please help!! ABCD is a square. If AC = 26, Find BC
a. 7
b. 26
c. 12
d. 18.4
![please help ABCD is a square If AC 26 Find BC a 7 b 26 c 12 d 184 class=](https://us-static.z-dn.net/files/d3e/15a3d7da6aa64ebbc09687cf6bc306d7.png)
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Let AC = x.
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As given figureis square, therefore all sides are equal :-
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So now instead of square focus on triangle ABC.
where :-
[tex] \small \rm \angle B = 90 \degree[/tex]
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Equation formed:-
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[tex] \rm \dashrightarrow AC {}^{2} = AB {}^{2} + BC {}^{2} [/tex]
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[tex] \rm \dashrightarrow 26 {}^{2} = x{}^{2} + x {}^{2} [/tex]
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[tex] \rm \dashrightarrow 26 {}^{2} =2x {}^{2} [/tex]
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[tex] \rm \dashrightarrow 26 \times 26 =2x {}^{2} [/tex]
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[tex] \rm \dashrightarrow2x {}^{2} = 26 \times 26 [/tex]
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[tex] \rm \dashrightarrow x {}^{2}=\dfrac{ 26 \times 26 }{2}[/tex]
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[tex] \rm \dashrightarrow x {}^{2}=\dfrac{ 26 \times \cancel{26 }}{\cancel{2}}[/tex]
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[tex] \rm \dashrightarrow x {}^{2}=\dfrac{ 26 \times 13}{1}[/tex]
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[tex] \rm \dashrightarrow x {}^{2}=26 \times 13[/tex]
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[tex] \rm \dashrightarrow x = \sqrt{26 \times 13} [/tex]
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[tex] \rm \dashrightarrow x = \sqrt{338} [/tex]
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[tex] \bf \dashrightarrow x = 18.38[/tex]
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Formula used:-
[tex] \bigstar \boxed{\tt Hypotenuse^2 = Base^2+Perpendicular^2}[/tex]
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Therefore BC is equal to 18.38 cm.
⭑Related Concept⭑
Property of square:-