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Let-X-be-a-point-inside-a-square-ABCD-such-that-XA-10-cm-XB-6-cm-and-XC-14-cm-Find-the-area-of-the-square-

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Question Number 221255 by MATHEMATICSAM last updated on 28/May/25
Let X be a point inside a square ABCD, such that XA = 10 cm, XB = 6 cm and XC = 14 cm. Find the area of the square.
$$\mathrm{Let}\:\mathrm{X}\:\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{inside}\:\mathrm{a}\:\mathrm{square}\:\mathrm{ABCD},\: \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{XA}\:=\:\mathrm{10}\:\mathrm{cm},\:\mathrm{XB}\:=\:\mathrm{6}\:\mathrm{cm}\:\mathrm{and} \\ $$$$\mathrm{XC}\:=\:\mathrm{14}\:\mathrm{cm}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}. \\ $$
Commented by mr W last updated on 28/May/25
see also Q72294
$${see}\:{also}\:{Q}\mathrm{72294} \\ $$
Commented by MATHEMATICSAM last updated on 28/May/25
I solved it but it will be good if you give me only the ans of the qs. and also two answers will come and we need to reject one maybe.
$$\mathrm{I}\:\mathrm{solved}\:\mathrm{it}\:\mathrm{but}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be}\:\mathrm{good}\:\mathrm{if}\:\mathrm{you}\:\mathrm{give} \\ $$$$\mathrm{me}\:\mathrm{only}\:\mathrm{the}\:\mathrm{ans}\:\mathrm{of}\:\mathrm{the}\:\mathrm{qs}.\:\mathrm{and}\:\mathrm{also}\:\mathrm{two} \\ $$$$\mathrm{answers}\:\mathrm{will}\:\mathrm{come}\:\mathrm{and}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{reject}\: \\ $$$$\mathrm{one}\:\mathrm{maybe}. \\ $$
Commented by mr W last updated on 28/May/25
the area of the square is acc. to the formula given in that post s^2 =((10^2 +14^2 )/2)+(√(6^2 ×(10^2 +14^2 −6^2 )−(((10^2 −14^2 )/2))^2 )) =232 since the point is inside the square, there is only this one solution.
$${the}\:{area}\:{of}\:{the}\:{square}\:{is}\:{acc}.\:{to}\:{the} \\ $$$${formula}\:{given}\:{in}\:{that}\:{post} \\ $$$${s}^{\mathrm{2}} =\frac{\mathrm{10}^{\mathrm{2}} +\mathrm{14}^{\mathrm{2}} }{\mathrm{2}}+\sqrt{\mathrm{6}^{\mathrm{2}} ×\left(\mathrm{10}^{\mathrm{2}} +\mathrm{14}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} \right)−\left(\frac{\mathrm{10}^{\mathrm{2}} −\mathrm{14}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:=\mathrm{232} \\ $$$${since}\:{the}\:{point}\:{is}\:{inside}\:{the}\:{square}, \\ $$$${there}\:{is}\:{only}\:{this}\:{one}\:{solution}. \\ $$
Answered by mr W last updated on 29/May/25
Commented by mr W last updated on 29/May/25
cos α=((s^2 +6^2 −10^2 )/(2s×6))=((s^2 −64)/(12s)) cos β=((s^2 +6^2 −14^2 )/(2s×6))=((s^2 −160)/(12s))=sin α ⇒(s^2 −64)^2 +(s^2 −160)^2 =(12s)^2 s^4 −296s^2 +14848=0 ⇒s^2 =148+84=232 ✓
$$\mathrm{cos}\:\alpha=\frac{{s}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} }{\mathrm{2}{s}×\mathrm{6}}=\frac{{s}^{\mathrm{2}} −\mathrm{64}}{\mathrm{12}{s}} \\ $$$$\mathrm{cos}\:\beta=\frac{{s}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\mathrm{14}^{\mathrm{2}} }{\mathrm{2}{s}×\mathrm{6}}=\frac{{s}^{\mathrm{2}} −\mathrm{160}}{\mathrm{12}{s}}=\mathrm{sin}\:\alpha \\ $$$$\Rightarrow\left({s}^{\mathrm{2}} −\mathrm{64}\right)^{\mathrm{2}} +\left({s}^{\mathrm{2}} −\mathrm{160}\right)^{\mathrm{2}} =\left(\mathrm{12}{s}\right)^{\mathrm{2}} \\ $$$${s}^{\mathrm{4}} −\mathrm{296}{s}^{\mathrm{2}} +\mathrm{14848}=\mathrm{0} \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\mathrm{148}+\mathrm{84}=\mathrm{232}\:\checkmark \\ $$
Commented by MATHEMATICSAM last updated on 29/May/25
Tell me how to reject the second solution by using this method.
$$\mathrm{Tell}\:\mathrm{me}\:\mathrm{how}\:\mathrm{to}\:\mathrm{reject}\:\mathrm{the}\:\mathrm{second}\:\mathrm{solution} \\ $$$$\mathrm{by}\:\mathrm{using}\:\mathrm{this}\:\mathrm{method}. \\ $$
Commented by mr W last updated on 29/May/25
for point X inside the square, 0°< α, β <90° that means cos α >0 and cos β >0. s^2 −160 >0 ⇒ s^2 >160. since s^2 =148−84=64 <160, it must be rejected.
$${for}\:{point}\:{X}\:{inside}\:{the}\:{square}, \\ $$$$\mathrm{0}°<\:\alpha,\:\beta\:<\mathrm{90}° \\ $$$${that}\:{means}\:\mathrm{cos}\:\alpha\:>\mathrm{0}\:{and}\:\mathrm{cos}\:\beta\:>\mathrm{0}. \\ $$$${s}^{\mathrm{2}} −\mathrm{160}\:>\mathrm{0}\:\Rightarrow\:{s}^{\mathrm{2}} \:>\mathrm{160}. \\ $$$${since}\:{s}^{\mathrm{2}} =\mathrm{148}−\mathrm{84}=\mathrm{64}\:<\mathrm{160},\:{it}\:{must} \\ $$$${be}\:{rejected}. \\ $$
Commented by mr W last updated on 29/May/25
the second root represents the case where the point X is outside the square:
$${the}\:{second}\:{root}\:{represents}\:{the}\:{case} \\ $$$${where}\:{the}\:{point}\:{X}\:{is}\:{outside}\:{the} \\ $$$${square}: \\ $$
Commented by mr W last updated on 29/May/25

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