SOLUTION: Express the Area A of an isoceles triangle with a height of 8 inches and a base of b inches as a function of the length s of one of its two equal sides.
I got this far
A=1/2b
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-> SOLUTION: Express the Area A of an isoceles triangle with a height of 8 inches and a base of b inches as a function of the length s of one of its two equal sides.
I got this far
A=1/2b
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Question 17689: Express the Area A of an isoceles triangle with a height of 8 inches and a base of b inches as a function of the length s of one of its two equal sides.
I got this far
A=1/2bh. b=A/4
P=s+s+b, 2s+A/4
now how do I find the A as a function of s?
thanks Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! Express the Area A of an isoceles triangle with a height of 8 inches and a base of b inches as a function of the length s of one of its two equal sides
let the isocelles triangle be PQR with QR as base =b and equal sides PQ=PR=s,Let PS=h be the altitude from P to QR.since PQR is isocelles triangle QS=SR=b/2.further PQS is a right angled triangle.hence
PQ^2=QS^2+PS^2
s^2=(b/2)^2+h^2 or.s^2-h^2=b^2/4..or..b=Square root of [4(s^2-64)]
area =A = (1/2)*b*h=8*b/2=4b = 4*Square root of [4(s^2-64)]...........
A=8*Square root of [(s^2-64)]=
