By the Pythagorean Theorem, the distance from A to C, which we will call $\displaystyle D$, is equal to 

$\displaystyle D = \sqrt{ 3^{2}+ 4^{2}} = \sqrt{ 9+ 16} = \sqrt{ 25} = 5$

The perimeter of the triangle is $\displaystyle 3 + 4 + 5 = 12$. The insect has traveled $\displaystyle 3 + 5 = 8$ units out of 12, which is 

$\displaystyle \frac{8}{12} \times 100 = 66 \frac{2}{3} \%$ of the perimeter.

By the Pythagorean Theorem, the distance from A to C, which we will call $\displaystyle D$, is equal to 

$\displaystyle D = \sqrt{ 3^{2}+ 4^{2}} = \sqrt{ 9+ 16} = \sqrt{ 25} = 5$

The perimeter of the triangle is $\displaystyle 3 + 4 + 5 = 12$. The insect has traveled $\displaystyle 3 + 4 = 7$ units, or $\displaystyle \frac{7}{12}$ of the perimeter.

By the Pythagorean Theorem, the distance from B to C, which we will call $\displaystyle D$, is equal to 

$\displaystyle D = \sqrt{13^{2}- 5^{2}} = \sqrt{169- 25 }= \sqrt{144} = 12$

The perimeter of the triangle is 

$\displaystyle 5+12 + 13 = 30$.

The insect has traveled $\displaystyle 12+13 = 25$ units, which is 

$\displaystyle \frac{25}{30} = \frac{25 \div 5}{30 \div 5} = \frac{5}{6}$

of the perimeter.

By the Pythagorean Theorem, the distance from B to C, which we will call $\displaystyle D$, is

$\displaystyle D = \sqrt{13^{2}- 5^{2}} = \sqrt{169- 25 }= \sqrt{144} = 12$.

The perimeter of the triangle is 

$\displaystyle 5+12 + 13 = 30$.

The insect has traveled $\displaystyle 5 + 12 = 17$ units, or 

$\displaystyle \frac{17}{30} \times 100 = 56 \frac{2}{3} \%$ of the perimeter.

Use the area formula for a triangle, setting $\displaystyle A = 4200,b=80$:

$\displaystyle A = \frac{1}{2} bh$

$\displaystyle 4200 = \frac{1}{2}\cdot 80 \cdot h = 40h$

$\displaystyle h = 4200 \div 40 = 105$ inches

The legs of an isosceles right triangle have equal length, so, if the sum of their lengths is one meter, which is equal to 100 centimeters, each leg measures half of this, or 

$\displaystyle 100 \div 2 = 50$ centimeters.

The area of a triangle is half the product of its height and base; for a right triangle, the legs serve as height and base, so the area of the triangle is

$\displaystyle \frac{1}{2} \times 50 \times 50 = 1,250$ square centimeters.

Since $\displaystyle X$, $\displaystyle Y$, and $\displaystyle Z$ are the midpoints of $\displaystyle \overline{AD}$, $\displaystyle \overline{BC}$, and $\displaystyle \overline{AB}$, if we call $\displaystyle s$ the length of each side of the square, then 

$\displaystyle AX = AZ = \frac{1}{2}s$

The area of $\displaystyle \bigtriangleup AZX$ is half the product of the lengths of its legs:

$\displaystyle \frac{1}{2} \cdot AX \cdot AZ$

$\displaystyle = \frac{1}{2} \cdot \frac{1}{2}s \cdot \frac{1}{2}s$

$\displaystyle = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\cdot s \cdot s$

$\displaystyle = \frac{1 \cdot 1 \cdot 1}{2\cdot 2 \cdot 2}\cdot s ^{2}$

$\displaystyle = \frac{1}{8} s ^{2}$

The area of the square is the square of the length of a side, which is $\displaystyle s^{2}$. This is eight times the area of $\displaystyle \bigtriangleup AZX$, so the correct choice is $\displaystyle 8N$

The area of a right triangle is half the product of the lengths of its legs, which here are 25 and 60. So

$\displaystyle A = \frac{1}{2} \cdot 25 \cdot 60 = 750$

which is less than 800.

The area of a right triangle is half the product of the lengths of its legs, which here are $\displaystyle A$ feet and $\displaystyle B$ feet.

Multiply each length by 12 to convert to inches - the lengths become $\displaystyle 12A$ and $\displaystyle 12 B$. The area in square inches is therefore

$\displaystyle \frac{1}{2} (12A )(12 B) = \frac{1}{2} \cdot 12 \cdot 12 \cdot A \cdot B = 72AB$ square inches.

Square $\displaystyle ABCD$ has area 1,600, so the length of each side is $\displaystyle \sqrt{1,600 }= 40$.

Since $\displaystyle AZ = 16$,

$\displaystyle BZ = AB - AZ = 40 - 16 = 24$

Therefore, $\displaystyle BZ > AZ$.

$\displaystyle \bigtriangleup AZX$ has as its area $\displaystyle \frac{1}{2} \cdot AX \cdot AZ$; $\displaystyle \bigtriangleup BZY$ has as its area $\displaystyle \frac{1}{2} \cdot BX \cdot BZ$.

Since $\displaystyle BZ > AZ$ and $\displaystyle AX = BX$, it follows that

$\displaystyle BX \cdot BZ > AX \cdot AZ$

and

$\displaystyle \frac{1}{2} \cdot BX \cdot BZ >\frac{1}{2} \cdot AX \cdot AZ$

$\displaystyle \bigtriangleup BZY$ has greater area than $\displaystyle \bigtriangleup AZX$.