NCERT Solution for Constructions

In each of the following, give the justification of the construction also:

I. Draw a line segment AB = 7.6 cm.

II. Draw a ray AX making an acute angle with AB.

III. Mark 13 (8 + 5) equal points on AX, and mark them as X_{1}, X_{2}, X_{3}, ........, X_{13}.

IV. Join 'point X_{13}' and B.

V. From 'point X_{5}', draw X_{5}C || X_{13}B, which meets AB at C.

Thus, C divides AB in the ratio 5 : 8

On measuring the two parts, we get:

C = 4.7 cm and BC = 2.9 cm

In Δ ABX_{13} and Δ ACX_{5}, we have

CX_{5} || BX_{13}

I. Draw a Δ ABC such that BC = 6 cm, AC = 5 cm and AB = 4 cm.

II. Draw a ray BX making an acute angle ∠CBX.

III. Mark three points X_{1}, X_{2}, X_{3} on BX such that BX_{1} = X_{1}X_{2} = X_{2}X_{3}.

IV. Join X_{3} C.

V. Draw a line through X_{2} such that it isss parallel to X_{3} C and meets BC at C.

VI. Draw a Iine through C parallel to CA to intersect BA at A'.

Thus, A'BC is the required triangle.

I. Construct a Δ ABC such that AB = 5 cm, BC = 7 cm and AC = 6 cm.

II. Draw a ray BX such that ∠CBX is an acute angle.

III. Mark 7 points of X_{1}, X_{2}, X_{3}, X_{4}, X_{5}, = X_{5}X_{6} = X_{6}X_{7}

IV. Join X_{5} to C.

V. Draw a line through X_{7} intersecting BC (produced) at C' such that X_{5}C || X_{7}C'

VI. Draw a line through C' parallel to CA to intersect BA (produced) at A'. Thus, ΔA'BC' is the required triangle.

I. Draw BC = 8 cm

II. Draw the perpendicular bisector of BC which intersects BC at D.

III. Mark a point A on the above perpendicular such that DA = 4cm.

IV. Join AB and AC.

Thus, Δ ABC is the required isosceles triangle.

V. Now, draw a ray BX such that ∠CBX is an acute angle.

Vl. On BX, mark three points X_{1}, X_{2} and X_{3} such that:

BX_{1} = X_{1}X_{2} = X_{2}X_{3}

VII. Join X_{2} to C.

VIII. Draw a line through X_{3} parallel to X_{2} C and intersecting BC (extended) to C

IX. Draw a line through C' parallel to CA intersecting BA (extended) at A', thus, Δ A'BC' is the required triangle.

We have C'A' || CA [By construction]

∴ Using AA similarity, Δ ABC ~ Δ A' BC'

I. Construct a Δ ABC such that BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

II. Draw a ray such that ∠CBX is an acute angle.

III. Mark four points X_{1}, X_{2}, X_{3} and X_{4} on BX such that

BX_{1} = X_{1}X_{2} = X_{2}X_{3} = X_{3}X_{4}

IV. Join X_{4}C and draw X_{3}C' II X_{4}C such that C' is on BC.

V. Also draw another line through C' and parallel to CA to intersect BA at A'.

Thus, Δ A'BC' is the required triangle.

I. Construct a a ABC such that BC = 7 cm, LB = 45° and LA = 105°.

II. Draw a ray BX making an acute angle ∠CBX with BC.

III. On BX, mark four points X_{1}, X_{2}, X_{3} and X_{4} such that

BX_{1} = X_{1}X_{2} = X_{2}X_{3} = X_{3}X_{4}.

IV. Join X_{3} to C.

V. Draw X_{4}C || X_{3}C such that C' lies on BC (extended).

VI. Draw a line through C' parallel to CA intersecting the extended line segment BA at A'.

Thus, Δ A' BC' is the required triangle.

I. Construct the right triangle ABC such that ∠B = 90°, BC = 4 cm and BA = 3 cm.

II. Draw a ray BX such that an acute angle ∠CBX is formed.

III. Mark 5 points X_{1}, X_{2}, X_{3}, X_{4} and X_{5} on BX such that

BX_{1} = X_{1}X_{2} = X_{2}X_{3} = X_{3}X_{4} = X_{4}X_{5}.

IV. Join X_{3} to C.

V. Draw a line through X_{5} parallel to X_{3} C, intersecting the extended line segment BC at C'.

VI. Draw another line through C' parallel to CA intersecting the extended line segment BA at A'.

Thus, Δ A' BC' is the required triangle.

