Right triangle ABC has angle C=90º, perimeter p=7, and hypotenuse c=3. What is area of the triangle?

Answer 1

7/4.

The legs are of length x and 4-x, by the perimeter concerns.

Then

x^2 + (4-x)^2 = 3^2 = 9
...
2x^2 - 8x + 7 = 0

Solutions are
(4 +or- sqrt(2))/2.

The two sides are (4-sqrt(2))/2, (4+sqrt(2))/2.
So the area is

(1/2)[(4-sqrt(2))/2][(4+sqrt(2))/2] = (1/8)(16-2) = 14/8 = 7/4.

Answer 2

we know that a^2 + b^2 = c^2 from Pythagoras
therefore a^2 + b^2 = 9. Also we know that a + b + c = p
a + b + 3 = 7 => a + b = 4 => a = 4-b

Substituting a = 4-b into the equation a^2 + b^2 = 9 gives
(4-b)^2 + b^2 = 9
16 - 8b + b^2 + b^2 = 9
2b^2 - 8b + 7 = 0
b = 2 +- sqrt(2)/2

substituting b back into the equation for a reveals that if
b = 2 + sqrt(2)/2 then a = 2 - sqrt(2)/2 and also if b = 2 - sqrt(2) then a = 2 + sqrt(2)/2 so we can use 2+sqrt(2)/2 and 2 - sqrt(2)/2 as the two other lenghts.

The area of a triangle is given using the equation
a = 1/2 * base * perpendicular height
a = 1/2 * {2 + sqrt(2)/2} * { 2 - sqrt(2)/2}
"difference of 2 squares!"
a = 1/2 * (4 - 2/4)
a = 1/2 * (7/2)

a = 7/4

Answer 3

Let AB=c=3 AC=b and CB=a

The perimeter p = a + b + c = 7 and c=3
Therefore, a + b = 7 - 3 = 4 (i)

The angle C=90 therefore by Pythagore, a^2 + b^2 = c^2
therefore, a^2 + b^2 = 9 (ii)

Based on (i), b = 4 -a

If we substitute b in (ii) we obtain a^2 + (4-a)^2 = 9
Developped, this equation becomes 2a^2 - 8a + 7 = 0

The 2 possible solutions for this equation are 2 - (1/2)sqrt(2) and 2 + (1/2)sqrt (2)

These are the values of a and b or b and a.

The area is (1/2)*a*b = 7/4

Answer 4

OK

This one is a little tricky.

From Pythagorian theorem:

x^2 + y^2 = 3^2
x^2 + y^2 = 9

We also know that x+y = 4. Substitute

(4-y)^2 + y^2 = 9
16-8y+y^2+y^2= 9
7-8y+2y^2 = 0

8 +-sqrt(64 -(4)(2)(7)) / 2(2)
8 +-sqrt(64-56) /4
8+-sqrt(8) /4

2+1/2(sqrt(2)) and 2-1/2(sqrt(2))

2+.707 and 2-.707

y = 2.707 and or 1.293

So the two sides are 2.707 and 1.293

Now area = 1/2 (2.707)(1.293) = 1.75 sq units

Check

(2.707)^2 + (1.293)^2 = 9??
7.328 + 1.672 = 9??
9 = 9 YES!!

Hope that helps

Answer 5

Let's establish some definitions first.

If p=7, then a+b+c=7, which means a+b+3=7.

Therefore, a=4-b
Area=.5bh=.5ba=.5b(4-b)=2b-.5b^2

In a right triangle, a^2+b^2=c^2, so:

(4-b)^2+b^2=9
16-8b+2b^2=9
2b^2-8b+7=0

Quadratic formula:

(8±sqrt(64-56))/4=(8±2sqrt(2))/4=2±sqrt(2)/2=b

So a=4-2±sqrt(2)/2=2±sqrt(2)/2

Therefore, a=b. So the area is:

.5*(2±sqrt(2)/2)^2=.5*(4.5±2sqrt(2))=2.25±sqrt(2)

This means that from the given information, it can be either 2.25+sqrt(2) which is about 3.664, or it can be 2.25-sqrt(2), about 0.8358.

We need more information before we can determine which of these it is.

-IMP ;) :)

Answer 6

perimeter p = 7, hypotenuse c = 3
sum of sides (a + b) = p -- c = 7 -- 3 = 4
area = product of sides / 2 = ab/2
whence 0 < area = ab/2 < 4 = sum of sides
area may be say, 0.5*3.5/2 = 0.875 minimum side
and may be say, 2*2/2 = 2.000 maximum side.

Answer 7

a^2 + b^2 = 9
a + b + 3 = 7

This gives the lengths of the other 2 sides as
[8 + sqrt(8)]/4 and [8-sqrt(8)]/4

The area would be 1/2 x a x b, which would give (64+8)/32 = 2.25

Answer 8

let one of the leg = x
the other leg = 4-x
(4-x)^2 + x^2 = 3^2
16 - 8x + x^2 + x^2 = 9
2x^2 -8x + 7 = 0
x = 2.7 or 1.3
Area = 1/2 * 2.7 * 1.3 = 1.75

Answer 9

area = 2

other sides would have to add up to 4 and if its 1 and 3 then c isnt hypotenuse, so we have left the sides being 2 and 2
2*2=4 4/2=2 so area is 2