Solved Examples: Polygons

Let the:
base = b
perpendicular height = h
area = A

$ A = \dfrac{bh}{2} \\[5ex] (a.) \\[3ex] b = 9\;cm \\[3ex] h = 7\;cm \\[3ex] A = \dfrac{7 * 9}{2} \\[5ex] A = 31.5\;cm^2 \\[5ex] (b.) \\[3ex] b = 9\;cm \\[3ex] h = 4\;cm \\[3ex] A = \dfrac{9 * 4}{2} \\[5ex] A = 18\;cm^2 $

Let:
Area = A
Length = L
Wdith = W

$ A = 1815\:yd^2 \\[3ex] L = 50\:m \\[3ex] W =...to\:\:be\:\:in\:\:m \\[3ex] Convert\:\:1815\:\:yd^2\:\:to\:\:m^2 \\[3ex] 1815\:\:yd * yd * \dfrac{1\:m * 1\:m}{1.1\:yd * 1.1\:yd} \\[5ex] A = \dfrac{1815m^2}{1.21} \\[5ex] A = 1500\:m^2 \\[3ex] W = \dfrac{A}{l} \\[5ex] W = \dfrac{1500\:m^2}{50\:m} \\[5ex] W = 30\:m $

$ (A.) \\[3ex] a = 9cm \\[3ex] b = 40cm \\[3ex] c = 41cm \\[3ex] s = \dfrac{a + b + c}{2} \\[5ex] s = \dfrac{9 + 40 + 41}{2} \\[5ex] s = \dfrac{90}{2} \\[5ex] s = 45cm \\[5ex] Area = \sqrt{s(s - a)(s - b)(s - c)} \\[3ex] Area = \sqrt{45(45 - 9)(45 - 40)(45 - 41)} \\[3ex] Area = \sqrt{45 * 36 * 5 * 4} \\[3ex] Area = \sqrt{32400} \\[3ex] Area = 180\;cm^2 \\[5ex] (B.) \\[3ex] a = 7cm \\[3ex] b = 24cm \\[3ex] c = 25cm \\[3ex] s = \dfrac{a + b + c}{2} \\[5ex] s = \dfrac{7 + 24 + 25}{2} \\[5ex] s = \dfrac{56}{2} \\[5ex] s = 28cm \\[5ex] Area = \sqrt{s(s - a)(s - b)(s - c)} \\[3ex] Area = \sqrt{28(28 - 7)(28 - 24)(28 - 25)} \\[3ex] Area = \sqrt{28 * 21 * 4 * 3} \\[3ex] Area = \sqrt{7056} \\[3ex] Area = 84\;cm^2 $

$ Area\:\:of\:\:rectangle \\[3ex] = (x + 5)(x + 7) \\[3ex] = x^2 + 7x + 5x + 35 \\[3ex] = x^2 + 12x + 35 \\[3ex] Area\:\:of\:\:square \\[3ex] = x^2 \\[3ex] Remaining\:\:area\:\:of\:\:rectangle \\[3ex] = x^2 + 12x + 35 - x^2 \\[3ex] = 12x + 35 $

Because the diagonal of the first rectangle is 6 times the diagonal of the second rectangle:
This implies that the sides of the first rectangle is also 6 times the sides of the second rectangle.
Compare the Pythagorean triple: 30 – 24 – 18 with 5 – 4 – 3 ...because the diagonal of a rectangle divides the rectangle into two right triangles
Let x = 5m be the length of the second rectangle
6x = 6(5) = 30m will be the length of the first rectangle



$ \underline{Pythagorean\;\;Theorem} \\[3ex] hyp^2 = leg^2 + leg^2 \\[3ex] 30^2 = 24^2 + 18^2 \\[3ex] 5^2 = 4^2 + 3^2 \\[3ex] 5 * 6 = 30 \\[3ex] 4 * 6 = 24 \\[3ex] 3 * 6 = 18 \\[3ex] $ So, the sides of the first rectangle is 6 times the size of the second rectangle
This also implies that the perimeter of the first rectangle is 6 times the perimeter of the second rectangle


$ Scale\;\;Factor \\[3ex] = \dfrac{Perimeter\;\;of\;\;1st\;\;Rectangle}{Perimeter\;\;of\;\;2nd\;\;Rectangle} \\[5ex] = \dfrac{30 + 24 + 18}{5 + 4 + 3} \\[5ex] = \dfrac{72}{12} \\[5ex] = \dfrac{6}{1} \\[5ex] $ Scale Factor, Perimeter Ratio, and Area Ratio of Similar Figures Theorem
Applies to all similar figures.
It states that for two similar figures, the ratio of the perimeters is the same as the scale factor; and the ratio of the areas is the ratio of the square of the scale factor.

