The area of the right triangle shown is 24 square feet. which equations can be used to find the lengths of the legs of the triangle? check all that apply. a x2 + 2x – 24 = 0 b x(x + 2) = 24 c x2 + (x + 2)2 = 100 d 0.5(x)(x + 2) = 24 e x2 + 2x – 48 = 0 f x2 + (x + 2)2 = 24

Answer with explanation:

It is given that area of right triangle is 24 square feet

Since Sides of any triangle is always Positive.So, we have to find those equation among six options , which gives two positive Integral values,as area of right triangle is equal to product of base length and Corresponding Altitude.

Starting from option 1

⇒1. x² +2 x - 24 =0

 Splitting the middle term

→x²+6 x - 4 x -24=0

→x×(x+6)-4×(x+6)=0  

→(x-4)(x+6)

By substituting the expression equal to , 0, we will get one positive zero and one negative Zero so , it can't be the equation which can be sides of right triangle

Option B:

→x×(x+2)=0

Gives, two zeroes which are equal to 0 and , -2 So, it is also not that expression which gives sides of right triangle, when area =24 square feet

Option C:

x²+ (x+2)×2=100

x²+2 x +4-100=0

x²+2 x-96=0

It will give one Positive zero and one negative zero ,so it can't be the equation which gives length of legs of right triangle.

Option D:

0.5 × x×(x+2)=24,

as 0.5 is equal to

x²+2 x -48=0

(x-6)(x+8)=0

The two zeroes are , such that one of them is positive and other is negative.So this is not the equation which can be used to find the lengths of the legs of the right triangle.

⇒Option E.

x²+2 x - 48=0

Same as option D

It cant be the expression which can be used to find the lengths of the legs of the triangle.

⇒Option F:

x²+(x+2)×2=24

x²+2 x +4-24=0

x²+ 2 x -20=0

It will give one positive zero and one negative zero , so it can't be the equation which can be used to find the lengths of the legs of the triangle.

→→None, of the option can be used to find the length of sides of right  triangle when area is equal to 24 square feet because none of them gives two positive Integral value.

let's cover over the formula of the vertex:

y = a(x – h)^2 + k

in this formula, (h,k) are the vertex, always remember that the positive or negative value of the vertex is based on the sign it has.

for example, if the vertex is -3,-6 then both the x and y value would have the positive sign attached to it and if it were a positive x and y value, then a negative sign would be attached to it.

with that being said, for our problem the x value is a positive one so we are looking for a number with a negative sign next to it, in this case the first and second choice.

however, for the first answer choice it isn't quite part of the formula so we know that the answer choice correctly displays the x value.

to confirm, we can see if the -1 fits correctly with the second answer choice, and as mentioned above a negative number will have a positive sign attached to it. so, for our problem we can see that -1 has been changed to +1, so we can confirm that the second answer choice is correct.

hence, our final answer is y=(x - 1)^2 + 1