PUMPA - SMART LEARNING

எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

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A scale factor is the ratio of similar figures' corresponding sides.
In the above theoretical material, we deal with similar triangles theoretically. Let us discuss how to construct a similar triangle using the concept of the scale factor. There are $$2$$ cases.

Let us understand the cases using examples.
Example:
Case 1: If the scale factor is less than $$1$$.

Construct a triangle similar to the given triangle $$PQR$$ with its sides equal to $$\frac{2}{5}$$ of the corresponding sides of the triangle $$PQR$$.

Solution:

Given a triangle $$PQR$$. We are required to construct another triangle whose sides are $$\frac{2}{5}$$ of the corresponding sides of the triangle $$PQR$$.

Construction:

Step 1: Construct a triangle $$PQR$$ with any measurement.

Step 2: Draw a ray $$QX$$ making an acute angle with $$QR$$ on the side opposite to vertex $$P$$.

Step 3: Locate $$5$$ (the greater of $$2$$ and $$5$$ in $$\frac{2}{5}$$) points. $$Q_1$$, $$Q_2$$, $$Q_3$$, $$Q_4$$ and $$Q_5$$ on $$QX$$ so that $$QQ_1 = Q_1Q_2 = Q_2Q_3 = Q_3Q_4 = Q_4Q_5$$.

Step 4: Join $$Q_5R$$ and draw a line through $$Q_2$$ (the second point, $$2$$ being smaller of $$2$$ and $$5$$ in $$\frac{2}{5}$$) parallel to $$Q_5R$$ to intersect $$QR$$ at $$R'$$.

Step 5: Draw a line through $$R'$$ parallel to the line $$RP$$ to intersect $$QP$$ at $$P'$$. Then, $$P'QR'$$ is the required triangle, each of whose side is two - fifths of the corresponding sides of $$\triangle PQR$$.

Let us consider the construction of triangle $$ABC$$ using the scale factor $$\frac{3}{4}$$.

Case 2: If the scale factor is greater than $$1$$.

Construct a triangle similar to a given triangle $$PQR$$ with its sides equal to $$\frac{3}{2}$$ of the corresponding sides of the triangle $$PQR$$.

Solution:

Given a triangle $$PQR$$. We are required to construct an another triangle whose sides are $$\frac{3}{2}$$ of the corresponding sides of the triangle $$PQR$$.

Construction:

Step 1: Construct a triangle $$PQR$$ with any measurement.

Step 2: Draw a ray $$QX$$ making an acute angle with $$QR$$ on the side opposite to vertex $$P$$.

Step 3: Locate $$3$$ (the greater of $$2$$ and $$3$$ in $$\frac{3}{2}$$) points. $$Q_1$$, $$Q_2$$, and $$Q_3$$ on $$QX$$ so that $$QQ_1 = Q_1Q_2 = Q_2Q_3$$.

Step 4: Join $$Q_2$$ (the 2nd point, 2 being smaller of $$2$$ and $$3$$ in $$\frac{3}{2}$$) to $$R$$ and draw a line through $$Q_3$$ parallel to $$Q_2R$$, intersecting the extended line segment $$QR$$ at $$R'$$.

Step 5: Draw a line through $$R'$$ parallel to the line $$RP$$ intersecting the extended line segment $$QP$$ at $$P'$$. Then, $$P'QR'$$ is the required triangle, each of whose side is three - twos of the corresponding sides of $$\triangle PQR$$.

Let us consider the construction of triangle $$ABC$$ using the scale factor $$\frac{5}{3}$$.