Exercise 2 Page 270

We want to find the area of the given triangle.

The area of a triangle is one half the product of a base b and its corresponding height h. In our case b= 3sqrt(2)+sqrt(5) and h= 2sqrt(2)+sqrt(5). Area=1/2bh ⇕ Area=1/2( 3sqrt(2)+sqrt(5))( 2sqrt(2)+sqrt(5))We can calculate the product of the two binomials using the Distributive Property.

Area=1/2(3sqrt(2)+sqrt(5))(2sqrt(2)+sqrt(5))

Distr

Distribute (3sqrt(2)+sqrt(5))

Area=1/2((3sqrt(2)+sqrt(5))2sqrt(2)+(3sqrt(2)+sqrt(5))sqrt(5))

▼

Distribute 2sqrt(2) & sqrt(5)

Distr

Distribute 2sqrt(2)

Area=1/2((3sqrt(2))2sqrt(2)+(sqrt(5))2sqrt(2)+(3sqrt(2)+sqrt(5))sqrt(5))

Distr

Distribute sqrt(5)

Area=1/2((3sqrt(2))2sqrt(2)+(sqrt(5))2sqrt(2)+(3sqrt(2))sqrt(5)+(sqrt(5))sqrt(5))

CommutativePropMult

Commutative Property of Multiplication

Area=1/2(3*2*sqrt(2)*sqrt(2)+2*sqrt(5)*sqrt(2)+3*sqrt(2)*sqrt(5)+sqrt(5)*sqrt(5))

Let's recall the Multiplication Property of Square Roots. sqrt(a) * sqrt(b)=sqrt(ab), for a≥ 0,b≥ 0 Thanks to this property, we can continue simplifying our expression.

Area=1/2(3*2*sqrt(2)*sqrt(2)+2*sqrt(5)*sqrt(2)+3*sqrt(2)*sqrt(5)+sqrt(5)*sqrt(5))

MultSqrt

sqrt(a)* sqrt(a)= a

Area=1/2(3*2*2+2*sqrt(5)*sqrt(2)+3*sqrt(2)*sqrt(5)+5)

ProdSqrt

sqrt(a)*sqrt(b)=sqrt(a* b)

Area=1/2(3*2*2+2*sqrt(5*2)+3*sqrt(2*5)+5)

Multiply

Multiply

Area=1/2(12+2sqrt(10)+3sqrt(10)+5)

Now, we will add the like terms together.

Area=1/2(12+2sqrt(10)+3sqrt(10)+5)

▼

Simplify right-hand side

CommutativePropAdd

Commutative Property of Addition

Area=1/2(12+5+2sqrt(10)+3sqrt(10))

FactorOut

Factor out sqrt(10)

Area=1/2(12+5+(2+3)sqrt(10))

AddTerms

Add terms

Area=1/2(17+5sqrt(10))

Distr

Distribute 1/2

Area=1/2(17)+1/2(5sqrt(10))

MoveRightFacToNum

a/c* b = a* b/c

Area=17/2+5sqrt(10)/2

CalcQuot

Calculate quotient

Area=8.5+2.5sqrt(10)

The area of the given triangle is 8.5+2.5sqrt(10) units squared. This result corresponds with option D.