Determine what shape is formed for the given coordinates for abcd and then find the perimeter and area as an exact value and rounded to the nearest 10th

Square; 68.0; 289.0  

Step-by-step explanation:

1. Shape

Your shape is plotted in the graph below.

It looks like a square.

2. Proof

To prove that the object is a square , we must show that adjacent sides are perpendicular and that the diagonals are perpendicular.

(a) Sides

If the slopes of thee lines are negative reciprocals, the lines are perpendicular, and the angles are 90 °.

The formula for the slope (rate of change) m of a straight line is  

m = Δy/Δx = (y₂ -y₁)/(x₂ - x₁)

(i) AB

m = [-11 - (-19)]/(25 - 10) = (-11 +19)/15 =  8/15

(ii) BC

m = [-4 - (-19)]/(2 - 10) = (-4 +19)/(-8) = -15/(-8)= -15/8

The slopes of AB and BC are negative reciprocals, so ∠B = 90 °.

(iii) CD

m = [4 - (-4)]/(17 - 2) = (4 +4)/15 = 8/15

The slopes of BC and CD are negative reciprocals, so ∠C = 90 °.

(iv) AD

m = [4 - (-11)]/(17 - 25) = (4 +11)/(-8) = 15/(-8)= -15/8

The slopes of CD and AD are negative reciprocals, so ∠D = 90 °.

Also, the slopes of AD and AB are negative reciprocals, so ∠ = 90 °.

All four angles are 90°.

(b)Diagonals

AC: m = [-11 - (-4)]/(25- 2) = (-11 +4)/23 = -7/23

BD: m = [4 - (-19)]/(17 - 10) = (4 +19)/7 = 23/7

The slopes are negative reciprocals, so the diagonals are perpendicular.

The adjacent sides of quadrilateral ABCD have negative reciprocal slopes, 8/15 and -15/8, so, they form right angles and the slopes of the diagonals are negative reciprocals, -7/23 and 23/7, so they form right angles.  Therefore the quadrilateral is a square.

3. Perimeter

The formula for the perimeter of a square is

P = 4s

where s is the side length

AB² = 8² + 15²

  s² = 64 + 225

  s² = 289.0  

    s = 17.0

P = 4 × 17.0 = 68.0

4. Area

The formula for the area of a square is

A = s²

A = 289.0

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

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step-by-step explanation: