Here is a test to how good your math is, in ∆abc , a = ( 1 , 2 ) ; b = ( 5 , 5 ) and < acb = 90° . if area of ∆abc is to be 6.5 square units , then the possible number of points for c is ?

Zero

Step-by-step explanation:

The distance AB is ...

... √((5-1)²+(5-2)²) = √(16+9) = 5

The largest right triangle that can be constructed with AB as the hypotenuse is one with an altitude of 5/2 = 2.5 units. Its area will be ...

... (1/2)·5·2.5 = 6.25 . . . . square units

It is not possible to construct the triangle ABC described.

In order to achieve the given area, ∠C would need to be 87.75° or smaller. It could not be 90°.

There are no points for C to make this triangle

Step-by-step explanation:

We need to find the distance from point A to B

d = sqrt((x2-x1)^2 + (y2-y1)^2 )

 = sqrt((5-1)^2 + (5-2)^2

= sqrt(4^2 + 3^2)

= sqrt(16+9)

= sqrt(25)

= 5

We know the hypotenuse only

We can find the base in terms of the hypotenuse using the Pythagorean theorem

c^2 = a^2 + b^2

c^2 - a^2 = b^2

Taking the square root of each side

sqrt( c^2 - a^2) = sqrt(b^2)

sqrt( c^2 - a^2) = b

c = 5

sqrt(25 - a^2)=b

The area of a triangle is

A = 1/2 b* h  where b is the base and a is the height

6.5 = 1/2 (sqrt(25-a^2)) *a

Multiply each side by 2

2*6.5 = 2*1/2 (sqrt(25-a^2)) *a

13 =  (sqrt(25-a^2)) *a

Divide each side by a

13/a = sqrt(25-a^2)

Square each side

169/a^2 = 25-a^2

Multiply each side by a^2

169 = 25a^2 -a^4

Subtract 169 from each side

0= -a^4 +25a^2 -169

Divide by -1

0= a^4 -25a^2 +169

Using the discriminant

b^2 -4ac

25^2 - 4 * 1 * 169

625 - 676

-51

a^2 is imaginary

There is no solution

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