$ Ratio\;\;of\;\;their\;\;areas \\[3ex] = Square\;\;of\;\;the\;\;Scale\;\;Factor \\[3ex] = \left(\dfrac{6}{1} \right)^2 \\[5ex] = \dfrac{6^2}{1^2} \\[5ex] = \dfrac{36}{1} \\[5ex] = 36:1 \\[3ex] $ OR

$ Ratio\;\;of\;\;their\;\;areas \\[3ex] = \dfrac{Area\;\;of\;\;1st\;\;Rectangle}{Area\;\;of\;\;2nd\;\;Rectangle} \\[5ex] = \dfrac{24 * 18}{4 * 3} \\[5ex] = \dfrac{6 * 6}{1 * 1} \\[5ex] = \dfrac{36}{1} \\[5ex] = 36:1 $

Because it is a right triangle, the hypotenuse is 17 meters (longest side)
This implies that the other two sides are the base and the height.
Any of those two sides could be either the base or the height. No information in the question suggets otherwise.

$ \underline{Short\:\:Method:\:\:Recommended\:\:for\:\:the\:\:ACT} \\[3ex] A = \dfrac{1}{2} * b * h \\[5ex] A = \dfrac{1}{2} * 8 * 15 \\[5ex] A = 4 * 15 \\[3ex] A = 60\:m^2 \\[5ex] \underline{Long\:\:Method:\:\:Hero's\:\:Formula} \\[3ex] Semiperimeter = \dfrac{height + base + hypotenuse}{2} \\[5ex] Semiperimeter = \dfrac{15 + 8 + 17}{2} = \dfrac{40}{2} = 20\:cm \\[5ex] first\:\:difference = Semiperimeter - height = 20 - 15 = 5\:cm \\[3ex] second\:\:difference = Semiperimeter - base = 20 - 8 = 12\:cm \\[3ex] third\:\:difference = Semiperimeter - hypotenuse = 20 - 17 = 3\:cm \\[3ex] Area = \sqrt{Semiperimeter * first\:\:difference * second\:\:difference * third\:\:difference} \\[3ex] Area = \sqrt{20 * 5 * 12 * 3} \\[3ex] Area = \sqrt{3600\:m^4} \\[3ex] Area = 60\:m^2 $

Find the area of each smaller rectangle.
Then add these areas.

$ (a.) \\[3ex] Area\;\;of\;\;the\;\;square \\[3ex] = (3 + 5) * (3 + 5) \\[3ex] = 8 * 8 \\[3ex] = 64\;units \\[5ex] (b.) \\[3ex] Area\;\;of\;\;the\;\;rectangle \\[3ex] = 5 * (5 + 8) \\[3ex] = 5 * 13 \\[3ex] = 65\;units \\[3ex] $ (c.) The lengths of the smaller sides of the triangles in the square and the rectangle are not equivalent.

(a.) Start at opposite vertices.
Use the same​ distance, say d from the vertices to make points on the opposite sides of the rectangle.
Connect these points.
Different trapezoids are formed with different values of d.

(b.) Congruent figures have the same shape and same size.
Values about figures can be​ equal, like the lengths of sides of a triangle or their​ areas, but two figures cannot be congruent unless all of the corresponding sides and angles are exactly the same.

(c.) No, because the sum of the measures of four acute angles is less than 360°.

(d.) Parallelogram ABCD is a rhombus.
Construction: Draw lines from A perpendicular to $\overline{BC}$ and $\overline{CD}$ and call the points of intersection E and F respectively.



