Exercise 33 Page &nbsp; 505

a= 1, b= 1

1^2+ 1^2=c^2

Calculate power

1+1=c^2

Add terms

2=c^2

sqrt(LHS)=sqrt(RHS)

sqrt(2)=c

We have found that the legs of the triangle are equal to 1 unit and sqrt(2) units.

Let's draw the triangle formed.

Let's plug in the values for the legs of this triangle into the Pythagorean Theorem to find the length of d, the diagonal of the cube.

a^2+b^2=c^2

a= 1, b= sqrt(2)

1^2+( sqrt(2))^2=c^2

Calculate power

1+2=c^2

Add terms

3=c^2

sqrt(LHS)=sqrt(RHS)

sqrt(3)=c

The length of the diagonal with edges of length 1 unit is equal to sqrt(3) units.

b Let's repeat the process from Part A with edges of length 2 units this time. We will start by finding the diagonal of the base, which is the length of the other leg.

a^2+b^2=c^2

a= 2, b= 2

2^2+ 2^2=c^2

Calculate power

4+4=c^2

Add terms

8=c^2

sqrt(LHS)=sqrt(RHS)

sqrt(8)=c

We have found that the legs of the triangle are equal to 2 units and sqrt(8) units. Finally, let's plug in these values into the Pythagorean Theorem to find the length of d, the diagonal of the cube.

a^2+b^2=c^2

a= 2, b= sqrt(8)

2^2+( sqrt(8))^2=c^2

Calculate power

4+8=c^2

Add terms

12=c^2

sqrt(LHS)=sqrt(RHS)

sqrt(12)=c

Write as a product

sqrt(4 * 3)=c

SqrtProd

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

sqrt(4) * sqrt(3)=c

CalcRoot

Calculate root

2sqrt(3)=c

The length of the diagonal with edges of length 2 units is equal to 2sqrt(3) units.

c Let's repeat the same process from Parts A and B with edges of length s units this time. We will start by finding the length of the other leg.

a^2+b^2=c^2

a= s, b= s

s^2+ s^2=c^2

Add terms

2s^2=c^2

sqrt(LHS)=sqrt(RHS)

sqrt(2s^2)=c

We have found that the legs of the triangle are equal to s units and sqrt(2s^2) units. Finally, let's plug in these values into the Pythagorean Theorem to find the length of d, the diagonal of the cube.

a^2+b^2=c^2

a= s, b= sqrt(2s^2)

s^2+( sqrt(2s^2))^2=c^2

Calculate power

s^2+2s^2=c^2

Add terms

3s^2=c^2