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10.1 DIVISION OF A LINE SEGENT :

In order to divide a line segment internally is a given ratio $m$ : $n$, where both $m$ and $n$ are positive integers, we follow the following steps:

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Step of construction :

(i) Draw a line segment $A B$ of given length by using a ruler.

(ii) Draw and ray $A X$ making an acute angle with $A B$.

(iii) Along $A X$ mark off $(m+n)$ points $A_1, A_2, \ldots, A_{m+n}$ such that $A A_1=A_1 A_2=\ldots=A_{m+n+n} A_{m+n}$.

(iv) Join $B A_{m+n}$

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(v) Through the point $A_m$ draw a line parallel to $A_{m+n} B$ by making an angle equal to $\angle A A_{m+n} B$ at $A_m$. Suppose this line meets $A B$ at a point $P$.

The point $P$ so obtained is the required point which divides $A B$ internally in the ratio $m: n$.

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Ex. 1 Divide a line segment of length $12 cm$ internally in the ratio $3: 2$.

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Sol. Following are the steps of construction.

Step of construction :

(i) Draw a line segment $A B=12 cm$ by using a ruler.

(ii) Draw any ray making an acute angle $\angle B A X$ with $A B$.

(iii) Along $A X$, mark-off $5(=3+2)$ points $A_1, A_2, A_3, A_4$ and $A_5$ such that $A A_1=A_1 A_2=A_2 A_3=A_3 A_4=$ $A_4 A_5$.

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(iv) Join $B A_5$

(v) Through $A_3$ draw a line $A_3 P$ parallel to $A_5 B$ by making an angle equal to $\angle A A_5 B$ at $A_3$ intersecting $A B$ at a point $P$.

The point $P$ so obtained is the required point, which divides $A B$ internally in the ratio $3: 2$.

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10.2 ALTERNATIVE METHOD FOR DIVISION OF A LINE SEGMENT INTERNALLY IN A GIVEN RATIO :

Use the following steps to divide a given line segment $A B$ internally in a given ration $m: n$, where $m$ and natural members.

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Steps of Construction :

(i) Draw a line segment $A B$ of given length.

(ii) Draw any ray $A Z$ making an acute angle $\angle B A X$ with $A B$.

(iii) Draw a ray $B Y$, on opposite side of $A X$, parallel to $A X$ making an angle $\angle A B Y$ equal to $\angle B A X$.

(iv) Mark off a points $A_1, A_2, \ldots . A_m$ on $A X$ and $n$ points $B_1, B_2, \ldots B_n$ on $B Y$ such that $A A_1=A_1 A_2=$ $A_{m-1} A_m=B_1 B_2=\ldots . B_{n-1} B_n$.

(v) Join $A_m B_n$. Suppose it intersect $A B$ at $P$.

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The point $P$ is the required point dividing $A B$ in the ratio $m: n$.

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Ex. 2 Decide a line segment of length $6 cm$ internally in the ratio $3: 4$.

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Sol. Follow the following steps :

Steps of Construction :

(i) Draw a line segment $A B$ of length $6 cm$.

(ii) Draw any ray $A X$ making an acute angle $\angle B A X$ with $A B$.

(iii) Draw a ray $B Y$ parallel to $A X$ by making $\angle A B Y$ equal to $\angle B A X$.

(iv) Mark of three point $A_1, A_2, A_3$ on $A X$ and 4 points $B_1, B_2 m B_3$, $B_4$ on $B Y$ such that $A A_1=A_1 A_2=A_2 A_3$ $=BB_1=B_1 B_2=B_2 B_3=B_2 B_4$.

(v) Join $A_3 B_4$. Suppose it intersects $A B$ at a point $P$.

Then, $P$ is the point dividing $A B$ internally in the ratio 3:4.

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10.3 CONTRUCTION OF A TRIANGLE SIMILAR TO A GIVEN TRIANGLE :

Scale Factor: The ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle is known as their scale factor.

