this is a challenge for all .answer this 42 question with steps compulsory and i will guarantee to mark your answer as brainlist
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Let
sides of triangle be a,b and c resp
area of triangle be ∆
by heron's formula
area of triangle ∆
= √s(s-a)(s-b)(s-c)
where s = a+b+c/2
each side is increased by 200
thus new sides became a+200/100*a = 3a
à=3a
same as 3b,3c
and new s = 3 old s
thus
new area = √3s(3s-3a)(3s-3b)(3s-3c)
= 9√s(s-a)(s-b)(s-c)
= 9∆
let area increased by x%
∆+X/100*∆ = 9∆
thus X = 800%
area of triangle increased by 800%
sides of triangle be a,b and c resp
area of triangle be ∆
by heron's formula
area of triangle ∆
= √s(s-a)(s-b)(s-c)
where s = a+b+c/2
each side is increased by 200
thus new sides became a+200/100*a = 3a
à=3a
same as 3b,3c
and new s = 3 old s
thus
new area = √3s(3s-3a)(3s-3b)(3s-3c)
= 9√s(s-a)(s-b)(s-c)
= 9∆
let area increased by x%
∆+X/100*∆ = 9∆
thus X = 800%
area of triangle increased by 800%
Answered by
68
Answer:
Let
sides of triangle be a,b and c resp
area of triangle be ∆ by heron's formula
area of triangle ∆
= √s(s-a)(s-b)(s-c)
where s = a+b+c/2
each side is increased by 200
thus new sides became a+200/100*a = 3a
à=3a
same as 3b,3c
and new s = 3 old s
thus
new area = √3s(3s-3a)(3s-3b)(3s-3c)
= 9√s(s-a)(s-b)(s-c)
= 9∆
let area increased by x%
∆+X/100*∆ = 9∆
thus X = 800%
area of triangle increased by 800%
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