Each piece of tape has a pair of parallel opposite​ sides, hence two sets of parallel sides are formed with two pieces of tape crossing.​
Therefore, they form Parallelogram ABCD.

A rhombus is a parallelogram with two adjacent congruent sides.
To prove that it is a​ rhombus:
Since the two pieces of tape have the same​ width, AE = AF.
Since ∠AEB and ∠AFD are both right​ angles, and ∠ABE and ∠ADF are alternate interior​ angles, and are​ congruent, ▵ABE and ▵ADF are congruent by AAS (Angle – Angle – Side).​
Therefore, AB = AD.

The reasons why it is not a square or rectangle:
When the pieces of tape cross each other​ perpendicularly, they form a​ square, which is also a rectangle.​
However, it cannot be determined from the figure that the pieces of tape are crossing each other perpendicularly.

(e.) Yes.
To construct such a​ kite, construct a right triangle in which no angle measures 45° and reflect it along the hypotenuse.

(f.) That is correct.
This is because if you know four parts then you have enough information to use at least one of​ these congruencies:
SSS (Side – Side –)
SAS (Side – Angle – Side)
ASA (Angle – Side – Angle)
AAS (Angle – Angle – Side).

(g.) That is incorrect.
This is because AAA (Angle – Angle – Angle) is insufficient to prove congruency.

(h.) That is not correct.
Although all squares have four congruent 90° ​angles, any congruent squares must also have the same side​ lengths, which is not true for all squares.

(i.) Extend the shorter side and construct the perpendicular line segment from a vertex not on the​ side, to the line containing the side.

(j.)

(k.)

(l.)

(m.) That is correct.
This is because if $\overline{AC} = \overline{BC}$, then the triangles are congruent by SSS (Side — Side — Side).

(n.)

(o.)

(p.)

(q.)

(r.)

(s.)

(t.)

(u.)

(v.)

(w.)

(x.)

(y.)

(z.)

The square feet of carpeting needed to cover the entire floor is the area of the composite shape.
The area of the composite shape is the sum of the area of section 1 and the area of section 2

$ Area\:\:of\:\:section\:1 = 12(12) = 144\:ft^2 \\[3ex] Area\:\:of\:\:section\:2 = 6(21) = 126\:ft^2 \\[3ex] Area\:\:of\:\:composite\:\:shape = 144 + 126 = 270\:ft^2 \\[3ex] $ 270 square feet of carpeting are needed to cover the entire floor.

(a.) The three angles in a triangle sum to 180°.
Therefore the 90° angle in a right triangle is the largest.
Since the hypotenuse is opposite the 90° ​angle, it must be the largest side.

(b.) The distance is the length of the perpendicular segment from the point to the line.
Any other segment will be a hypotenuse of a right triangle in which the perpendicular segment is a leg and hence shorter than the hypotenuse.

Construction: Draw an equilateral triangle ABC of side: x unit and perpendicular height: h unit from point B to meet line segment $\overline{AC}$ at point D
The perpendicular line bisects the base.



$ \underline{\triangle ABC} \\[3ex] Area = \dfrac{1}{2} * base * height \\[5ex] Area = \dfrac{1}{2} * x * h \\[5ex] $ We need to solve for x because the question wants us to determine the area in terms of h $ \underline{\triangle BDC} \\[3ex] hyp^2 = leg^2 + leg^2 ...Pythagorean\;\;Theorem \\[3ex] x^2 = h^2 + \left(\dfrac{x}{2}\right)^2 \\[5ex] x^2 = h^2 + \dfrac{x^2}{4} \\[5ex] Solve\;\;for\;\;x \\[3ex] \dfrac{x^2}{1} - \dfrac{x^2}{4} = h^2 \\[5ex] \dfrac{4x^2 - x^2}{4} = h^2 \\[5ex] \dfrac{3x^2}{4} = h^2 \\[5ex] 3x^2 = 4h^2 \\[3ex] x^2 = \dfrac{4h^2}{3} \\[5ex] x = \sqrt{\dfrac{4h^2}{3}} \\[5ex] x = \dfrac{\sqrt{4h^2}}{\sqrt{3}} \\[5ex] x = \dfrac{2h}{\sqrt{3}} \\[5ex] Rationalize\;\;the\;\;denominator \\[3ex] x = \dfrac{2h}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\[5ex] x = \dfrac{2h\sqrt{3}}{3} \\[5ex] $ Let us get back to finding the area in terms of h