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Steps of Construction when $\mathbf{m}<\mathbf{n}$ :

(i) Construct the given triangle $ABC$ by using the given data.

(ii) Take any one of the three side of the given triangle as base. Let $A B$ be the base of the given triangle.

(iii) At one end, say $A$, of base $A B$. Construct an acute angle $\angle B A X$ below the base $A B$.

(iv) Along $A X$ mark of $n$ points $A_1, A_2, A_3, \ldots . . A_n$ such that $A A_1=A_1 A_2=\ldots . .=A_{n-1} A_n$.

(v) Join $A_n B$.

(vi) Draw $A_m B^{\prime}$ parallel to $A_n B$ which meets $A B$ at $B^{\prime}$. (vii) From $B^{\prime}$ draw $B^{\prime} C^{\prime} | C B$ meeting $A C$ at $C^{\prime}$.

Triangle $A B^{\prime} C^{\prime}$ is the required triangle each of whose side is $(\frac{m}{n})^{\text {th }}$ of the corresponding side of $\triangle A B C$.

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Ex. 3 Construction a $\triangle ABC$ in which $AB=5 cm, BC=6 cm$ and $AC=7 cm$. Now, construct a triangle similar to $\triangle ABC$ such that each of its side is two-third of the corresponding side of $\triangle ABC$.

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Sol. Steps of Construction

(i) Draw a line segment $A B=5 cm$.

(ii) With $A$ as centre and radius $AC=7 cm$, draw an arc.

(iii) With $B$ as centre and $B C=6 cm$, draw another arc, intersecting the arc draw in step (ii) at $C$.

(iv) Join $A C$ and $B C$ to obtain $\triangle A B C$.

(v) Below $A B$, make an acute angle $\angle B A X$.

(vi) Along $A X$, mark off three points (greater of 2 and 3 in $\frac{2}{3}$ ) $A_1, A_2, A_3$ such that $A A_1=A_1 A_2=A_2 A_3$.

(vii) Join $A_3 B$.

(viii) Draw $A_2 B^{\prime} || A_3 B$, meeting $A B$ at $B^{\prime}$.

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(iv) From $B^{\prime}$, draw $B^{\prime} C^{\prime} || B C$, meeting $A C$ at $C^{\prime}$.

$A B^{\prime} C^{\prime}$ is the required triangle, each of the whose sides is two-third of the corresponding sides of $\triangle A B C$.

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Steps of Construction when $\mathbf{m}>\mathbf{n}$ :

(i) Construct the given triangle by using the given data.

(ii) Take any of the three sides of the given triangle and consider it as the base. Let $A B$ be the base of the given triangle.

(iii) At one end, say $A$, of base $AB$ construct an acute angle $\angle BAX$ below base $AB$ i.e. on the composite side of the vertex C.

(iv) Along $A X$, mark-off $m$ (large of $m$ and $n$ ) points $A_1, A_2, \ldots . . A_m$ on $A X$ such that $A A_1=A_1 A_2=\ldots . A_{m-1} A_m$.

(v) Join $A_n$ to $B$ and draw a line through $A_m$ parallel to $A_n B$, intersecting the extended line segment $A B$ at $B$ ‘.

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(vi) Draw a line through $B^{\prime}$ parallel to $B C$ intersecting the extended line segment $A C$ at $C^{\prime}$.

(vii) $\triangle AB^{\prime} C^{\prime}$ so obtained is the required triangle.

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Ex. 4 Draw a triangle $ABC$ with side $BC=7 cm, \angle B=45^{\circ}, \angle A=150^{\circ}$ Construct a triangle whose side are $(4 / 3)$ times the corresponding side of $\triangle ABC$.

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Sol. In order to construct $\triangle ABC$, follow the following steps :

(i) Draw $BC=7 cm$.

(ii) At B construct $\angle CBX=45^{\circ}$ and at $C$ construct $\angle BCY=180^{\circ}-(45^{\circ}+105^{\circ})=30^{\circ}$ Suppose $B C$ and $C Y$ intersect at $A$. $\triangle A B C$ so obtained is the given triangle.

