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There are two ways to divide a given line segment in a given ratio.

**First method: **

Let AB be the given line segment.

Step 1: Draw a segment AC of any convenient length making an acute angle with the given line segment AB. Mark a few points at an equal distance from each other on AC. The number of points depends on the ratio in which you have to divide the given line segment. If the ratio is x is to y, the number of points is x + y.

Step 2: Now, using compass to any small convenient length mark four points say X_{1}, X_{2}, X_{3} and X_{4} on AC, so that AX_{1} = X_{1}X_{2} = X_{2}X_{3} = X_{3}X_{4}. Make sure that the distance to which the compass is open is not disturbed as you mark the points.

Step 3: Draw a line to join the points X_{4} and B.

Step 4: Using a pair of set-squares draw a line parallel to X_{4}B from the point X_{1} to intersect the given line segment AB at point P.

Now the point P divides the line segment AB in the ratio x is to y.

**Second Method:**

Let AB be the given line segment

Step1:Draw a segment AC of a convenient length, making an acute angle with the given line segment AB.

Step2: Draw a segment BD of any convenient length making the same angle with AB as AC on the opposite side of AC.

Step 3: Now, using compass to any small convenient length mark x number points on AC and y number points on BD or vice versa such that AX_{1} = X_{1} X_{2} = X_{2}X_{3} =--- = BY_{1} = Y_{1}Y_{2} = Y_{2}Y_{3} = ---

Step4: Join X_{x} to Y_{y} to intersect AB at point P. Now Point P divides the line segment AB in the ratio x is to y.

You will find a scale factor on most maps. A scale factor determines how large or small an object is in comparison with another object.

**Constructing a triangle similar to a given triangle:**

Let ABC be a given triangle.

Step1: Draw a segment of a convenient length AX making an acute angle, with a side AB of the given triangle.

Step2: Now, using compass you have to mark a few points say X_{1}, X_{2}, X_{3} ... at an equal distance from each other on AX. The number of points depends on the given scale factor. If the scale factor is x by y,

the number of points is equal to the larger value amongst x and y.

Step 3: Draw a line to join the points X_{3} and B.

Step 4: Draw a line parallel to X_{3}B from X_{2} to intersect AB at P.

Step 5: Draw a line parallel to side BC at P to intersect side AC at Q.

Now, the triangle APQ is similar to the given triangle ABC.

**Construction of Tangent to a Circle at a given point:**

Step 1: Take a point O and draw a circle of given radius.

Step 2: Take a point P on the circle.

Step 3: Join O and P i.e OP.

Step 4: Construct ∠OPQ = 90°

Step 5: Produce QP to R and to get QPR is the required tangent.

These are the required tangent.

**Construction of Tangents from an External Point to a Circle:**

Let P be a point outside the circle having centre O.

Step 1: Join OP.

Step 2: Draw perpendicular bisector of OP.

Step 3: Draw a circle with a centre M and radius MO to intersect the given circle at points A and B.

Step 4: Join AP and BP.

These are the two required tangents.

Let AB be the given line segment.

Step 1: Draw a segment AC of any convenient length making an acute angle with the given line segment AB. Mark a few points at an equal distance from each other on AC. The number of points depends on the ratio in which you have to divide the given line segment. If the ratio is x is to y, the number of points is x + y.

Step 2: Now, using compass to any small convenient length mark four points say X

Step 3: Draw a line to join the points X

Step 4: Using a pair of set-squares draw a line parallel to X

Now the point P divides the line segment AB in the ratio x is to y.

Let AB be the given line segment

Step1:Draw a segment AC of a convenient length, making an acute angle with the given line segment AB.

Step2: Draw a segment BD of any convenient length making the same angle with AB as AC on the opposite side of AC.

Step 3: Now, using compass to any small convenient length mark x number points on AC and y number points on BD or vice versa such that AX

Step4: Join X

You will find a scale factor on most maps. A scale factor determines how large or small an object is in comparison with another object.

Let ABC be a given triangle.

Step1: Draw a segment of a convenient length AX making an acute angle, with a side AB of the given triangle.

Step2: Now, using compass you have to mark a few points say X

the number of points is equal to the larger value amongst x and y.

Step 3: Draw a line to join the points X

Step 4: Draw a line parallel to X

Step 5: Draw a line parallel to side BC at P to intersect side AC at Q.

Now, the triangle APQ is similar to the given triangle ABC.

Step 1: Take a point O and draw a circle of given radius.

Step 2: Take a point P on the circle.

Step 3: Join O and P i.e OP.

Step 4: Construct ∠OPQ = 90°

Step 5: Produce QP to R and to get QPR is the required tangent.

These are the required tangent.

Let P be a point outside the circle having centre O.

Step 1: Join OP.

Step 2: Draw perpendicular bisector of OP.

Step 3: Draw a circle with a centre M and radius MO to intersect the given circle at points A and B.

Step 4: Join AP and BP.

These are the two required tangents.

