[Maths Class Notes] on Similar Triangle Construct Pdf for Exam

Construction of Similar Triangles Class 10

Here, construction of similar triangles is given as per scale factor. Scale factor refers to the ratio of the sides of the triangle to be drawn with the corresponding sides of the given triangle.

The construction of similar triangle involves two different situations:

(i) The triangle to be drawn is smaller than the given triangle; here scale factor is less than 1.

(ii) The triangle to be drawn is larger than the given triangle, here scale factor is greater than 1.

Understanding Construction of Similar Triangles Using Real-Life Example

To understand the concept of similarity or similarity of triangles, imagine Eiffel Tower. Now imagine in your mind a mini version of it. Did you understand? The mini version is actually just a scaled-down version of the original monument. The shape remains the same just the size alters. Same is the case with the Triangles. This below figure will help you understand better. 

Let’s get to know how to check the similarity of triangles using the solve questions.

AA Similarity Theorem to Test the Similarity of Triangles

Angle Angle (AA) similarity theorem implies that two triangles must be similar to one another provided they consist of two corresponding angles in a manner that they are equal or congruent in measurement.

Applying this theorem, there will be no requirement of displaying that all three corresponding angles belonging to two triangles are in congruence for the purpose of proving that they are similar.

Pythagoras Theorem to Test and Prove Similarity of Triangles

In a right angled triangle the square of the longest side i.e. the hypotenuse is equivalent to the sum of squares of the remaining two sides.

Solved Examples

Example1: ABC is a right-angle triangle right angled at B. MN is parallel to BC. AB = 6 cm and AM: MB = 1: 2. Find out the lengths of AC and BC.

Solution:

Given, 

AM / MB = ½

MB / AM = 2

AM + MB / AM = 2+1

AB/AM = 3

In △ ABC and △ AMN

∠MAN=∠BAC (Common angle)

∠AMN=∠ABC =90⁰ (MN II BC)

∠MNA=∠BCA (3rd angle)

Thus, △ABC∼△AMN (Application of AAA rule)

Therefore, 

AM/AB = MN/BC​= AN/AC (Corresponding sides)

AN/AC = 3

Thus, AC = 3 × AC = 3×4 = 12cm

Now, in △ABC,

AB² + BC² = AC² (Applying the Pythagoras theorem)

6² + BC² =12²

BC² = 144−36

BC =√108

BC =10.392cm

 

Example 2: Draw a triangle similar to a given triangle PQR in a way that each of its sides is (5/7)th of the corresponding sides of ∆PQR. It is given that

PQ = 4 cm, PR = 5 cm and QR = 6 cm.

Solution:

Follow these Steps of Construction in an orderly way:

Step I: Construct a line segment QR = 6 cm.

Step II: With Q as centre point and radius = PQ = 4 cm, make an arc.

Step III : With R as centre point and radius = PR = 5 cm, draw another arc, bisecting the arc drawn in step II at P.

Step IV: Join PQ and PR to get the triangle PQR.

Step V: Below base QR, draw an acute angle ∠RQX.

Step VI: Along QX, mark 7 points Q1, Q2, Q3, Q4, Q5, Q6, Q7 such that

QQ1 = Q1Q2 = …… = Q6Q7.

Step VII: Connect Q7R.

Step VIII: Because we have to construct a triangle each of whose sides is (5/7)th of the corresponding sides of ∆PQR. So take 5 parts out of 7 equal parts on QX i.e. from Q5, Draw Q5R´ || Q7R, bisecting QR at R´.

Step IX: From R´, draw R´P´ || RP, meeting QP at P´.

∆P´QR´ is the needed triangle each of whose sides is (5/7)th of the corresponding sides of ∆PQR.