Question Number 222546 by Tawa11 last updated on 29/Jun/25
Commented by Tawa11 last updated on 29/Jun/25
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{AB}. \\ $$
Commented by fantastic last updated on 29/Jun/25
$${AC}=\mathrm{17},{AD}=\mathrm{25},{CD}=\mathrm{28}\:{AB}=?? \\ $$
Answered by fantastic last updated on 29/Jun/25
Commented by fantastic last updated on 29/Jun/25
![ACBD is a kite in △ACD and△CBD AC=CB DB=DA CD common So △ACD≅△CBD(S−S−S) So ∡ACD=∡BCD in △ACO and△COB AC=CB ∡ACO=∡BCO CO common ∴△ACO≅△CBO(S−A−S) so ∡AOC=∡COB So AO=OB and∡AOC=90^0 [∡AOC+∡BOC=180^0 or 2∡AOC=180^0 or ∡AOC=90^0 ] Now △ACD=(√(s(s−a)(s−b)(s−c)))[s=((a+b+c)/2)] =(√(35(35−17)(35−25)(35−28)))=210 let AO=x So (1/2)×28×x=210 or x=15 So AO=15 AB=2×15=30 determinant (((Final Answer AB=30)))](https://www.tinkutara.com/question/Q222553.png)
$${ACBD}\:{is}\:{a}\:{kite} \\ $$$${in}\:\bigtriangleup{ACD}\:{and}\bigtriangleup{CBD} \\ $$$${AC}={CB} \\ $$$${DB}={DA} \\ $$$${CD}\:{common} \\ $$$${So}\:\bigtriangleup{ACD}\cong\bigtriangleup{CBD}\left({S}−{S}−{S}\right) \\ $$$${So}\:\measuredangle{ACD}=\measuredangle{BCD} \\ $$$${in}\:\bigtriangleup{ACO}\:{and}\bigtriangleup{COB} \\ $$$${AC}={CB} \\ $$$$\measuredangle{ACO}=\measuredangle{BCO} \\ $$$${CO}\:{common} \\ $$$$\therefore\bigtriangleup{ACO}\cong\bigtriangleup{CBO}\left({S}−{A}−{S}\right) \\ $$$${so}\:\measuredangle{AOC}=\measuredangle{COB} \\ $$$${So}\:{AO}={OB}\:{and}\measuredangle{AOC}=\mathrm{90}^{\mathrm{0}} \left[\measuredangle{AOC}+\measuredangle{BOC}=\mathrm{180}^{\mathrm{0}} \:\:\:{or}\:\mathrm{2}\measuredangle{AOC}=\mathrm{180}^{\mathrm{0}} \:{or}\:\measuredangle{AOC}=\mathrm{90}^{\mathrm{0}} \right] \\ $$$${Now} \\ $$$$\bigtriangleup{ACD}=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}\left[{s}=\frac{{a}+{b}+{c}}{\mathrm{2}}\right] \\ $$$$=\sqrt{\mathrm{35}\left(\mathrm{35}−\mathrm{17}\right)\left(\mathrm{35}−\mathrm{25}\right)\left(\mathrm{35}−\mathrm{28}\right)}=\mathrm{210} \\ $$$${let}\:{AO}={x} \\ $$$${So}\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{28}×{x}=\mathrm{210} \\ $$$${or}\:{x}=\mathrm{15} \\ $$$${So}\:{AO}=\mathrm{15} \\ $$$${AB}=\mathrm{2}×\mathrm{15}=\mathrm{30} \\ $$$$\begin{array}{|c|}{{Final}\:{Answer}\:{AB}=\mathrm{30}}\\\hline\end{array} \\ $$
Commented by Tawa11 last updated on 29/Jun/25
$$\mathrm{Sir},\:\mathrm{I}\:\mathrm{think}\:\:\mathrm{s}\:\:=\:\:\frac{\mathrm{a}\:\:+\:\:\mathrm{b}\:\:+\:\:\mathrm{c}}{\mathrm{2}} \\ $$$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by fantastic last updated on 29/Jun/25
$${Oh}\:!\:{sorry} \\ $$
Answered by fantastic last updated on 29/Jun/25
$${AB}=\mathrm{30} \\ $$
Answered by mr W last updated on 29/Jun/25
$${a}=\mathrm{17},\:{b}=\mathrm{25},\:{c}=\mathrm{28} \\ $$$${AB}={d}=\mathrm{2}{h} \\ $$$$\Delta{ACD}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$$$\Delta{ACD}=\frac{{ch}}{\mathrm{2}}=\frac{{cd}}{\mathrm{4}} \\ $$$$\frac{{cd}}{\mathrm{4}}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$$$\Rightarrow{d}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{{c}} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{840}}{\mathrm{28}}=\mathrm{30}\:\checkmark \\ $$
Commented by Tawa11 last updated on 29/Jun/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$