Respuesta :

cubstellationoupyb6
The area of a traingle is [tex] \frac{1}{2}bh[/tex]
So we need to know the base and the height
To work out the sides of a right angle triangle we can use a²+b²=c
c is the hypotenuse and a and b are the other sides
Because this triangle is isoceles we know that a and b are the same

a²+b²=c²
a²+a²=c²
2a²=c²
2a²=(x√6)²
2a²=(x√6)(x√6)
2a²=x²+x√6+x√6+6
2a²=x²+x√6+x√6+6
a²=[tex] \frac{x^{2}+x \sqrt{6}+x \sqrt{6}+6 }{2} [/tex]
a=[tex] \sqrt{\frac{x^{2}+x\sqrt{6}+x\sqrt{6}+6}{2}}[/tex]

A=[tex] \frac{1}{2}bh[/tex]
A=[tex] \frac{1}{2}a*a[/tex]
A=[tex] \frac{1}{2} [/tex]×[tex] \sqrt{\frac{x^{2}+x\sqrt{6}+x\sqrt{6}+6}{2}}[/tex]×[tex] \sqrt{\frac{x^{2}+x\sqrt{6}+x\sqrt{6}+6}{2}}[/tex]
A=[tex] \frac{1}{2} [/tex]×[tex] \frac{x^{2}+x \sqrt{6}+x \sqrt{6}+6 }{2} [/tex]
A=[tex]\frac{x^{2}+x \sqrt{6}+x \sqrt{6}+6 }{4}[/tex]

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