Sagot :
Given: 35 cm = the measure of the segment
3 : 4 = the ratio of the two shorter segments divided from 35 cm
Asked: How long is each segment?
Mathematical sentence:
[tex]35 cm [/tex] x [tex] \frac{3}{3 + 4} [/tex] = length of the shorter segment
[tex]35 cm[/tex] x [tex] \frac{4}{3 + 4} [/tex] = length of the longer segment
Solution:
Length of the shorter segment =
[tex]35 cm [/tex] x [tex] \frac{3}{3 + 4} [/tex]
[tex]35 cm [/tex] x [tex] \frac{3}{7} [/tex]
[tex] \frac{105 cm}{7} [/tex]
15 cm
Length of the shorter segment = 15 cm
Length of the longer segment:
[tex]35 cm[/tex] x [tex] \frac{4}{3 + 4} [/tex]
[tex]35 cm[/tex] x [tex] \frac{4}{7} [/tex]
[tex] \frac{140 cm}{7} [/tex]
20 cm
Length of the longer segment = 20 cm
Check: 15 : 20 or [tex] \frac{15}{20} [/tex] = 3 : 4 or [tex] \frac{3}{4} [/tex]
Cross multiply:
15 x 4 = 20 x 3
60 = 60, CORRECT
Therefore:
Length of the shorter segment = 15 cm
Length of the longer segment = 20 cm
3 : 4 = the ratio of the two shorter segments divided from 35 cm
Asked: How long is each segment?
Mathematical sentence:
[tex]35 cm [/tex] x [tex] \frac{3}{3 + 4} [/tex] = length of the shorter segment
[tex]35 cm[/tex] x [tex] \frac{4}{3 + 4} [/tex] = length of the longer segment
Solution:
Length of the shorter segment =
[tex]35 cm [/tex] x [tex] \frac{3}{3 + 4} [/tex]
[tex]35 cm [/tex] x [tex] \frac{3}{7} [/tex]
[tex] \frac{105 cm}{7} [/tex]
15 cm
Length of the shorter segment = 15 cm
Length of the longer segment:
[tex]35 cm[/tex] x [tex] \frac{4}{3 + 4} [/tex]
[tex]35 cm[/tex] x [tex] \frac{4}{7} [/tex]
[tex] \frac{140 cm}{7} [/tex]
20 cm
Length of the longer segment = 20 cm
Check: 15 : 20 or [tex] \frac{15}{20} [/tex] = 3 : 4 or [tex] \frac{3}{4} [/tex]
Cross multiply:
15 x 4 = 20 x 3
60 = 60, CORRECT
Therefore:
Length of the shorter segment = 15 cm
Length of the longer segment = 20 cm
[tex]|^{\underline{~~~~~~~~~~~~~~X~~~~~~~~~~~~~~~~~}}|^{\underline{~~~~~~~~~~~~Y~~~~~~~~~~~~}}| \\
X+Y=35~cm \\
\frac{Y}{X}= \frac{3}{4} \Longrightarrow 4Y=3X \Longrightarrow X= \frac{4Y}{3} \\
\frac{4Y}{3}+^{3)}Y=^{3)}35 \\
\frac{4Y}{3}+ \frac{3Y}{3}= \frac{105}{3}|*3 \\
7Y=105 \\
\boxed{Y=15~cm} \\
X=35-15 \\
\boxed{X=20~cm} [/tex]