(iii) Construct an acute angle $\angle C B Z$ at $B$ on opposite side of vertex $A$ of $\triangle A B C$.

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(iv) Mark-off four (greater of 4 and 3 in $\frac{4}{3}$ ) points, $B_1, B_2, B_3, B_4$ on $B Z$ such that $B_2-B_1 B_2=V_2 B_3=B_3 B_4$.

(v) Join $B_3$ ( the third point) to $C$ and draw a line through $B_4$ parallel to $B_3 C$, intersecting the extended line segment $B C$ at $C^{\prime}$.

(vi) Draw a line through $C^{\prime}$ parallel to $C A$ intersecting the extended line segment $B A$ at $A^{\prime}$ Triangle $A^{\prime} B C$ ’ so obtained is the required triangle such that $\frac{A^{\prime} B^{\prime}}{A B}=\frac{B C^{\prime}}{B C}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{4}{3}$

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10.4 CONSTRCUTION OF TANGENT TO A CIRCLE :

10.4 (a)To Draw the Tangent to a Circle at a Given Point on it, When the Centre of the Circle is Known :

Given: A circle with centre $O$ and a point $P$ and it.

Required : To draw the tangent to the circle at $P$.

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Steps of Construction.

(i) Join OP.

(ii) Draw a line AB perpendicular to $O P$ at the point $P$. APB is the required tangent at $P$.

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Ex. 5 Draw a circle of diameter $6 cm$ with centre O. Draw a diameter AOB. Through A or B draw tangent to the circle.

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Sol. Given : A circle with centre $O$ and a point $P$ on it.

Required : To draw tangent to the circle at $B$ or $A$.

Steps of Construction.

(i) With $O$ as centre and radius equal to $3 cm$ ( $6 \div 2$ ) draw a circle.

(ii) Draw a diameter $AOB$.

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(iii) Draw $C D \perp A B$.

(iv) So. CD is the required tangent.

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10.4 (b)To Draw the Tangent to a Circle at a Given Point on it, When the Centre of the Circle is not Known :

Given : A circle and a point $P$ on it.

Required : To draw the tangent to the circle at $P$.

Steps of Construction

(i) Draw any chord $P Q$ and Joint $P$ and $Q$ to a point $R$ in major arc $P Q$ (or minor arc $P Q$ ).

(ii) Draw $\angle QPB$ equal to $\angle PRQ$ and on opposite side of chord $PQ$.

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The line BPA will be a tangent to the circle at $P$.

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Ex. 6 Draw a circle of radius $4.5 cm$. Take a point $P$ on it. Construct a tangent at the point $P$ without using the centre of the circle. Write the steps of construction.

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Sol. Given : To draw a tangent to a circle at $P$.

Steps of Construction

(i) Draw a circle of radius $=4.5 cm$.

(ii) Draw a chord $P Q$, from the given point $P$ on the circle.

(iii) Take a point $R$ on the circle and joint $P R$ and $Q R$.

(iv) Draw $\angle QPB=\angle PRQ$ on the opposite side of the chord $PQ$.

(v) Produce BP to A. Thus, APB is the required tangent.

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10.4 (c) To Draw the Tangent to a Circle from a Point Outside it (External Point) When its Centre is known

Given : $A$ circle with centre $O$ and a point $P$ outside it.

Required : To construct the tangents to the circle from $P$.

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Steps of Construction :

(i) Join $OP$ and bisect it. Let $M$ be the mid point of OP.

(ii) Taking $M$ as centre and $M O$ as radius, draw a circle to intersect $C(O, r)$ in two points, say $A$ and $B$

(iii) Join PA and $PB$. These are the required tangents from $P$ to $C(O, r)$

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Ex. 7 Draw a circle of radius $2.5 cm$. From a point $P, 6 cm$ apart from the centre of a circle, draw two tangents to the circle.

