Constructing similar triangle as per scale factor - Scale factor < 1
Constructing similar triangle as per scale factor - Scale factor < 1
Last updated at April 16, 2024 by Teachoo
Question 7 Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle First we draw a rough sketch To construct it, We follow these steps Steps to draw Ξ ABC Draw base BC of side 3 cm Draw β B = 90Β° 3. Taking B as center, 4 cm as radius, we draw an arc Let the point where arc intersects the ray be point A 4. Join AC β΄ Ξ ABC is the required triangle Now, we need to make a triangle which is 5/3 times its size β΄ Scale factor = 5/3 > 1 Steps of construction Draw any ray BX making an acute angle with BC on the side opposite to the vertex A. Mark 5 (the greater of 5 and 3 in 5/3 ) points π΅_1, π΅_2, π΅_3,π΅_4,π΅_5 on BX so that γπ΅π΅γ_1=π΅_1 π΅_2=π΅_2 π΅_3=π΅_3 π΅_4=π΅_4 π΅_5 Join π΅_3 πΆ (3rd point as 3 is smaller in 5/3) and draw a line through π΅_5 parallel to π΅_3 πΆ, to intersect BC extended at Cβ². Draw a line through Cβ² parallel to the line AC to intersect AB extended at Aβ². Thus, Ξ Aβ²BCβ² is the required triangle Justification Since scale factor is 5/3, we need to prove (π¨^β² π©)/π¨π©=(π¨^β² πͺ^β²)/π¨πͺ=(π©πͺ^β²)/π©πͺ =π/π By construction, BC^β²/π΅πΆ=(π΅π΅_5)/(π΅π΅_3 )=5/3 Also, AβCβ is parallel to AC So, they will make the same angle with line BC β΄ β AβCβB = β ACB Now, In Ξ AβBCβ and Ξ ABC β B = β B β AβCβB = β ACB Ξ AβBCβ βΌ Ξ ABC Since corresponding sides of similar triangles are in the same ratio (π΄^β² π΅)/π΄π΅=(π΄^β² πΆ^β²)/π΄πΆ=(π΅πΆ^β²)/π΅πΆ So, (π¨^β² π©)/π¨π©=(π¨^β² πͺ^β²)/π¨πͺ=(π©πͺ^β²)/π©πͺ =π/π Thus, our construction is justified