$ Area = \dfrac{1}{2} * x * h \\[5ex] Area = \dfrac{1}{2} * \dfrac{2h\sqrt{3}}{3} * h \\[5ex] Area = \dfrac{h^2\sqrt{3}}{3} $

Area\:\;in\:\:square\:\:feet = 90\:feet * 30\:feet = 2700\:square\:feet \\[3ex] Convert\:\: 2700\:feet^2 \:\:to\:\: yard^2 \\[3ex] \underline{First\:\:Method:\:\:Unity\:\:Fraction\:\:Method} \\[3ex] 2700\:ft^2\:\:to\:\:yd^2 \\[3ex] 2700\:ft * ft * \dfrac{.....yd}{.....ft} * \dfrac{.....yd}{.....ft} \\[5ex] 2700\:ft * ft * \dfrac{1\:yd}{3\:ft} * \dfrac{1\:yd}{3\:ft} \\[5ex] = \dfrac{2700 * 1 * 1}{3 * 3} \\[5ex] = \dfrac{2700}{9} \\[5ex] = 300\:yd^2 \\[3ex] $ Second Method: Proportional Reasoning Method

Let $c$ = converted unit from $ft^2$ to $yd^2$

$ 3\:ft = 1\:yd \\[3ex] Square\:\:both\:\:sides \\[3ex] (3\:ft)^2 = (1\:yd)^2 \\[3ex] 3^2 * ft^2 = 1^2 * yd^2 \\[3ex] 9\:ft^2 = 1\:yd^2 \\[3ex] $

$ft^2$	$yd^2$
$9$	$1$
$2700$	$c$

$ \dfrac{c}{1} = \dfrac{2700}{9} \\[5ex] c = 300\:yd^2 \\[3ex] \therefore 2700\:ft^2 = 300\:yd^2 \\[3ex] $ The area of the rectangular stage is 300 square yards

Let us do some constructions to help us understand the problem.
Given: Parallel lines dividing $\overline{AB}$ into three congruent segments implies:



Draw a vertical perpendicular line through the vertex, B to bases: $\overline{FG}$ and $\overline{AC}$
This implies that:
Perpendicular Bisector Theorem: If a line drawn from the vertex of a triangle to a base is perpendicular to the base, then it bisects the base.
Also, because the parallel lines dividing $\overline{AB}$ into three congruent segments implies: $h_1 = h_2 = h_3$



Let us determine the perpendicular height, h



$ (3a)^2 = h^2 + \left(\dfrac{3a}{2}\right)^2 ...Pythagorean\;\;Theorem \\[5ex] 9a^2 = h^2 + \dfrac{9a^2}{4} \\[5ex] h^2 + \dfrac{9a^2}{4} = 9a^2 \\[5ex] h^2 = 9a^2 - \dfrac{9a^2}{4} \\[5ex] h^2 = \dfrac{36a^2 - 9a^2}{4} \\[5ex] h^2 = \dfrac{27a^2}{4} \\[5ex] h = \sqrt{\dfrac{27a^2}{4}} \\[5ex] h = \dfrac{\sqrt{27a^2}}{\sqrt{4}} \\[5ex] h = \dfrac{\sqrt{27} * \sqrt{a^2}}{2} \\[5ex] h = \dfrac{\sqrt{9 * 3} * a}{2} \\[5ex] h = \dfrac{3a\sqrt{3}}{2} \\[5ex] h = h_1 + h_2 + h_3 \\[3ex] But:\;\; h_1 = h_2 = h_3 ...congruent\;\;segments \\[3ex] \implies \\[3ex] h = h_1 + h_1 + h_1 \\[3ex] h = 3h_1 \\[3ex] 3h_1 = \dfrac{3a\sqrt{3}}{2} \\[5ex] h_1 = \dfrac{1}{3} * \dfrac{3a\sqrt{3}}{2} \\[5ex] h_1 = \dfrac{a\sqrt{3}}{2} \\[5ex] \therefore h_1 = h_2 = h_3 = \dfrac{a\sqrt{3}}{2} $

