Constructions

Construction of Triangle Similar to given Triangle

Consider a triangle ABC. Let us construct a triangle similar to ∆ABC such that each of its sides is ||of ${\left(\frac{m}{n}\right)}^{th}$the corresponding sides of  ABC.

Steps of constructions when m < n:

Step 1: Construct the given triangle ABC by using the given data.

Step 2: Take any one of the three side of the given triangle as base. Let AB be the base of the given triangle.

Step 3: At one end, say A, of base AB. Construct an acute angle BAX below the base AB.

Step 4: Along AX mark off n points A₁, A₂, A₃,………, A_n such that

AA₁ = A₁A₂ = ……… = A_n-1 A_n

Step 5: Join A_nB

Step 6: Draw AmB’ parallel to AnB which meets AB at B’.

Step 7: From B’ draw B’C’||BC meeting AC at C’.

Triangle AB’C’ is the required triangle each of whose sides is ${\left(\frac{m}{n}\right)}^{th}$of the corresponding side of ΔABC.

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Justification

Since A_m B’|| A_n B. Therefore

$\frac{AB‘}{B‘B}=\frac{{AA}_m}{A_m{A}_n}$[by basic proportionality theorem]

$\frac{AB‘}{B‘B}=\frac{m}{n-m}$

$\frac{B‘B}{AB‘}=\frac{n-m}{m}$

Now, $\frac{AB}{AB‘}=\frac{A{B}^{‘}+B‘B}{AB‘}$

$\frac{AB}{AB‘}=1+\frac{B‘B}{AB‘}=1+\frac{n-m}{m}+\frac{n}{m}$

$\frac{AB‘}{AB}=\frac{m}{n}$

In triangles ABC and AB’C’, we have

∠BAC = ∠B’ AC’

And ∠ABC = ∠AB’ C’

So, by AA similarity criterion, we have

∆AB’C’ ̴ ∆ABC

$\frac{AB‘}{AB}=\frac{B‘C‘}{BC}=\frac{AC‘}{AC}$

$\frac{AB‘}{AB}=\frac{B‘C‘}{BC}=\frac{AC‘}{AC}=\frac{m}{n}$

Steps of construction when m > n:

Step 1: Construct the given triangle by using the given data.

Step 2: Take any one of the three sides of the given triangle and consider it as the base. Let AB be the base of the given triangle.

Step 3: At one end, say A, of base AB. Construct an acute angle ∠BAX below base AB i.e., on the opposite side of the vertex C.

Step 4: Along AX mark off m (large of m and n) points A1, A2, A3,………Am of AX such that

AA₁ = A₁A₂ = ………= A_m-1A_m.

Step 5: Join A_nB to B and draw a line through A_m parallel to A_nB, intersecting the extended line segment AB at B’.

Step 6: Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. Step 7: ΔAB’C’ so obtained is the required triangle.

Justification

Consider triangle ABC and AB’ C’. We have:

∠BAC = ∠B‘ AC ‘

∠ABC = ∠AB ‘C ‘

So, by AA similarity criterion,

ΔABC ̴ ΔAB ‘C’

$\frac{AB}{AB‘}=\frac{BC}{B‘C‘}=\frac{AC}{AC‘}$

ΔA A_m B’, A_nB || A_mB’

$\therefore \frac{AB}{BB‘}=\frac{{AA}_n}{A_n{A}_m}$

$\frac{BB‘}{AB}=\frac{A_n{A}_m}{{AA}_n}$

$\frac{BB‘}{AB}=\frac{m-n}{n}$

$\frac{{AB}^{‘}-AB}{AB}=\frac{m-n}{n}$

$\frac{{AB}^{‘}}{AB}-1=\frac{m-n}{n}$

$\frac{{AB}^{‘}}{AB}=\frac{m}{n}$

From (i) and (ii), we have

$\frac{AB‘}{AB}=\frac{B‘C‘}{BC}=\frac{AC‘}{AC}=\frac{m}{n}$

The tangent to a circle is a line that intersects the circle at exactly one point.

Tangent to a circle is perpendicular to the radius through the point of contact.

Construction of Triangle to a Circle from a point outside the Circle

Construction of a tangent to a circle from a point outside the circle, when its centre is known

The steps of constructions are as follows:

Step 1: Join the centre O of the circle to the point P.

Step 2: Draw perpendicular bisector of OP intersecting OP at Q.

Step 3: With Q as centre and radius OQ, draw a circle. This circle has OP as its diameter.

Step 4: Let this circle intersect the first circle at two points T and T’. Join PT and P T’ .

PT and P T’ are the two tangents to the given circle from the point P.

