Answer:
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Step-by-step explanation:
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[tex] \sf \large { \overline{RS}} \: and \: { \overline{ST}} \: and \: { \overline{RT}} \: are \: midsegments \: of \: ∆DEF.[/tex]
[tex] \sf \large{ \overline{TF}} = 4.5 \: cm \\ \\ \sf \large{ \overline{DR}} = 4.5 \: cm[/tex]
what is the length of [tex] \sf \large{ \overline{DF}} \sf?[/tex]
what is the length of [tex] \sf \large{ \overline{DF}}[/tex]?
[tex] \sf \large{ \overline{RD}} = { \overline{RE}}⇒{ \overline{DE}} = 4.5 \times 2 = 9 \: \: and \\ \\ \sf \large{ \overline{ET}} = { \overline{TF}}⇒{ \overline{EF}} = 4.5 \times 2 = 9 \: \: \: \: \: \: \: \: \: \: [/tex]
[tex] \sf \large { \overline{DF}} = \sqrt{{( { \overline{DF}}) {}^{2} + ({ \overline{DF}}) {}^{2} }} [/tex]
[tex] \sf \large = \sqrt{{(9 ){}^{2} +( 9 ){}^{2} }} = 9 \sqrt{2} \: cm \: (or) \: 12.728 \: cm[/tex]
[tex] \sf \large \color{red} \therefore The \:length \: of \: { \overline{DF}} = 9 \sqrt{2} \: cm.[/tex]
Hope you have satisfied. ⚘
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Answers & Comments
Answer:
4.255 is right now
Step-by-step explanation:
I hope help you
Verified answer
Question:-
[tex] \sf \large { \overline{RS}} \: and \: { \overline{ST}} \: and \: { \overline{RT}} \: are \: midsegments \: of \: ∆DEF.[/tex]
[tex] \sf \large{ \overline{TF}} = 4.5 \: cm \\ \\ \sf \large{ \overline{DR}} = 4.5 \: cm[/tex]
what is the length of [tex] \sf \large{ \overline{DF}} \sf?[/tex]
Given:-
[tex] \sf \large{ \overline{TF}} = 4.5 \: cm \\ \\ \sf \large{ \overline{DR}} = 4.5 \: cm[/tex]
To Find:-
what is the length of [tex] \sf \large{ \overline{DF}}[/tex]?
Solution:-
Here,
[tex] \sf \large{ \overline{RD}} = { \overline{RE}}⇒{ \overline{DE}} = 4.5 \times 2 = 9 \: \: and \\ \\ \sf \large{ \overline{ET}} = { \overline{TF}}⇒{ \overline{EF}} = 4.5 \times 2 = 9 \: \: \: \: \: \: \: \: \: \: [/tex]
[tex] \sf \large { \overline{DF}} = \sqrt{{( { \overline{DF}}) {}^{2} + ({ \overline{DF}}) {}^{2} }} [/tex]
[tex] \sf \large = \sqrt{{(9 ){}^{2} +( 9 ){}^{2} }} = 9 \sqrt{2} \: cm \: (or) \: 12.728 \: cm[/tex]
Answer:-
[tex] \sf \large \color{red} \therefore The \:length \: of \: { \overline{DF}} = 9 \sqrt{2} \: cm.[/tex]
Hope you have satisfied. ⚘