Let us calculate the area of the triangles

$ Area\;\;of\;\;\triangle DBE \\[3ex] = \dfrac{1}{2} * \left(\dfrac{a}{2} + \dfrac{a}{2}\right) * h_1 \\[5ex] = \dfrac{1}{2} * a * \dfrac{a\sqrt{3}}{2} \\[5ex] = \dfrac{a^2\sqrt{3}}{4} \\[5ex] Area\;\;of\;\;\triangle FBG \\[3ex] = \dfrac{1}{2} * (a + a) * (h_1 + h_2) \\[5ex] = \dfrac{1}{2} * 2a * \left(\dfrac{a\sqrt{3}}{2} + \dfrac{a\sqrt{3}}{2}\right) \\[5ex] = a * a\sqrt{3} \\[3ex] = a^2\sqrt{3} \\[5ex] Area\;\;of\;\;\triangle ABC \\[3ex] = \dfrac{1}{2} * \left(\dfrac{3a}{2} + \dfrac{3a}{2}\right) * (h_1 + h_2 + h_3) \\[5ex] = \dfrac{1}{2} * 3a * h \\[5ex] = \dfrac{1}{2} * 3a * \dfrac{3a\sqrt{3}}{2} \\[5ex] = \dfrac{9a^2\sqrt{3}}{4} $

Let us calculate the area of the trapezoids

$ Area\;\;of\;\;trapezoid\;DEGF \\[3ex] = \dfrac{1}{2} * \left[\left(\dfrac{a}{2} + \dfrac{a}{2}\right) + (a + a) \right] * h_2 \\[5ex] = \dfrac{1}{2} * (a + 2a) * \dfrac{a\sqrt{3}}{2} \\[5ex] = \dfrac{1}{2} * 3a * \dfrac{a\sqrt{3}}{2} \\[5ex] = \dfrac{3a^2\sqrt{3}}{4} \\[5ex] Area\;\;of\;\;trapezoid\;FGCA \\[3ex] = \dfrac{1}{2} * \left[(a + a) + \left(\dfrac{3a}{2} + \dfrac{3a}{2}\right)\right] * h_3 \\[5ex] = \dfrac{1}{2} * (2a + 3a) * \dfrac{a\sqrt{3}}{2} \\[5ex] = \dfrac{1}{2} * 5a * \dfrac{a\sqrt{3}}{2} \\[5ex] = \dfrac{5a^2\sqrt{3}}{4} \\[5ex] Area\;\;of\;\;trapezoid\;DECA \\[3ex] = \dfrac{1}{2} * \left[\left(\dfrac{a}{2} + \dfrac{a}{2}\right) + \left(\dfrac{3a}{2} + \dfrac{3a}{2}\right)\right] * (h_2 + h_3) \\[5ex] = \dfrac{1}{2} * (a + 3a) * \left(\dfrac{a\sqrt{3}}{2} + \dfrac{a\sqrt{3}}{2}\right) \\[5ex] = \dfrac{1}{2} * 4a * \dfrac{2a\sqrt{3}}{2} \\[5ex] = 2a * a\sqrt{3} \\[3ex] = 2a^2\sqrt{3} $

