Question

given a triangle with side AB = 8 centimetre to get a line segment AB' is equals to 3/ 4 of AB find the ratio in which the line segment AB is divided. PLZ EXPLAIN COMPLETELY.

Answer 1

Let me rewrite your question first:

You wish to mark a point B' on the side AB of triangle ABC, such that AB' = 3/4 AB. AB = 8 unit. So, where should the point B' be located?

Let me explain now, AB' is 3/4 AB , means , if you divide AB into 4 equal parts. Then AB' will have 3 parts out of them.

So, first of all divide AB into 4 equal parts. & wherever 3rd part ends, mark there B' .

So ratio of the length of AB' to the length of AB = 3/4

Measure wise if you calculate the length of each part = 8/4 = 2 unit. Then 3 parts length = 6 unit. So ratio = 6/8 = 3/4

Answer 2

Answer:

Line segment AB is divided in ration 3:1 .

Step-by-step explanation:

Explanation:

Given, a triangle with side AB = 8cm to get a line segment AB' is equal to $\frac{3}{4}$ of AB.

Ab = 8cm and AB' = $\frac{3}{4}$  of AB.

Step 1:

Therefore, AB' = $\frac{3}{4}$ × 8 = 6 cm .

and BB' = AB - AB'

⇒BB' = 8cm - 6cm = 2cm .

Now the ratio of AB': BB' = 3 :1.

Final answer:

Hence, the ratio of line segment AB is divided in 3:1.

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