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The sides of triangle are 16,30,34 cm. Find are of triangle.
I wanna recheck my answer.
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Answered by
11
Heya!
Here is yr answer.....
Given,
sides of triangle are 16 , 30, 34
a = 16 , b = 30 , c = 34
Perimeter S of triangle = a+b+c/2
= 16+30+34/2
= 80/2
= 40cm
According to heron's formua -----
=> SS - aS - bS - c
=> 4040-1640-3040-34
=> 40×24×10×6
=> 56700
=> 240
Area of triangle = 240cm²
Hope it hlpz..
Here is yr answer.....
Given,
sides of triangle are 16 , 30, 34
a = 16 , b = 30 , c = 34
Perimeter S of triangle = a+b+c/2
= 16+30+34/2
= 80/2
= 40cm
According to heron's formua -----
=> SS - aS - bS - c
=> 4040-1640-3040-34
=> 40×24×10×6
=> 56700
=> 240
Area of triangle = 240cm²
Hope it hlpz..
ShuchiRecites:
Sorry but I want area
Please read ques carefully
ohhk..
I asked to takeout area of triangle
u typed... are
kk wait
Sides of a triangle are a = 16 cm, b = 30 cm and c = 34 cm
s = (16 + 30 + 34)/2 = 80/2 = 40 cm
Area of triangle = √[s(s – a)(s – b)(s – c)]
= √[40(40 – 16)(40 – 30)(40 – 34)
= √[40 × 24 × 10 × 6]
= 240 sq cm
s = (16 + 30 + 34)/2 = 80/2 = 40 cm
Area of triangle = √[s(s – a)(s – b)(s – c)]
= √[40(40 – 16)(40 – 30)(40 – 34)
= √[40 × 24 × 10 × 6]
= 240 sq cm
Answered by
0
Suppose the sides of triangle are a, b and c
Where a = 16 cm; b = 30 cm and c = 34 cm
And S the semi perimeter of triangle.
So, S = a+b+c2 = 16+30+342 = 802 = 40
According to the heron's formula we have;
Area; A = S(S−a)(S−b)(S−c)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√
= 40(40−16)(40−30)(40−34)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=40×24×10×6‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=57600‾‾‾‾‾‾√=240
So, the area of triangle is 240 square centimeter
Where a = 16 cm; b = 30 cm and c = 34 cm
And S the semi perimeter of triangle.
So, S = a+b+c2 = 16+30+342 = 802 = 40
According to the heron's formula we have;
Area; A = S(S−a)(S−b)(S−c)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√
= 40(40−16)(40−30)(40−34)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=40×24×10×6‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=57600‾‾‾‾‾‾√=240
So, the area of triangle is 240 square centimeter
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