Another PRE-CAL question for genuises!!?

Question

Express the area (A) of an isosceles right triangle as a function of the length (s) of one of its two equal sides!!
I looked in the back and the answer was A= s squared/2!!! Need steps!!

Answer 1

This one is very simple

Since the triange has two same sides and it is right, the equation for it's area is s*s/2 (comes from (b)(h)/2). Now if we compare that to s ( which is one of its side) we can do several functions to to make it equal to the area.

1. s*s= s^2
2. s^2/2= A

Therefore the equation is A=s^2/2

Dude, this is all theory. It's not like regular math. You really can't learn much by looking at ways people do the problem because there are no numbers. You have to think about them. There is also more than one answer to some of these problems. For this one you could use trig.

Answer 2

The area of any triangle is 1/2 x base x height. In an isosceles right triangle, the base and height are equal, so call them "s". There you have 1/2 x s x s = exactly what the answer said.

Answer 3

sparkling up utilising the bigger/messier side (left side) first to compare the different: 9 sec^2(?) - 5 tan^2(?) = 5 + 4 sec^2(?) --------------------------------------... you comprehend yet another identity maximum in all probability: sin^2 + cos^2 = a million And dividing each and each of those through (cos^2) you get: sin^2/cos^2 + cos^2/cos^2 = a million/cos^2 tan^2 + a million = sec^2 which is often written with one on the different side: tan^2 = sec^2 - a million utilising this at the same time: 9 sec^2(?) - 5 tan^2(?) = 5 + 4 sec^2(?) 9 sec^2(?) - 5 tan^2(?) -->replace tan^2 as (sec^2-a million) 9 sec^2(?) - 5(sec^2(?) - a million) -->distribute the 5. 9 sec^2(?) - 5 sec^2(?) + 5 -->upload like factors. 4 sec^2(?) + 5 So utilising the left side you manipulated it into searching like the right side. this is a proved identity now.