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2 years ago

In this figure triangle BAE is congruent to triangle DEA. Which statement is true by CPCTC?

Mathematics

2 answers:

Hatshy [7]2 years ago

8 0

Answer: ∠BAF≅∠DEF

Step-by-step explanation:

In the given figure triangle BAE is congruent to triangle DEA i.e. ΔBAE ≅ ΔDEA.

We know that if two triangles are congruent then their corresponding angles are congruent.

Since, ΔBAE ≅ ΔDEA, the by CPCTC [congruent parts of congruent triangles are congruent]

∠BAE≅∠DEA

Since F is a point on AE

So, ∠BAE≅∠DEA

⇒ ∠BAF≅∠DEF

Vsevolod [243]2 years ago

5 0

The correct answer is the last option

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Unit 7 polygons & quadrilaterals homework 3: rectangles Gina Wilson answer key

Irina-Kira [14]

Answer:

In the image attached you can find the Unit 7 homework.

We need to findt he missing measures of each figure.

Notice that the first figure is a rectangle, which means opposite sides are congruent so,

VY = 19

WX = 19

YX = 31

VW = 31

To find the diagonals we need to use Pythagorean's Theorem, where the diagonals are hypothenuses.

$VX^{2}=19^{2}+31^{2}\\ VX=\sqrt{361+961}=\sqrt{1322} \\VX \approx 36.36$

Also, $YW \approx 36.36$ , beacuse rectangles have congruent diagonals, which intercect equally.

That means, $ZX = \frac{VX}{2} \approx \frac{36.36}{2}\approx 18.18$

Figure number two is also a rectangle.

If GH = 14, that means diagonal GE = 28, because diagonals intersect in equal parts.

Now, GF = 11, because rectangles have opposite sides congruent.

DF = 28, because in a reactangle, diagonals are congruent.

HF = 14, because its half of a diagonal.

To find side DG, we need to use Pythagorean's Theorem, where GE is hypothenuse

$GE ^{2}=11^{2}+DG^{2}\\28^{2}-11^{2}=DG^{2}\\DG=\sqrt{784-121}=\sqrt{663}\\ DG \approx 25.75$

This figure is also a rectangle, which means all four interior angles are right, that is, equal to 90°, which means angle 11 and the 59° angle are complementary, so

$\angle 11 +59\°=90\°\\\angle 11=90\°-59\°\\\angle 11=31\°$

Now, angles 11 and 4 are alternate interior angles which are congruent, because a rectangle has opposite congruent and parallel sides.

$\angle 4 = 31\°$

Which means $\angle 3 = 59\°$ , beacuse it's the complement for angle 4.

Now, $\angle 6 = 59\°$ , because it's a base angle of a isosceles triangle. Remember that in a rectangle, diagonals are congruent, and they intersect equally, which creates isosceles triangles.

$\angle 9=180-59-59=62$ , by interior angles theorem.

$\angle 8 =62$ , by vertical angles theorem.

$\angle 10 = 180- \angle 9=180-62=118\°$ , by supplementary angles.

$\angle 7 = 118\°$ , by vertical angles theorem.

$\angle 5=90-59=31$ , by complementary angles.

$\angle 2 = \angle 5 = 31\°$ , by alternate interior angles.

$\angle 1 = 59\°$ , by complementary angles.

$m\angle BCD=90\°$ , because it's one of the four interior angles of a rectangle, which by deifnition are equal to 90°.

$m\angle ABD = 6\° = m\angle BDC$ , by alternate interior angles and by given. $m\angle CBE=90-6=84$ , by complementary angles.

$m\angle ADE=90-6=84$ , by complementary angles.

$m\angle AEB=180-6-6=168\°$ , by interior angles theorem.

$m\angle DEA=180-168=12$ , by supplementary angles.

$m\angle JMK=180-126=54$ , by supplementary angles.

$m\angle JKH=\frac{180-54}{2}=\frac{126}{2}=63$ , by interior angles theorem, and by isosceles triangle theorem.

$m\angle HLK=90\°$ , by definition of rectangle.

$m\angle HJL=\frac{180-126}{2}=27$ , by interior angles theorem, and by isosceles triangle theorem.

$m\angle LHK=90-27=63$ , by complementary angles.

$m\angle = JLK= m\angle HJL=27$ , by alternate interior angles.

The figure is a rectangle, which means its opposite sides are equal, so

$WZ=XY\\7x-6=3x+14\\7x-3x=14+6\\4x=20\\x=\frac{20}{4}\\ x=5$

Then, we replace this value in the expression of side WZ

$WZ=7x-6=7(5)-6=35-6=29$

Therefore, side WZ is 29 units long.

We know that the diagonals of a rectangle are congruent, so

$SQ=PR\\11x-26=5x+28\\11x-5x=28+26\\6x=54\\x=\frac{54}{6}\\ x=9$

Then,

$PR=5x+28=5(9)+28=45+28=73$

Therefore, side PR is 73 units long.

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Hey there!

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