Find the area of a triangle, 2 sides of which are 18 cm and 10 cm and the perimeter is 42 cm. Also find the length of the altitude corresponding to the third side.
URGENT!!!!
Answers
Answered by
4
a = 18 cm b = 10 cm
perimeter a + b + c = 42 cm
semi-perimeter = 21 cm
c = 42 cm - 18 - 10 cm = 14 cm
Area of triangle =
[tex] \sqrt{21(21-18)(21-10)(21-14)} \\ \\ \sqrt{7*3*11*7} \\ 7 \sqrt{33} cm^{2} \\ \\ Altitude = \frac{2* 7 \sqrt{33} }{14} \\ \sqrt{33} \\ [/tex]
area: 7√33 cm and altitude √33 cm
perimeter a + b + c = 42 cm
semi-perimeter = 21 cm
c = 42 cm - 18 - 10 cm = 14 cm
Area of triangle =
[tex] \sqrt{21(21-18)(21-10)(21-14)} \\ \\ \sqrt{7*3*11*7} \\ 7 \sqrt{33} cm^{2} \\ \\ Altitude = \frac{2* 7 \sqrt{33} }{14} \\ \sqrt{33} \\ [/tex]
area: 7√33 cm and altitude √33 cm
Answered by
5
a=18;b=10;c=?
perimeter=42
area=?
a+b+c=perimeter
18+10+c=42
28+c=42
c=14
area=√s(s-a)(s-b)(s-c)
s=(a+b+c)/2
s=42/2
s=21
=√21(21-18)(21-10)(21-14)
=√21*3*11*7
=√7*3*7*3*11
=7*3√11
=21√11
area=21√11
altitude is same method done by Kvnmurty
perimeter=42
area=?
a+b+c=perimeter
18+10+c=42
28+c=42
c=14
area=√s(s-a)(s-b)(s-c)
s=(a+b+c)/2
s=42/2
s=21
=√21(21-18)(21-10)(21-14)
=√21*3*11*7
=√7*3*7*3*11
=7*3√11
=21√11
area=21√11
altitude is same method done by Kvnmurty
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