Let us now answer the questions that we were asked

$ (a.) \\[3ex] Ratio\;\;of\;\;the\;\;areas\;\;of\;\;\triangle DBE\;\;and\;\;Trapezoid\;DEGF \\[3ex] \dfrac{Area\;\;of\;\;\triangle DBE}{Area\;\;of\;\;Trapezoid\;DEGF} \\[5ex] = \dfrac{a^2\sqrt{3}}{4} \div \dfrac{3a^2\sqrt{3}}{4} \\[5ex] = \dfrac{a^2\sqrt{3}}{4} * \dfrac{4}{3a^2\sqrt{3}} \\[5ex] = \dfrac{1}{3} \\[5ex] (b.) \\[3ex] Ratio\;\;of\;\;the\;\;areas\;\;of\;\;\triangle DBE\;\;and\;\;Trapezoid\;FGCA \\[3ex] \dfrac{Area\;\;of\;\;\triangle DBE}{Area\;\;of\;\;Trapezoid\;FGCA} \\[5ex] = \dfrac{a^2\sqrt{3}}{4} \div \dfrac{5a^2\sqrt{3}}{4} \\[5ex] = \dfrac{a^2\sqrt{3}}{4} * \dfrac{4}{5a^2\sqrt{3}} \\[5ex] = \dfrac{1}{5} \\[5ex] (c.) \\[3ex] Ratio\;\;of\;\;the\;\;areas\;\;of\;\;Trapezoid\;DEGF\;\;and\;\;Trapezoid\;FGCA \\[3ex] \dfrac{Area\;\;of\;\;Trapezoid\;DEGF}{Area\;\;of\;\;Trapezoid\;FGCA} \\[5ex] = \dfrac{3a^2\sqrt{3}}{4} \div \dfrac{5a^2\sqrt{3}}{4} \\[5ex] = \dfrac{3a^2\sqrt{3}}{4} * \dfrac{4}{5a^2\sqrt{3}} \\[5ex] = \dfrac{3}{5} \\[5ex] (d.) \\[3ex] Ratio\;\;of\;\;the\;\;areas\;\;of\;\;Trapezoid\;DEGF\;\;and\;\;\triangle ABC \\[3ex] \dfrac{Area\;\;of\;\;Trapezoid\;DEGF}{Area\;\;of\;\;\triangle\;ABC} \\[5ex] = \dfrac{3a^2\sqrt{3}}{4} \div \dfrac{9a^2\sqrt{3}}{4} \\[5ex] = \dfrac{3a^2\sqrt{3}}{4} * \dfrac{4}{9a^2\sqrt{3}} \\[5ex] = \dfrac{1}{3} \\[5ex] (e.) \\[3ex] Ratio\;\;of\;\;the\;\;areas\;\;of\;\;Trapezoid\;FGCA\;\;and\;\;\triangle ABC \\[3ex] \dfrac{Area\;\;of\;\;Trapezoid\;FGCA}{Area\;\;of\;\;\triangle\;ABC} \\[5ex] = \dfrac{5a^2\sqrt{3}}{4} \div \dfrac{9a^2\sqrt{3}}{4} \\[5ex] = \dfrac{5a^2\sqrt{3}}{4} * \dfrac{4}{9a^2\sqrt{3}} \\[5ex] = \dfrac{5}{9} \\[5ex] (f.) \\[3ex] Ratio\;\;of\;\;the\;\;areas\;\;of\;\;\triangle ABC\;\;and\;\;Trapezoid\;DECA \\[3ex] \dfrac{Area\;\;of\;\;\triangle\;ABC}{Area\;\;of\;\;Trapezoid\;DECA} \\[5ex] = \dfrac{9a^2\sqrt{3}}{4} \div 2a^2\sqrt{3} \\[5ex] = \dfrac{9a^2\sqrt{3}}{4} * \dfrac{1}{2a^2\sqrt{3}} \\[5ex] = \dfrac{9}{8} \\[5ex] (g.) \\[3ex] Ratio\;\;of\;\;the\;\;areas\;\;of\;\;\triangle DBE\;\;and\;\;\triangle\;ABC \\[3ex] \dfrac{Area\;\;of\;\;\triangle\;DBE}{Area\;\;of\;\;\triangle\;ABC} \\[5ex] = \dfrac{a^2\sqrt{3}}{4} \div \dfrac{9a^2\sqrt{3}}{4} \\[5ex] = \dfrac{a^2\sqrt{3}}{4} * \dfrac{4}{9a^2\sqrt{3}} \\[5ex] = \dfrac{1}{9} $

The vertex of one square is at the center of the other square.
Do not let the titled square confuse you.
If you stand it directly, it has the same area as when it is titled because it's vertex is at the center of the other square.
This implies that half of the length of the titled square is at the center of the other square.
The 2 squares in the figure below have the same dimensions.
This implies that $\dfrac{1}{2} * 8 = 4\:inches$ of the titled square is at the center of the other square.

