Question

find the area of the triangle with the length of the side 9 cm , 10 cm , and 17 cm.​

Answer 1

Answer:

Hope it helps you..........

Answer 2

Answer=The area of a triangle whose side lengths are a, b,a,b, and cc is given by

The area of a triangle whose side lengths are a, b,a,b, and cc is given byA=√{s(s-a)(s-b)(s-c)},

{s(s-a)(s-b)(s-c)},A=

{s(s-a)(s-b)(s-c)},A= s(s−a)(s−b)(s−c)

{s(s-a)(s-b)(s-c)},A= s(s−a)(s−b)(s−c)

{s(s-a)(s-b)(s-c)},A= s(s−a)(s−b)(s−c) ,

{s(s-a)(s-b)(s-c)},A= s(s−a)(s−b)(s−c) ,where s=\dfrac{(\text{perimeter of the triangle})}{2}=\dfrac{a+b+c}{2}s=

{s(s-a)(s-b)(s-c)},A= s(s−a)(s−b)(s−c) ,where s=\dfrac{(\text{perimeter of the triangle})}{2}=\dfrac{a+b+c}{2}s= 2

{s(s-a)(s-b)(s-c)},A= s(s−a)(s−b)(s−c) ,where s=\dfrac{(\text{perimeter of the triangle})}{2}=\dfrac{a+b+c}{2}s= 2(perimeter of the triangle)

{s(s-a)(s-b)(s-c)},A= s(s−a)(s−b)(s−c) ,where s=\dfrac{(\text{perimeter of the triangle})}{2}=\dfrac{a+b+c}{2}s= 2(perimeter of the triangle)

{s(s-a)(s-b)(s-c)},A= s(s−a)(s−b)(s−c) ,where s=\dfrac{(\text{perimeter of the triangle})}{2}=\dfrac{a+b+c}{2}s= 2(perimeter of the triangle) =

{s(s-a)(s-b)(s-c)},A= s(s−a)(s−b)(s−c) ,where s=\dfrac{(\text{perimeter of the triangle})}{2}=\dfrac{a+b+c}{2}s= 2(perimeter of the triangle) = 2

{s(s-a)(s-b)(s-c)},A= s(s−a)(s−b)(s−c) ,where s=\dfrac{(\text{perimeter of the triangle})}{2}=\dfrac{a+b+c}{2}s= 2(perimeter of the triangle) = 2a+b+c

{s(s-a)(s-b)(s-c)},A= s(s−a)(s−b)(s−c) ,where s=\dfrac{(\text{perimeter of the triangle})}{2}=\dfrac{a+b+c}{2}s= 2(perimeter of the triangle) = 2a+b+c

{s(s-a)(s-b)(s-c)},A= s(s−a)(s−b)(s−c) ,where s=\dfrac{(\text{perimeter of the triangle})}{2}=\dfrac{a+b+c}{2}s= 2(perimeter of the triangle) = 2a+b+c , semi-perimeter of the triangle

S=(9+10+17)/2

36/2=18