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tatuchka [14]
3 years ago
5

Which statement accurately describes the two congruent triangles? Triangles B A E and B C D where point B is an apex of both tri

angles. Line segments A B and B C are the same length, line segments C D and A E are the same length, and line segments B E and B D are the same length. Angles A E B and B D C have the same measurement, angles D C B and B A E have the same measurement, and angles A B E and C B D have the same measurement. Question 1 options: triangle A B E is congruent to triangle B C D triangle E A B is congruent to triangle D B C triangle A B E is congruent to triangle C B D triangle A E B is congruent to triangle C B D
Mathematics
1 answer:
miss Akunina [59]3 years ago
4 0

Answer:

Triangle ABE is congruent to triangle CBD.

Step-by-step explanation:

Given two congruent triangles BAE and BCD where B is an apex of both triangles.

Given AB=BC, CD=AE, BE=BD

& also ∠AEB=∠BDC, ∠DCB=∠BAE, ∠ABE=∠CBD

By CPCT i.e. Corresponding parts of corresponding triangles

Correspondimg sides and angles area in the same position or spot in two different triangles.

Hence, Triangle ABE is congruent to triangle CBD.


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equivalent ratios.

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3 years ago
What is the solution of
kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

\sqrt{2x+4}-\sqrt{x}=2

Isolating √(2x+4): Addind √x both sides of the equation:

\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}

Squaring both sides of the equation:

(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}

Simplifying on the left side, and applying on the right side the formula:

(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}

2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}

Squaring both sides of the equation:

(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

x^{2}-16x=16x-16x\\ x^{2}-16x=0

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2

x=0 is a solution


2) x=16

\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2

x=16 is a solution


Then the solutions are x=0 and x=16


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