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Sol. Given: A point $P$ is at a distance of $6 cm$ from the centre of a circle of radius $2.5 cm$

Required : To draw two tangents to the circle from the given point P.

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Steps of Construction :

(i) Draw a circle of radius $2.5 cm$. Let it centre be $O$.

(ii) Join OP and bisect it. Let $M$ be mid-point of OP.

(iii) Taking $M$ as centre and $M O$ as radius draw a circle to intersect $C$ in two points, say $A$ and $B$.

(iv) Join PA and PB. These are the required tangents from $P$ to $C$.

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10.4 (d) To Draw Tangents to a Circle From a Point Outside it (When its Centre is not Known):

Given : $P$ is a point outside the circle.

Required : To draw tangents from a point $P$ outside the circle.

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Steps of Construction :

(i) Draw a secant $PAB$ to intersect the circle at $A$ and $B$.

(ii) Produce AP to a point $C$, such that $P A=P C$.

(iii) With $B C$ as a diameter, draw a semicircle.

(iv) Draw PD $\perp CB$, intersecting the semicircle at $D$.

(v) Taking PD as radius and $P$ as centre, draw arcs to intersect the circle at $T$ and $T$ ‘.

(iv) Join PT and PT’. Then, PT and PT’ are the required tangents.

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Ex. 8 Draw a circle of radius $3 cm$. From a point $P$, outside the circle draw two tangents to the circle without using

the centre of the circle.

Given : A point $P$ is outside the circle of radius $3 cm$.

Required : To draw two tangents to the circle from the point $P$, without the use of centre.

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Steps of constructing

(i) Draw a circle of radius $3 cm$.

(ii) Take a point $P$ outside the circle and draw a secant $P A B$, intersecting the circle at $A$ and $B$.

(iii) Produce AP to $C$ such that $A P=C P$.

(iv) Draw a semicircle, wit $C B$ as a diameter.

(v) Draw PD $\perp A B$, intersecting the semi-circle AT D.

(vi) With $PD$ as radius and $P$ as centre draw two arcs to intersect the given circle at $T$ and $T$ ‘.

(vii) Joint PT and PT’. Which are the required tangents.

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DAILY PRATICE PROBLEMS 10

SUBEJCTIVE DPP -10.1

1. Draw a circle of radius $2.5 cm$. Take a point $P$ on it. Draw a tangent to the circle at the point $P$.

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2. From a point $P$ on the circle of radius $4 cm$, draw a tangent to the circle without using the centre. Also, write steps of construction.

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3. Draw a circle of radius $3.5 cm$. Take a point $P$ on it. Draw a tangent to the circle at the point $P$, without using the centre of the circle.

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4. Draw a circle of radius $3 cm$. Take a point $P$ at a distance of $5.6 cm$ from the centre of the circle. From the point $P$, draw two tangents to the circle.

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5. Draw a circle of radius $4.5 cm$. Take point $P$ outside the circle. Without using the centre of the circle, draw two tangents to the circle from the point $P$.

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6. Construct a triangle $ABC$, similar to a given equilateral triangle $PQR$ with side $5 cm$. such that each of its side is 6/7th of the corresponding side of the $\triangle PQR$.

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7. Construct a triangle $A B C$. similar to a given isosceles triangle $P Q R$ with $Q R=5 cm, P R=P Q=cm$, such that each of its side is $5 / 3$ of the corresponding sides of the $\triangle PQR$.

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8. Draw a line segment $A B=7 cm$. Divide it externally in the ratio of (i) $3: 5$ (ii) $5: 3$

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9. Draw a $\triangle ABC$ with side $BC=6 cm, AB=5 cm$ and $\angle ABC=60^{\circ}$. Construct a $\triangle AB^{\prime} C^{\prime}$ similar to $\triangle ABC$ such that sides of $\triangle A^{\prime} C^{\prime}$ are $\frac{3}{4}$ of the corresponding sides of $\triangle A B C$.

[CBSE - 2008]