CBSE Class 10-Mathematics: Chapter – 11 Constructions Part 15 (For CBSE, ICSE, IAS, NET, NRA 2023)

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Question 26:

Draw a circle of radius . Take two points P and Q on one of its extended diameter each at a distance of from its centre. Draw tangents to the circle from these two points P and Q.

Answer:

To construct: A circle of radius and take two points P and Q on one of its extended diameter each at a distance of from its centre and then draw tangents to the circle from these two points P and Q.

Steps of Construction:

Bisect PO. Let be the mid-point of PO.

(b) Taking M as center and MO as radius, draw a circle. Let it intersects the given circle at the points A and B.

Then PA and PB are the required two tangents.

(d) Bisect QO. Let N be the mid-point of QO.

(e) Taking N as center and NO as radius, draw a circle. Let it intersects the given circle at the points C and D.

(f) Join QC and QD.

Then QC and QD are the required two tangents.

Justification: Join OA and OB

Then is an angle in the semicircle and therefore

Since OA is a radius of the given circle, PA has to be a tangent to the circle. Similarly, PB is also a tangent to the circle.

Again, join OC and OD.

Then is an angle in the semicircle and therefore

Since OC is a radius of the given circle, QC has to be a tangent to the circle. Similarly, QD is also a tangent to the circle.

Question 27:

Draw a line segment AB of length . Taking A as centre, draw a circle of radius and taking B as centre, draw another circle of radius . Construct tangents to each circle from the centre of the other circle.

Answer:

To construct: A line segment of length and taking A as centre, to draw a circle of radius and taking B as center, draw another circle of radius . Also, to construct tangents to each circle from the center to the other circle

Steps of Construction:

(a) Bisect BA. Let be the mid-point of BA.

(b) Taking M as center and MA as radius, draw a circle. Let it intersects the given circle at the points P and Q.

(d) Again, let be the mid-point of AB.

(e) Taking M as center and MB as radius, draw a circle. Let it intersects the given circle at the points R and S.

(f) Join AR and AS. Then, AR and AS are the required two tangents from A to the circle with center B

Justification: Join BP and BQ.

Then being an angle in the semicircle is .

Since AP is a radius of the circle with center A, BP has to be a tangent to a circle with center A. Similarly, BQ is also a tangent to the circle with center A.

Again, join AR and AS.

Then being an angle in the semicircle is .

Since BR is a radius of the circle with center B, AR has to be a tangent to a circle with center B. Similarly, AS is also a tangent to the circle with center B