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3 years ago

Congruent Triangle Combination

Mathematics

1 answer:

Svetlanka [38]3 years ago

3 0

Answer:

1) ΔABC ≅ ΔDBC by Side-Angle-Side (SAS)

2) ΔABC ≅ ΔCDB by Side-Side-Side (SSS)

3) ΔABC ≅ ΔDBE by Angle-Angle-Side (AAS)

4) m∠Y = 53°

5) y = 3

6) ΔABC ≅ ΔEDC by Side-Angle-Side (SAS)

7) ΔABC ≅ ΔADC by Angle-Angle-Side (AAS)

8) ΔABC ≅ ΔADC by Side-Angle-Side (SSS)

Step-by-step explanation:

1) Side AC ≅ Side BD ${}$ given

Side BC ≅ Side BC ${}$ reflexive property

∠DBC ≅ ∠ACB ${}$ given

Therefore ΔABC ≅ ΔDBC by Side-Angle-Side (SAS)

2) Side AC ≅ Side BD ${}$ given

Side DC ≅ Side BA ${}$ given

Side BC ≅ Side BC ${}$ reflexive property

Therefore, ΔABC ≅ ΔCDB by Side-Side-Side (SSS)

3) ∠BAC ≅ ∠DEB ${}$ given

Side BC ≅ Side BD ${}$ given

∠ABC ≅ ∠DBE ${}$ vertically opposite angles

Therefore, ΔABC ≅ ΔDBE by Angle-Angle-Side (AAS)

4) ΔXYZ ≅ ΔABC ${}$ given

m∠A = 44°, and m∠C = 83°

m∠B = 180 - (44 + 83) = 53° ${}$ Sum of angles in a triangle

m∠Y = m∠B = 53° ${}$ Corresponding Parts of Congruent Triangles are Congruent (CPCTC)

5) Given that Triangle GHK is congruent to Triangle TZK

GH ≅ XT, KG ≅ KT, and HK ≅ ZK by the definition of congruency

Therefore, GH = 4 = XT = 3y - 2

4 = 3y - 2

3y - 2 = 4

3y = 4 + 2 = 6

y = 6/2 = 3

y = 3

6) Side BC ≅ Side EC ${}$ given

Side AC ≅ Side DC ${}$ given

∠ACB ≅ ∠DCE ${}$ vertically opposite angles

Therefore, ΔABC ≅ ΔEDC by Side-Angle-Side (SAS)

7) ∠BAC ≅ ∠DAC ${}$ given

∠ADC ≅ ∠ABC ${}$ given

Side AC ≅ Side AC ${}$ reflexive property

Therefore, ΔABC ≅ ΔADC by Angle-Angle-Side (AAS)

8) Side AD ≅ Side BC ${}$ given

Side AC ≅ Side AC ${}$ reflexive property

Side AB ≅ Side DC ${}$ given that the included angle between the two legs side AD and side AC and side AC and side BC are both acute, the third sides side AB and side DC are equal

Therefore, ΔABC ≅ ΔADC by Side-Angle-Side (SSS)

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A bee flies at 9 feet per second directly to a flower bed from its hive. The bee stays at the flowerbed for 15 minutes, and then

Ivan

Complete question is;

A bee flies at 9 feet per second directly to a flower bed from its hive. The bee stays at the flowerbed for 15 minutes, and then flies directly back to the hive at 6 feet per second. It is away from the hive for a total of 18 minutes.

a. What equation can you use to find the distance of the flowerbed from the hive?

b. How far is the flowerbed from the hive?

Answer:

A) (d/9) + 900 + (d/6) = 1080

B) d = 648 ft

Step-by-step explanation:

A) We are told that A bee flies at 9 feet per second directly to a flower bed from its hive.

Time taken is given as;

t = distance/speed

If distance is d from the hive to the flower bird.

Then, t1 = d/9

The bee stays at the flowerbed for 15 minutes. Thus, time spent at the flowerbird in seconds is; t2 = 15 × 60 = 900 seconds.

We are told that the bee flew directly back to the hive at 6 feet per second.

Thus,time is;

t3 = d/6

Since total time spent away from hive is 18 minutes, then converting to seconds, we have; t = 18 × 60 = 1080 s

Thus;

(d/9) + 900 + (d/6) = 1080

B) Since we have derived;

(d/9) + 900 + (d/6) = 1080

Then we can find d.

Multiply through by 18 to get;

2d + (900 × 18) + 3d = 1080 × 18

2d + 16200 + 3d = 19440

5d = 19440 - 16200

5d = 3240

d = 3240/5

d = 648 ft

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kodGreya [7K]

Hello,

g(x)=x²+3
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Explainations:

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You have 2+2 in each 2 there are 1's so

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Oksana_A [137]

<h2>Hello!</h2>

The answers are:

$FirstCarSpeed=41mph\\SecondCarSpeed=55mph$

To calculate the speed of the cars, we need to write two equations, one for each car, in order to create a relation between the two speeds and be able to calculate one in function of the other.

So,

Tet be the first car speed "x" and the second car speed "y", writing the equations we have:

For the first car:

$x_{FirstCar}=x_o+v*t$

For the second car:

We know that the speed of the second car is the speed of the first car plus 14 mph, so:

$x_{SecondCar}=x_o+(v+14mph)*t$

Now, from the statement that both cars met after 2 hours and 45 minutes, and the distance between to cover (between A and B) is 264 miles, so, we can calculate the relative speed between them:

If the cars are moving towards each other the relative speed will be:

$RelativeSpeed=FirstCarSpeed-(-SecondCarspeed)\\\\RelativeSpeed=x-(-x-14mph)=2x+14mph$

Then, since we know that they covered a combined distance equal to 264 miles in 2 hours + 45 minutes, we have:

$2hours+45minutes=120minutes+45minutes=165minutes\\\\\frac{165minutes*1hour}{60minutes}=2.75hours$

Writing the equation:

$264miles=(2x+14mph)*t\\\\264miles=(2x+14mph)*2.75hours\\\\2x+14mph=\frac{264miles}{2.75hours}\\\\2x=96mph-14mph\\\\x=\frac{82mph}{2}=41mph$

So, the speed of the first car is equal to 41 mph.

Now, for the second car we have that:

$SecondCarSpeed=FirstCarSpeed+14mph\\\\SecondCarSpeed=41mph+14mph=55mph$

We have that the speed of the second car is equal to 55 mph.

Hence, the answers are:

$FirstCarSpeed=41mph\\SecondCarSpeed=55mph$

Have a nice day!

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