There are two ways to divide a given line segment in a given ratio.

**First method: **

Let AB be the given line segment.

Step 1: Draw a segment AC of any convenient length making an acute angle with the given line segment AB. Mark a few points at an equal distance from each other on AC. The number of points depends on the ratio in which you have to divide the given line segment. If the ratio is x is to y, the number of points is x + y.

Step 2: Now, using compass to any small convenient length mark four points say X_{1}, X_{2}, X_{3} and X_{4} on AC, so that AX_{1} = X_{1}X_{2} = X_{2}X_{3} = X_{3}X_{4}. Make sure that the distance to which the compass is open is not disturbed as you mark the points.

Step 3: Draw a line to join the points X_{4} and B.

Step 4: Using a pair of set-squares draw a line parallel to X_{4}B from the point X_{1} to intersect the given line segment AB at point P.

Now the point P divides the line segment AB in the ratio x is to y.

**Second Method:**

Let AB be the given line segment

Step1:Draw a segment AC of a convenient length, making an acute angle with the given line segment AB.

Step2: Draw a segment BD of any convenient length making the same angle with AB as AC on the opposite side of AC.

Step 3: Now, using compass to any small convenient length mark x number points on AC and y number points on BD or vice versa such that AX_{1} = X_{1} X_{2} = X_{2}X_{3} =--- = BY_{1} = Y_{1}Y_{2} = Y_{2}Y_{3} = ---

Step4: Join X_{x} to Y_{y} to intersect AB at point P. Now Point P divides the line segment AB in the ratio x is to y.

You will find a scale factor on most maps. A scale factor determines how large or small an object is in comparison with another object.

**Constructing a triangle similar to a given triangle:**

Let ABC be a given triangle.

Step1: Draw a segment of a convenient length AX making an acute angle, with a side AB of the given triangle.

Step2: Now, using compass you have to mark a few points say X_{1}, X_{2}, X_{3} ... at an equal distance from each other on AX. The number of points depends on the given scale factor. If the scale factor is x by y,

the number of points is equal to the larger value amongst x and y.

Step 3: Draw a line to join the points X_{3} and B.

Step 4: Draw a line parallel to X_{3}B from X_{2} to intersect AB at P.

Step 5: Draw a line parallel to side BC at P to intersect side AC at Q.

Now, the triangle APQ is similar to the given triangle ABC.

**Construction of Tangent to a Circle at a given point:**

Step 1: Take a point O and draw a circle of given radius.

Step 2: Take a point P on the circle.

Step 3: Join O and P i.e OP.

Step 4: Construct ∠OPQ = 90°

Step 5: Produce QP to R and to get QPR is the required tangent.

These are the required tangent.

**Construction of Tangents from an External Point to a Circle:**

Let P be a point outside the circle having centre O.

Step 1: Join OP.

Step 2: Draw perpendicular bisector of OP.

Step 3: Draw a circle with a centre M and radius MO to intersect the given circle at points A and B.

Step 4: Join AP and BP.

These are the two required tangents.

Let AB be the given line segment.

Step 1: Draw a segment AC of any convenient length making an acute angle with the given line segment AB. Mark a few points at an equal distance from each other on AC. The number of points depends on the ratio in which you have to divide the given line segment. If the ratio is x is to y, the number of points is x + y.

Step 2: Now, using compass to any small convenient length mark four points say X

Step 3: Draw a line to join the points X

Step 4: Using a pair of set-squares draw a line parallel to X

Now the point P divides the line segment AB in the ratio x is to y.

Let AB be the given line segment

Step1:Draw a segment AC of a convenient length, making an acute angle with the given line segment AB.

Step2: Draw a segment BD of any convenient length making the same angle with AB as AC on the opposite side of AC.

Step 3: Now, using compass to any small convenient length mark x number points on AC and y number points on BD or vice versa such that AX

Step4: Join X

You will find a scale factor on most maps. A scale factor determines how large or small an object is in comparison with another object.

Let ABC be a given triangle.

Step1: Draw a segment of a convenient length AX making an acute angle, with a side AB of the given triangle.

Step2: Now, using compass you have to mark a few points say X

the number of points is equal to the larger value amongst x and y.

Step 3: Draw a line to join the points X

Step 4: Draw a line parallel to X

Step 5: Draw a line parallel to side BC at P to intersect side AC at Q.

Now, the triangle APQ is similar to the given triangle ABC.

Step 1: Take a point O and draw a circle of given radius.

Step 2: Take a point P on the circle.

Step 3: Join O and P i.e OP.

Step 4: Construct ∠OPQ = 90°

Step 5: Produce QP to R and to get QPR is the required tangent.

These are the required tangent.

Let P be a point outside the circle having centre O.

Step 1: Join OP.

Step 2: Draw perpendicular bisector of OP.

Step 3: Draw a circle with a centre M and radius MO to intersect the given circle at points A and B.

Step 4: Join AP and BP.

These are the two required tangents.