In a triangle with one of the angles 120^{cir | vedclass

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(A) Let the sides of the triangle be $a-d$,$a$,and $a+d$,where $d > 0$. The greatest side is $a+d = 7$. Since the angle opposite to the greatest side

Vedclass · Accepted Answer

$\frac{15 \sqrt{3}}{4} \ m^2$

Vedclass · Answer

(A) Let the sides of the triangle be $a-d$,$a$,and $a+d$,where $d > 0$. The greatest side is $a+d = 7$. Since the angle opposite to the greatest side is $120^{\circ}$,we use the Law of Cosines: $(a+d)^2 = (a-d)^2 + a^2 - 2a(a-d) \cos(120^{\circ})$. Substituting $a+d=7$,we have $d = 7-a$. $49 = (a-(7-a))^2 + a^2 - 2a(a-(7-a))(-1/2)$. $49 = (2a-7)^2 + a^2 + a(2a-7)$. $49 = 4a^2 - 28a + 49 + a^2 + 2a^2 - 7a$. $7a^2 - 35a = 0$. Since $a 
eq 0$,$7a = 35$,so $a = 5$. The sides are $5-(7-5) = 3$,$5$,and $7$. The area of the triangle is $\frac{1}{2} 	imes 	ext{side}_1 	imes 	ext{side}_2 	imes \sin(120^{\circ})$. Area $= \frac{1}{2} 	imes 3 	imes 5 	imes \frac{\sqrt{3}}{2} = \frac{15 \sqrt{3}}{4} \ m^2$.

In a triangle with one of the angles $120^{\circ}$,the lengths of the sides form an $A$.$P$. If the length of the greatest side is $7 \ m$,then the area of the triangle is

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