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Gwar [14]
3 years ago
14

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

Mathematics
1 answer:
vivado [14]3 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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<em><u>Solution:</u></em>

Let stamps be s and postcards be p

Given that,

The number of stamps was 4 more than twice the number of postcards

s = 4 + 2p -------- eqn 1

Jason bought both 41-cent stamps and 26-cent postcards and spent $10.28

41 cent = $ 0.41

26 cent = $ 0.26

Therefore,

s \times 0.41 + p \times 0.26 = 10.28

0.41s + 0.26p = 10.28 --------- eqn 2

Substitute eqn 1 in eqn 2

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1.08p = 10.28 - 1.64

1.08p = 8.64

Divide both sides by 1.08

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Substitute p = 8 in eqn 1

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Thus Jason bought 20 stamps and 8 post cards

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3 years ago
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Step-by-step explanation:

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then subtract 9 to the other side so 2n


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