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Justification

Join OT and OT’

It can be seen that ∠PTO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle.

∴∠ PTO = 90°

OT ꓕ PT

Since OT is the radius of the circle, PT has to be a tangent of the circle. Similarly, PT’ is a tangent of the circle.

Number of Tangents to a circle from a given point

If the point is in an interior region of the circle, any line through that point will be a secant. So, in this case, there is no tangent to the circle.

AB is a secant drawn through the point S

When the point lies on the circle, there is accurately only one tangent to a circle.

PQ is the tangent touching the circle at A

When the point lies outside of the circle, there are exactly two tangents to a circle.

PT1 and PT2 are tangents touching the circle at T1 and T2

Drawing tangents to a circle from a point outside the circle

To construct the tangents to a circle from a point outside it.

Consider a circle with centre O and let P be the exterior point from which the tangents to be drawn.

Step 1: Join the PO and bisect it. Let M be the midpoint of PO.

Step 2: Taking M as the centre and MO(or MP) as radius, draw a circle. Let it intersect the given circle at the points Q and R.

Step 3: Join PQ and PR

Step 4:PQ and PR are the required tangents to the circle.

Important Questions

Multiple Choice questions-

1. To divide a line segment AB in the ratio p : q (p, q are positive integers), draw a ray AX so that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is

(a) greater of p and q

(b) p + q

(c) p + q – 1

(d) pq

2. To draw a pair of tangents to a circle which are inclined to each other at an angle of 35°. It is required to draw tangents at the end points of those two radii of the circle, the angle between which is

(a) 105°

(b) 70°

(c) 140°

(d) 145°

3. To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn so that ∠BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is

(a) 8

(b) 10

(c) 11

(d) 12

4. To divide a line segment AB in the ratio 4 : 7, ray AX is drawn first such that ∠BAX is an acute angle and then points A₁, A₂, A₃,……… are located at equal distances on the ray AX and the point B is joined to

(a) A₁₂

(b) A₁₁

(c) A₁₀

(d) A₉

5. To divide a line segment AB in the ratio 5 : 6, draw a ray AX such that ∠BAX is an acute angle, then draw a ray B4 parallel to AX and the points A₁, A₂, A₃, …….. and B₁, B₂, B₃,………. are located at equal distances on ray AX and B4, respectively. Then the points joined are:

(a) A₅ and B₆

(b) A₆ and B₅

(c) A₄ and B₅

(d) A₅ and B₄

6. To construct a triangle similar to a given ΔABC with its sides $\frac{3}{7}$ of the corresponding sides of ΔABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B₁, B₂, B₃, on BX at equal distances and next step is to join

(a) B₁₀ to C

(b) B₃ to C

(c) B₇ to C

(d) B₄ to C

7. To construct a triangle similar to a given ΔABC with its sides $\frac{8}{5}$ of the corresponding sides of ΔABC draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect to BC. Then minimum number of points to be located at equal distances on ray BX is

(a) 5

(b) 8

(c) 13

(d) 3

8. To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be:

(a) 135°

(b) 90°

(c) 60°

(d) 120°

9. To construct a pair of tangents to a circle at an angle of 60° to each other, it is needed to draw tangents at endpoints of those two radii of the circle, the angle between them should be:

(a) 100°

(b) 90°

(c) 180°

(d) 120°

10. A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of ___________ from the centre.

(a) 3.5 cm

(b) 2.5 cm

(c) 5 cm

(d) 2 cm

  Very Short Questions:

Long Questions :

Case Study Questions:

Assertion Reason Questions-

Assertion: a, b and c are the lengths of three sides of a triangle, then a+b > c

Reason: The sum of two sides of a triangle is always greater than the third side.

Assertion: The side lengths 4cm, 4cm and 4cm can be sides of equilateral triangle.

Reason: Equilateral triangle has all its three sides equal.

Answer Key

Multiple Choice questions-

Very Short Answer :

Given AB = 8 cm

AB’ = $\frac{3}{4}$ of AB

= $\frac{3}{4}$ × 8 = 6 cm

BB’ = AB – AB’ = 8 – 6 = 2 cm.

⇒ AB’: BB’ = 6 : 2 = 3 : 1

Hence the required ratio is 3 : 1.

Hence, PA : PB = 3 : 4

∴ AP : PB = 3 : 2

Hence, AP : AB = 3 : 5 or $\frac{AP}{AB}=\frac{3}{5}$

Short Answer :

Long Answer :

Case Study Answer:

Assertion Reason Answer-

(a) Both A and R are true and R is the correct explanation of A.

(a) Both A and R are true and R is the correct explanation of A.

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