Question

Construct a triangle with side 6 cm, 8 cm, and 10cm. Construct a triangle similar to the triangle, whose sides are 3/5 of its corresponding sides

Answer 1

refer 2 the attachment ....
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Answer 2

Given: Sides of triangle Δ ABC are AC = 6 cm , CB = 8 cm and AB = 10 cm.

To construct: A similar triangle AB'C' with sides of ratio 3 : 5

First we construct ΔABC then ΔAB'C'

Step-by-step explanation:

Step 1: Draw a line segment AB of length 10 cm

Step 2: With A as center, draw an arc of radius 6 cm above Point A

Step 3: With B as center, draw an arc of radius 8 cm such that it intersect arc of step 2. Point of intersect of both arc is point C.

Step 4: Join AC & CB.

Step 5: Draw any arc AX making an acute angle with side AB on the side opposite to the vertex C.

Step 6: Locate 5 ( greater of 3 & 5 in 3 : 5 ) points $A_1\,,\,A_2\,,\,A_3\,,\,A_4\,,\,A_5$ on AX so that $AA_1=A_1A_2=A_2A_3=A_3A_4=A_4A_5$.

Step 7: Join $A_5B$ and draw a line through $A_3$  ( the 3rd point, 3 being smaller of 3 & 5 ) parallel to $A_5B$ to intersect AB at B'.

Step 8: Draw a line through B' parallel to line CB to intersect AC at C'.

Thus ΔAB'C' is the required triangle