$ Area\:\:of\:\:the\:\:shaded\:\:region \\[3ex] = 4(4) \\[3ex] = 16\:in^2 \\[3ex] $ Alternatively, one can say that the area of the shaded region (region of the titled square) is one-quarter of the area of the first square.

$ Area\:\:of\:\:the\:\:first\:\:square = 8(8) = 64\:in^2 \\[3ex] Area\:\:of\:\:the\:\:shaded\:\:region = \dfrac{1}{4} * 64 \\[3ex] = 16\:in^2 $

For any triangle to be a right triangle: any of these three conditions must be satisfied:
(1.) The question clearly states that the triangle is a right triangle
and/or
(2.) There is a square box (which indicates a right angle or an angle of 90°) inside the triangle
and/or
(3.) The Pythagorean Theorem applies: the square of one side of the triangle is equal to the sum of the squares of the other two sides

(a.) Numbers (1.) and (2.) does not apply.
So, let us try Number (3.)

$ (\sqrt{5})^2 = 5 \\[3ex] (\sqrt{4})^2 = 4 \\[3ex] 1^2 = 1 \\[3ex] 5 = 4 + 1...Pythagorean\;\;Theorem \\[3ex] $ (a.) is a right triangle

(b.) Numbers (1.), (2.), and (3.) does not apply.
(b.) is NOT a right triangle.

(c.) Numbers (1.) and (2.) does not apply.
So, let us try Number (3.)

$ (\sqrt{19})^2 = 19 \\[3ex] (\sqrt{17})^2 = 17 \\[3ex] 1^2 = 1 \\[3ex] 19 \ne 17 + 1...NOT\;\;Pythagorean\;\;Theorem \\[3ex] $ (c.) is NOT a right triangle

(d.) Numbers (1.), (2.), and (3.) does not apply.
(d.) is NOT a right triangle.

$ \underline{Square\;ABCD} \\[3ex] Side = 15 \\[3ex] Area = Side^2 \\[3ex] A_{square} = 15^2 = 225 \\[3ex] \underline{Rectangle} \\[3ex] width = 10\;meters \\[3ex] length = p \\[3ex] Area = length * width \\[3ex] A_{rectangle} = p * 10 = 10p \\[3ex] \underline{Question} \\[3ex] A_{rectangle} = A_{square} \\[3ex] 10p = 225 \\[3ex] p = \dfrac{225}{10} \\[5ex] p = 22.5 \\[3ex] $ The length of the rectangle is 22.5 meters

$ Angle:\;\; \angle ABC = \angle EDC = 90^\circ ...Diagram \\[3ex] Side:\;\; \overline{BC} = \overline{DC} = 80m...Diagram \\[3ex] Angle:\;\; \angle BCA = \angle DCE ...Vertical\;\;\angle s\;\;are\;\;congruent \\[3ex] \therefore \angle ABC \cong \angle EDC ...ASA \\[3ex] \implies \\[3ex] \overline{AB} \cong \overline{ED} = 92m...CPCTC $

Construction: A point E on $\overline{AB}$ such that $\overline{CE} \parallel \overline{BC}$



ABCE is a parallelogram
This implies that:
$\overline{AB} \cong \overline{CE}$...opposite sides of a parallelogram are congruent
But: $\overline{AB} \cong \overline{CD}$ ... Diagram
Hence, $\overline{CE} \cong \overline{CD}$...because $\overline{AB} \cong \overline{AB}$
This implies that $\overline{AB} \cong \overline{CD} \cong \overline{CE}$
This also implies that:
$\triangle DEC$ is an isosceles triangle... $\overline{CE} \cong \overline{CD}$
$\angle DEC \cong \angle EDC$ ...base angles of isosceles $\triangle DEC$
$\angle DEC \cong \angle DAB$ ...corresponding angles are congruent
This implies that:
$\overline{DAB} \cong \overline{EDC}$...because $\overline{DEC} \cong \overline{DEC}$
$\angle EDC = \angle D$ ... Diagram
$\angle DAB = \angle A$ ... Diagram
Therefore, $\overline{A} \cong \overline{D}$