This problem asks us to calculate the amount of space that the wallpaper will cover. The amount of space that something covers can be described as its area. In this case area is calculated by using the formula \(\displaystyle A=l \times w\)

\(\displaystyle A=5\times3\)

\(\displaystyle A=15ft^2\)

This problem asks us to calculate the amount of space that the wallpaper will cover. The amount of space that something covers can be described as its area. In this case area is calculated by using the formula \(\displaystyle A=l \times w\)

\(\displaystyle A=9\times7\)

\(\displaystyle A=63ft^2\)

The easiest way to answer the question is to locate \(\displaystyle Z\) on \(\displaystyle \overline{SQ}\) such that \(\displaystyle SZ = UY\):

Trapezoids \(\displaystyle SZYA\) and \(\displaystyle ZQUY\) have the same height, which is \(\displaystyle SA = QU\). Their bases, by construction, have the same lengths - \(\displaystyle AY = ZQ\) and \(\displaystyle SZ = YU\). Therefore, Trapezoids \(\displaystyle SZYA\) and \(\displaystyle ZQUY\) have the same area.

Since \(\displaystyle SX = UY + 1\), it follows that \(\displaystyle SX = SZ+ 1\), and \(\displaystyle SX > SZ\). It follows that Trapezoid \(\displaystyle SXYA\) is greater in area than Trapezoids \(\displaystyle SZYA\) and \(\displaystyle ZQUY\), and Trapezoid \(\displaystyle XQUY\) is less in area.

The length of the hypotenuse of a triangle with legs 5 feet and 12 feet is calculated using the Pythagorean Theorem, setting \(\displaystyle a = 5, b=12\):

\(\displaystyle c = \sqrt{a^{2}+b^{2}} = \sqrt{5^{2}+12^{2}} = \sqrt{25+144} = \sqrt{169} =13\)

The hypotenuse is 13 feet long. The perimeter is \(\displaystyle 5 + 12 + 13 = 30\) feet, which is equal to 10 yards.

The length of the second leg of the triangle can be calculated using the Pythagorean Theorem, setting \(\displaystyle a = 7,c =25\):

\(\displaystyle b = \sqrt{c^{2}-a^{2}} = \sqrt{25^{2}-7^{2}} = \sqrt{625-49}= \sqrt{576} = 24\)

The second leg has length 24 centimeters, so the perimeter of the triangle is 

\(\displaystyle 7 + 24 + 25 = 56\) centimeters.

One-half of a meter is one-half of 100 centimeters, or 50 centimeters, so (a) is greater.

Each figure has sides that are congruent, so in each case, multiply the sidelength by the number of sides.

(a) The triangle has perimeter \(\displaystyle 30 \times 3 = 90\) inches

(b) 2 feet are equal to 24 inches, so the square has sidelength \(\displaystyle 24 \times 4 = 96\) inches.

The square has the greater perimeter.

(a) The perimeter of the equilateral triangle is \(\displaystyle 25 \times 3 = 75\).

(b) \(\displaystyle CT = RE = 35\); \(\displaystyle EC, RT\) are of unknown value, but they are equal,  so we will call their common length \(\displaystyle N\).

Rectangle \(\displaystyle RECT\) has perimeter

\(\displaystyle 35 + 35 + N + N = 70 + 2N\).

Without knowing \(\displaystyle N\), it cannot be determined with certainty which figure has the longer perimeter. For example:

If \(\displaystyle N = 1\), then \(\displaystyle 70 + 2N = 70 + 2 \cdot 1 = 70 + 2 = 72 < 75\)

If \(\displaystyle N = 3\), then \(\displaystyle 70 + 2N = 70 + 2 \cdot 3 = 70 + 6 = 76>75\)

(a) As an isosceles triangle, \(\displaystyle \Delta ABC\), by definition, has two congruent sides. \(\displaystyle AB \neq BC\), so either :

\(\displaystyle AC = AB = 30\) 

in which case the perimeter of \(\displaystyle \Delta ABC\) is 

\(\displaystyle AB + AC + BC = 30 + 30 + 50 = 110\)

or

\(\displaystyle AC =BC = 50\)

in which case the perimeter of \(\displaystyle \Delta ABC\) is 

\(\displaystyle AB + AC + BC = 30 + 50 + 50 = 130\)

(b) \(\displaystyle \Delta DEF\) is an equilateral triangle, so, by definition, all of its sides are congruent; its perimeter is \(\displaystyle 35 \times 3 = 105\).

Regardless of the length of \(\displaystyle \overline{AC}\), \(\displaystyle \Delta ABC\) has the greater perimeter.

(a) \(\displaystyle \overline{AC}\) is the hypotenuse of \(\displaystyle \Delta ABC\), so by the Pythagorean Theorem, 

\(\displaystyle \left (AC \right )^{2} =\left ( AB \right ) ^{2} + \left ( B C\right )^{2}\)

\(\displaystyle \left (AC \right )^{2} =6^{2} + 8^{2}\)

\(\displaystyle \left (AC \right )^{2} =36 + 64 = 100\)

\(\displaystyle AC = \sqrt{100} = 10\)

(b) \(\displaystyle \overline{EF}\) is a leg of \(\displaystyle \Delta DEF\), whose hypotenuse is \(\displaystyle \overline{DF}\), so by the Pythagorean Theorem, 

\(\displaystyle \left (EF \right )^{2} =\left ( DF \right ) ^{2} - \left ( DE\right )^{2}\)

\(\displaystyle \left (EF \right )^{2} =26^{2} - 24^{2}\)

\(\displaystyle \left (EF \right )^{2} =676 -576 = 100\)

\(\displaystyle EF= \sqrt{100} = 10\)

\(\displaystyle AC = EF\)

The length of the hypotenuse of a triangle with legs 3 inches and 4 inches long is calculated using the Pythagorean Theorem, setting \(\displaystyle a = 3, b=4\):

\(\displaystyle c = \sqrt{a^{2}+b^{2}} = \sqrt{3^{2}+4^{2}} = \sqrt{9+16} = \sqrt{25} =5\)

The hypotenuse is 5 inches long. The perimeter is therefore \(\displaystyle 3 + 4 + 5 = 12\) inches, which is equal to one foot.