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 2 Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle. Letβs first construct Ξ ABC with sides 4 cm, 5 cm, 6 cm Steps to draw Ξ ABC Draw base AB of side 4 cm With A as center, and 5 cm as radius, draw an arc With B as center, and 6 cm as radius, draw an arc 3. Let C be the point where the two arcs intersect. Join AC & BC Thus, Ξ ABC is the required triangle Now, letβs make a similar triangle with Scale factor = 2/3 Steps of construction Draw any ray AX making an acute angle with AB on the side opposite to the vertex C. Mark 3 (the greater of 2 and 3 in 2/3 ) points π΄_1, π΄_2, π΄_3 on AX so that γπ΄π΄γ_1=π΄_1 π΄_2=π΄_2 π΄_3 Join π΄_3B and draw a line through π΄_2 (the 2nd point, 2 being smaller of 2 and 3 in 2/3) parallel to π΄_3 π΅, to intersect AB at Bβ². 4. Draw a line through Bβ² parallel to the line BC to intersect AC at Cβ². Thus, Ξ ABβCβ² is the required triangle Justification Since scale factor is 2/3, we need to prove (π¨π©^β²)/π¨π©=(π¨πͺ^β²)/π¨πͺ=(π©^β² πͺ^β²)/π©πͺ = π/π By construction, (π΄B^β²)/π΄π΅=(π΄π΄_2)/(π΄π΄_3 )= 2/3 Also, BβCβ is parallel to BC So, the will make the same angle with line AB β΄ β ABβCβ = β ABC (Corresponding angles) Now, In Ξ ABβCβ and Ξ ABC β A = β A (Common) β ABβCβ = β ABC (From (2)) Ξ ABβCβ βΌ Ξ ABC (AA Similarity) Since corresponding sides of similar triangles are in the same ratio (π΄π΅^β²)/π΄π΅=(π΄πΆ^β²)/π΄πΆ=(π΅^β² πΆ^β²)/π΅πΆ So,(π¨π©^β²)/π¨π©=(π¨πͺ^β²)/π¨πͺ=(π©^β² πͺ^β²)/π©πͺ =π/π. (From (1)) Thus, our construction is justified