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Gwar [14]
4 years ago
14

Prove that a triangle with the sides a - 1 cm to root under a ​

Mathematics
1 answer:
vivado [14]4 years ago
4 0

Question is Incomplete, Complete question is given below.

Prove that a triangle with the sides (a − 1) cm, 2√a cm and (a + 1) cm is a right angled triangle.

Answer:

∆ABC is right angled triangle with right angle at B.

Step-by-step explanation:

Given : Triangle having sides (a - 1) cm, 2√a and (a + 1) cm.  

We need to prove that triangle is the right angled triangle.

Let the triangle be denoted by Δ ABC with side as;

AB = (a - 1) cm

BC = (2√ a) cm

CA = (a + 1) cm  

Hence,

{AB}^2 = (a -1)^2 

Now We know that

(a- b)^2 = a^2+b^2 - 2ab

So;

{AB}^2= a^2 + 1^2 -2\times a \times1

{AB}^2 = a^2 + 1 -2a

Now;

{BC}^2 = (2\sqrt{a})^2= 4a

Also;

{CA}^2 = (a + 1)^2

Now We know that

(a+ b)^2 = a^2+b^2 + 2ab

{CA}^2= a^2 + 1^2 +2\times a \times1

{CA}^2 = a^2 + 1 +2a

{CA}^2 = AB^2 + BC^2

[By Pythagoras theorem]

a^2 + 1 +2a = a^2 + 1 - 2a + 4a\\\\a^2 + 1 +2a= a^2 + 1 +2a

Hence, {CA}^2 = AB^2 + BC^2

Now In right angled triangle the sum of square of two sides of triangle is equal to square of the third side.

This proves that ∆ABC is right angled triangle with right angle at B.

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masha68 [24]

Answer:

10.33

Step-by-step explanation:

\frac{31}{3} is equal to 31 divided by 3, which is 10.33333333333333333...

Rounded to the nearest hundredth we get 10.33.

Hope this helps!

5 0
4 years ago
I am offering Brainliest, A like on your answer, and a 5 star review. This should be simple, but I just don't understand.. Pleas
Oksana_A [137]

Answer:

(-4,2)

Step-by-step explanation:

Just had the test for this thing lol.

so what you do is find the point Y.

then subtract 1 from it's x axis then add 2 to it's y axis.

work

before: (-3,0)

-1. +2

after. (-4,2)

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6 0
3 years ago
Two possible solutions of √11-2x=√x^2+4x+4 are –7 and 1. Which statement is true? A) Only x = –7 is an extraneous solution.
11111nata11111 [884]

Answer:

Option D)Neither solution is extraneous.

Step-by-step explanation:

we have

\sqrt{11-2x}=\sqrt{x^{2}+4x+4}

we know that

two possible solutions are x=-7 and x=1

<u><em>Verify each solution</em></u>

Substitute each value of x in the expression above and interpret the results

1) For x=-7

\sqrt{11-2(-7)}=\sqrt{-7^{2}+4(-7)+4}

\sqrt{25}=\sqrt{25}

5=5 ----> is true

therefore

x=-7 is not a an extraneous solution

2) For x=1

\sqrt{11-2(1)}=\sqrt{1^{2}+4(1)+4}

\sqrt{9}=\sqrt{9}

3=3 ----> is true

therefore

x=1 is not a an extraneous solution

therefore

Neither solution is extraneous

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