Exercise 3 Page 642

For the given triangle, we are asked to find its area.

To find the area, we first need to find the missing base and height of the triangle.

Height

Let's first focus on finding the height of the triangle. Note that the missing length is one of the legs of the right triangle formed to the right.

For this right triangle, the lengths of the leg and the hypotenuse are 6m and 10m. Let's substitute these values in the Pythagorean Theorem and solve for the other leg b.

a^2+b^2=c^2

SubstituteII

a= 6, c= 10

6^2+b^2= 10^2

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Solve for c

CalcPow

Calculate power

36+b^2=100

SubEqn

LHS-36=RHS-36

b^2=64

SqrtEqn

sqrt(LHS)=sqrt(RHS)

b=sqrt(64)

CalcRoot

Calculate root

b=8

Please note that since b is the leg of a right triangle it must be non-negative, which is why we only kept the principal root when solving the equation. We found that the length of the other leg is b= 8 m. Keep in mind that this is also the height of the triangle for which we want to find the area.

Area

Recall that the area of a triangle is half the product of its base and its height. Note that the base is a sum of the segments.

In the given triangle, by the Segment Addition Postulate, we know that the base is 12m. What is more, we previously found that the height is 8m. We can substitute these two values in the formula for the area of a triangle and simplify.

A=1/2bh

SubstituteII

b= 12, h= 8

A=1/2(12)( 8)

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Evaluate right-hand side

Multiply

Multiply

A=1/2(96)

MoveRightFacToNumOne

1/b* a = a/b

A=96/2

CalcQuot

Calculate quotient

A=48

The area of the triangle is 48m^2.