to construct a triangle to a given triangle ABC with sides 8/5 of the corresponding sides of triangle ABC, draw a ray BX such that angle CBX is an acute angle and X is on the opposite side of A with respect to BC. the minimum number of points to be located at equal distances on ray BX is -----------------

a) 13 b) 8

explain the answer plzzzz.

what would the answer be if ratio was 5/8 instead of 8/5 ?

explain?

fast i'll give u a thumbs up!!!

1)After drawing triangle ABC ,draw an acute angle CBX below BC .

2)Along BX ,mark off 8 points at equal distances .

3) Join ${B}_{5}$ to C and draw a line ${B}_{8}$ parallel to ${B}_{5}C$ intersecting BC extended at C'.

4) Draw a line through C' parallel to CA intersecting BA extended at A'.

Thus A'BC' is the required triangle .

Explanation:-

$\u2206ABC~\u2206A\text{'}BC\text{'}\phantom{\rule{0ex}{0ex}}\therefore \frac{AB}{A\text{'}B}=\frac{AC}{A\text{'}C\text{'}}=\frac{BC}{BC\text{'}}\phantom{\rule{0ex}{0ex}}But\frac{BC}{BC\text{'}}=\frac{B{B}_{5}}{B{B}_{8}}=\frac{5}{8}\phantom{\rule{0ex}{0ex}}Hence\frac{BC}{BC\text{'}}=\frac{8}{5}andthus\frac{A\text{'}B}{AB}=\frac{A\text{'}C\text{'}}{AC}=\frac{BC\text{'}}{BC}=\frac{8}{5}$

Now if the ratio is $\frac{5}{8}$ instead of $\frac{8}{5}$ then also you have to mark off 8 points but the difference is that in this case you have to join ${B}_{8}$ to C

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