS and PT.