Exercise 38 Page 553

We want to find the area and the perimeter of the given triangle. In order to do that, we first need to find the length of its hypotenuse.

Hypotenuse

To find the missing side of the triangle, we will use the Pythagorean Theorem. a^2+b^2=c^2 In the formula, a and b are the legs and c is the hypotenuse of a right triangle. We are given a triangle with a=12 and b=16.

Let's substitute these values into the formula.

a^2+b^2=c^2

SubstituteII

a= 12, b= 16

12^2+ 16^2=c^2

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Solve for c

CalcPow

Calculate power

144+256=c^2

AddTerms

Add terms

400=c^2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

20=c

RearrangeEqn

Rearrange equation

c=20

Since a negative side length does not make sense, we only need to consider positive solutions.

Perimeter

The perimeter of a figure is the sum of the side lengths. For a triangle, this is the sum of sides a, b, and c. P=a+b+c To calculate P, we should substitute the given values a= 12, b= 16, and c= 20 into the formula and simplify.

P=a+b+c

SubstituteIII

Substitute a= 12, b= 16, c= 20

P= 12+ 16+ 20

AddTerms

Add terms

P=48

The triangle's perimeter is 48 units.

Area

Here we will use the formula for calculating the area of a triangle. A=1/2bh Now we substitute the value of the base, b= 16, and the value of the height, h= 12, into the formula to calculate A.

A=1/2bh

SubstituteII

b= 16, h= 12

A=1/2( 16)( 12)

MoveRightFacToNum

a/c* b = a* b/c

A=16/2* 12

CalcQuot

Calculate quotient

A=8*12

Multiply

Multiply

A=96

The area of the triangle is 96 square units.