By construction, we have:

Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

I. With O as centre and radius 6 cm, draw a circle.

II. Take a point P at 10 cm away from the centre.

III. Join O and P.

IV. Bisect OP at M.

V. Taking M as centre and MP or MO as radius, draw a circle.

VI. Let the new circle intersects the given circle at A and B.

VII.Join PA and PB.

Thus, PA and PB are the required two tangents.

By measurement, we have:

PA = PB = 9.6 cm.

Join OA and OB

Since PO is a diameter.

∴ ∠OAF = 90° = ∠OBP [Angles in a semicircle]

Also, OA and OR are radii of the same circle.

⇒ PA and PB are tangents to the circle.

I. Join PO and bisect it such that the mid point of PO is represented by M.

II. Taking M as centre and OM or MP as radius, draw a circle such that this circle intersects the circle (of radius 4 cm) at A and B.

III. Join A and P.

Thus, PA is the required tangent. By measurement, we have:

PA = 4.5 cm

Join OA such that

∠PAO = 90° [Angle in a semi-circle]

⇒ PA ⊥ OA

∵ OA is a radius of the inner circle.

∴ PA has to be a tangent to the inner circle.

I. Join P and O.

II. Bisect PO such that M be its mid-point.

III. Taking M as centre and MO as radius, draw a circle. Let it at A and B.

IV. Join PA and PB.

Thus, PA and PB are the two required tangents from P.

V. Now, join O and Q.

VI. Bisect OQ such that N is its mid point.

VII.Taking N as centre and NO as radius, draw a circle. Let it at C and D.

VIII. Join QC and QD.

Thus, QC and QD are the required tangents to the given

Join OA such that ∠OAP = 90° [Angle in a semi-circle]

⇒ PA ⊥ OA PA is a tangent.

Similarly, PB ⊥ OA ⇒ PB is a tangent

Now, join OC such that ∠QCO = 90° [Angle in a semi-circle]

⇒ QC ⊥ OC ⇒ QC is a tangent.

Similarly, QD ⊥ OC ⇒ QD is a tangent.

I. With centre O and radius = 5 cm, draw a circle.

II. Draw an angle ∠AOB = 120°.

III. Draw a perpendicular on OA at A.

IV. Draw another perpendicular on OB at B.

V. Let the two perpendiculars. meet at C.

CA and CB are the two required tangents to the given circle which are inclined to each other at 60°.

In a quadrilateral OACB, using angle sum property, we have:

120° + 90° + 90° + ∠ACE = 360°

⇒ 300 + ∠ACE = 360°

⇒ ∠ACB = 360° – 300° = 60°.

I. Bisect the line segment AB. Let its mid point be M.

II. With centre as M and MA (or MB) as radius, draw a circle such that it intersects the circle with centre A at the points P and Q.

III. Join BP and BQ.

Thus, BP and BQ are the required two tangents from B to the circle with centre A.

IV. Let the circle with centre M, intersects the circle with centre B at R and S.

V. Join RA and SA.

Thus, RA and SA are the required two tangents from A to the circle with centre B.

Let us join A and P.

∵ ∠APB = 90° [Angle in a semi circle]

∴ BP ⊥ AP

But AP is radius of the circle with centre A.

⇒ BP has to be a tangent to the circle with centre A.

Similarly, BQ has to be tangent to the circle with centre A.

Also, AR and AS have to be tangent to the circle with centre B.

I. Join AO (0 is the centre of the circle passing through B, C and D.)

II. Bisect AO. Let M be the mid point of AO.

III. Taking M as centre and MA as radius, draw a circle intersecting the given circle at B and E.

IV. Join AB and AE. Thus, AB and AE are the required two tangents to the given circle from A.

Join OE, then ∠AEO = 90° [Angle being in a semi circle]

∴ AE ⊥ OE.

But OE is a radius of the given circle.

⇒ AE has to be a tangent to the circle.

Similarly, AB is also a tangent to the given circle.

I. Draw the given circle using a bangle.

II. Take two non parallel chords PQ and RS of this circle.

III. Draw the perpendicular bisectors of PQ and RS such that they intersect at O. Therefore, O is the centre of the given circle.

IV. Take a point P' outside this circle.

V. Join OP' and bisect it. Let M be the mid point of OP'.

VI. Taking M as centre and OM as radius, draw a circle. Let it intersect the given circle at A and B.

VII.Join P'A and P'B. Thus, P'A and P'B are the required two tangents.

Join OA and OB.

Since ∠OAP = 90°

∴ PA ⊥ OA

Also OA is a radius.

∴ PA has to be a tangent to the given circle.

Similarly, PB is also a tangent to the given circle.