Chapter 11 Construction

Explanation:

Presented: A line segment of length 7.6 cm.

Need To construct: To divide it in the ratio 5: 8 and to measure the two parts.

Construction steps include:

(a)Every ray AX will form a sharp angle with AB.

(b) Locate 13 (=5 + 8) points on AX such that.

(c) Combine

(d) Draw a line through the point that is parallel to and intersects AB at point C.

AC: CB = 5: 8 then.

As measured, AC is 3.1 cm and CB is 4.5 cm.

Need to justify:

[ construction]

[using Basic Proportionality Theorem]

[Using construction]

Hence,

AC: CB = 5: 8

Explanation:

Need To construct: 

first build a triangle with sides of 4 cm, 5 cm, and 6 cm, and then another triangle with sides matching those of the first triangle.

Construction steps include:

(a) Sketch an ABC triangle with sides of 4, 5, and 6 cm.

(b) From any ray BX that forms a sharp angle with BC on the side that is diagonal from vertex A.

(c)Find three spots on the BX such that

(d) Connect the points, draw a line through them, and intersect BC at point C' with a line parallel to

(e) To intersect BA at A', draw a line through C' that is parallel to the line CA.

A'BC' is the necessary triangle as a result.

Need to justify:

[using construction]

[using Basic Proportionality Theorem]

[usingconstruction]

Hence,

………(i)

CA C’A’ [using construction]

BC’A’ BCA [AA similarity]

[by eq. (i)]

Explanation:

Need To construct: To build a triangle with sides of 5 cm, 6 cm, and 7 cm, and then another triangle with sides matching those of the first triangle.

Construction steps include:

(a) Sketch an ABC triangle with sides measuring 5, 6, and 7 cm.

(b) From any ray BX that forms a sharp angle with BC on the side that is diagonal from vertex A.

(c)Find seven spots on the BX such that

(d) Connect the points, draw a line through them, and intersect BC at point C' with a line parallel to BC.

(f) To intersect BA at A', draw a line through C' that is parallel to the line CA.

A'BC' is the necessary triangle as a result.

Need to Justify:

C’A’CA [using construction]

ABCA’BC’ [AA similarity]

[using Basic Proportionality Theorem]

[using construction]

[AA similarity]

[using construction]

Explanation:

Need To construct: To construct an isosceles triangle whose base is 8 cm and altitude is 4 cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle.

Construction steps include:

(a)To draw BC = 8 cm (a).

(b) Sketch the BC perpendicular bisector. Let's meet it at D and BC.

(c) Place an asterisk (*) at position A on the perpendicular bisector where AD = 4 cm.

(d) Admit AB and AC together. ABC is the necessary isosceles triangle as a result.

(e) From any ray BX that forms a sharp angle with BC on the side that is diagonal from vertex A.

(f)Find three points on the BX such that .

(g) Connect the point, draw a line across it, and intersect BC at point C' with a line parallel to

(h) To intersect BA at A', draw a line through C' that is parallel to the line CA.

A'BC' is the necessary triangle as a result.

Need to Justify:

C’A’CA [using construction]

ABCA’BC’ [using similarity]

[using Basic Proportionality Theorem]

[using construction]

[AA similarity]

But [using construction]

Explanation:

Need To construct: To build a triangle ABC with sides BC = 6 cm, AB = 5 cm, and ABC = , then another triangle with sides that are the same as those of the original triangle ABC.

Construction steps include:

(a)Sketch a triangle ABC in which the sides are BC = 6 cm, AB = 5 cm, and ABC =

(b) From any ray BX that forms a sharp angle with BC on the side that is diagonal from vertex A.

(c)Find four spots on the BX such that

(d) Connect the points, draw a line through them, and intersect BC at point C' with a line parallel to

(f) To intersect BA at A', draw a line through C' that is parallel to the line CA.

A'BC' is the necessary triangle as a result.

Need to justify:

[using construction]

[using Basic Proportionality Theorem]

But [using construction]

………(i)

CA C’A’ [using construction]

BC’A’ BCA [using similarity]

[By eq. (i)]

Explanation:

Need To construct: To build a triangle ABC with sides BC = 7 cm, B = and C = and after that another triangle with sides that are the same as those of the first triangle ABC.

