Given :
- ABC is a right angled triangle at A where AB is 6cm, AC is 8cm.
To Find :
Solution :
- Let the Base be AB cm, Perpendicular be AC cm and Hypotenuse be BC.
We know that,
[tex] \qquad \sf{ \pmb{(Hypotenuse ) {}^{2} = (Base) {}^{2} + (Perpendicular) {}^{2} }}[/tex]
Substituting the given values in the formula :
[tex] \qquad\sf{ \dashrightarrow{ ( BC ) {}^{2} = (AB) {}^{2} + (AC ) {}^{2} }}[/tex]
[tex] \qquad\sf{ \dashrightarrow{ ( BC ) {}^{2} = (6) {}^{2} + (8 ) {}^{2} }}[/tex]
[tex] \qquad\sf{ \dashrightarrow{ ( BC ) {}^{2} = 36 + 64 }}[/tex]
[tex] \qquad\sf{ \dashrightarrow{ ( BC ) {}^{2} = 100 }}[/tex]
[tex] \qquad\sf{ \dashrightarrow{ BC {} = \sqrt{100} }}[/tex]
[tex] \qquad\sf{ \dashrightarrow{ BC {} = 10 }}[/tex]
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