Construction steps include:

(a)A triangle ABC with sides BC = 7 cm, B = and C = should be drawn.

(b) From any ray BX that forms a sharp angle with BC on the side that is diagonal from vertex A.

 (c) Find four spots on the BX such that

(d) Connect the points, draw a line through them, and intersect BC at point C' with a line parallel to

(f) To intersect BA at A', draw a line through C' that is parallel to the line CA.

A'BC' is the necessary triangle as a result.

Need to Justify:

[using construction]

[using AA similarity]

[using Basic Proportionality Theorem]

But [using construction]

………(i)

CA C’A’ [using construction]

BC’A’ BCA [using AA similarity]

[By eq. (i)]

Explanation:

Need To construct: To build a right triangle with sides (other than the hypotenuse) that are 4 cm and 3 cm in length, and then another triangle that is comparable to it with sides that are the same lengths as the corresponding sides of the first triangle ABC.

Construction steps include:

(a) Sketch a right triangle with sides that are 4 cm and 3 cm long (apart from the hypotenuse).

(b) From any ray BX that forms a sharp angle with BC on the side that is diagonal from vertex A.

(c) Find five spots on the BX such that

(d) Connect the points, draw a line through them, and intersect BC at point C' with a line parallel to

(f) To intersect BA at A', draw a line through C' that is parallel to the line CA.

A'BC' is the necessary triangle as a result.

Need to Justify:

[using  construction]

[using  similarity]

[using Basic Proportionality Theorem]

But [using construction]

Therefore, ………(i)

CA C’A’ [using construction]

BC’A’ BCA [using AA similarity]

[By eq. (i)]

Explanation:

Presented: a circle with an O-shaped centre, a 6 cm radius, and a point P located 10 cm from the centre.

Need To construct: to create the two circle tangents and calculate their lengths.

Construction steps include:

(a) Connect PO and cut it in half. Let M represent PO's midpoint.

(b) Draw a circle with M as the centre and MO as the radius. At the locations Q and R, let it intersect the specified circle.

(c)Join PR and PQ.

The requisite two tangents are then PQ and PR.

By calculation, PQ = PR = 8 cm.

Need to justify: 

Integrate OQ and OR.

So the angles in semicircles are  OQP and ORP a

OQP = ORP

Moreover, PQ and PR will be the tangents to the circle at Q and R, respectively, as OQ and OR are the circle's radii.

We can observe that the supplied circle is increased by two points by the circle with OP as its diameter. There can only be two tangents drawn as a result.

Explanation:

Need To construct: To create a tangent to a circle with a radius of 4 cm from a point on a circle with a radius of 6 cm and measure the lengths of the tangent. Moreover, to confirm the measurements through accurate computation.

Construction steps include:

(a) Connect PO and cut it in half. Let M represent PO's midpoint.

(b) Draw a circle with M as the centre and MO as the radius. Let Q and R be the points where it intersects the provided circle.

(c)Combine PQ

PQ is then the necessary tangent.

By calculation, PQ = 4.5 cm

When calculated accurately,

PQ =

= =

= = 4.47 cm

Need to justify : 

Adopt OQ. PQO is thus an angle in the semicircle, hence,

PQO =

PQ OQ

PQ must be tangent to the circle because OQ is a radius of the supplied circle.

Explanation:

Need To construct: Draw tangents to the circle from the two points P and Q on a circle with a radius of 3 cm and an extended diameter with two points P and Q on each at a distance of 7 cm from the centre.

Construction steps include:

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

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

(c) Join PA and PB.

The PA and PB are the required two tangents.

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

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

(f) Join QC and QD.

The QC and QD are the required two tangents.

(a) Split the PO. Let M represent PO's midpoint.

(b) Draw a circle with M as the centre and MO as the radius. At positions A and B, let it intersect the specified circle.

(c) Combine PA and PB in (c).

The requisite two tangents are then PA and PB.

(d) Divide the QO. N should represent the middle of QO.

(e) Draw a circle with N as the centre and NO as the radius. At positions C and D, let it intersect the specified circle.

(f)Combine the QC and QD.

The requisite two tangents are then QC and QD.

Need to Justify: 

Combine OA and OB.

In such case, PAO is an angle in the semicircle, hence PAO =

PA OA

PA must be tangent to the circle because OA is a radius of the specified circle. Similar to PB, the circle is also tangent to PB.

Rejoin OC and OD now.

In such a case, QCO is an angle in the semicircle, and QCO =.

QC must be tangent to the circle because OC is the radius of the supplied circle. QD is likewise tangent to the circle in a similar way.

Explanation:

Need To construct: A set of tangents to a circle with a radius of 5 cm that is angled at an angle of

Construction steps include:

(a) Sketch an O-centered circle with a radius of 5 cm.

(b) Sketch an angle AOB of

(c) Sketch angles at A and B that converge at C.

The necessary tangents, AC and BC, are then inclined to one another at an angle of

Need to Justify:

OAC = and radius is OA.

[using construction]

A tangent to the circle is AC.

OBC = and radius is OB.

[using construction]

A tangent to the circle in BC.

Currently, in the OACB quadrilateral,

AOB + OAC + OBC + ACB =

[using Angle sum property of a quadrilateral]

+ ACB =

+ ACB =

ACB = 

Explanation:

Need To construct: Draw two circles, one with a radius of 4 cm and the other of 3 cm, using an 8cm line segment with A and B as the centres, respectively. Moreover, to create tangents from each circle's centre to the neighbouring circle.

Construction steps include:

(a)Cut BA Let M represent the centre of BA.

(b) Draw a circle with M as the centre and MA as the radius. Let's say it crosses the specified circle at P and Q.

(c) Join BQ and BP.

The needed two tangents from B to the circle with centre A are then BP and BQ.

(d) Once more, let M represent the centre of AB.

(e) Draw a circle with MB as the radius and M as the centre. At the locations R and S, let it intersect the specified circle.

(f)Join AS and AR.

The needed two tangents from A to the circle with centre B are then AR and AS.

Need to Justify: 

Combine BP and BQ.

APB is because it is an angle in the semicircle. 

BP AP

BP must be tangent to a circle with centre A because AP is a radius of the circle with centre A. Similar to how BQ is tangent to the circle with A as its centre.

Rejoin AS and AR.

AsARB is an angle within the semicircle, hence equal to

AR BR

As the circle with centre B has radius BR, the circle with centre B must have tangent AR. Similar to how AS is tangent to the circle with B as its centre.

Explanation:

To construct: The perpendicular from B on AC and the tangents from A to this circle form the right triangle ABC with AB = 6 cm, BC = 8 cm, and B =

Construction steps include:

(a) Draw a right triangle ABC with the following measurements: AB = 6 cm, BC = 8 cm, and B = Additionally, sketch a perpendicular BD on AC.

(b) Connect to AO and split it at M (here O is the centre of the circle through B, C, D).

(c) Draw a circle using M as the centre and MA as the radius. Let's say it crosses the provided circle at B and E.

(d) Align with AB and AE.

The requisite two tangents are then AB and AE.

Need to justify: Adopt OE.

AEO is then a semicircle angle.

AEO =

AE OE

AE must be a tangent to the circle since OE is a radius of the supplied circle. Similar to AB, the circle is also tangent to AB.

Explanation:

Need To construct: With a bangle, create a circle. Choose a location beyond the circle. Create the two tangents to the circle from this location.

Construction steps include:

(a) Use a bracelet to draw a circle.

(b) Choose the AB and CD chords of this circle, which are not parallel.

(c) Sketch AB and CD's perpendicular bisectors. Let O be where these cross. The drawn circle's center is then O.

(d) Bring point P outside of the circle.

(e) Combine PO and cut it in half. Let M represent PO's midpoint.

(f) Draw a circle using M as the centre and MO as the radius. At the locations Q and R, let it intersect the specified circle.

(g) Combine  PR and PQ.

The requisite two tangents are then PQ and PR.

Need to justify: 

Integrate OQ and OR.

PQO is then an angle within the semicircle.

PQO =

PQ OQ

PQ must be tangent to the circle because OQ is a radius of the supplied circle. PR is likewise tangent to the circle in a similar way.