A triangle with the sides 6,8 and 10 . what kind of triangle is it ?

2 answers:

Taya2010 [7]2 years ago

6 0

It would be right-angled Triangle.

You can check for that, by putting the values into Pythagoras Theorem:
H² = P² + B²
10² = 8² + 6²
100 = 64 + 36
100 = 100
L.H.S = R.H.S

As it obeys Pythagoras theorem, It would be a right-angled triangle only, with two small arms (Base & Perpendicular) and a long arm, opposite to perpendicular (Hypotenuse)!!!

Hope it helped.

Mrrafil [7]2 years ago

3 0

Any time you see a triangle with 6, 8 and 10 sides, just know it is a right angle triangle.

That is because in a right angled triangle, the sum of the squares of the two smaller sides = square of the greatest side. This law is called the Pythagoras theorem.

Let us check.

6² + 8²

36 + 64

100

Square of the greatest side: 10² = 100

So therefore : 6² + 8² = 10²

So it is a right angled

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The polynomial p(x)=x^3-19x-30 has a known factor of (x+2)

riadik2000 [5.3K]

Answer:

(x+2) (x+3) (x-5)

Step-by-step explanation:

x³-19x-30 = (x+2) (x²+ax-15) ... x³=x*(1*x²) while -30= (2)*(-15)

x³ +<u> 0</u>*x² - 19x -30 = x³ + (<u>2+a</u>)x² + (2a-15)x -30

2+a = 0

a = -2

x³-19x-30 = (x+2) (x²-2x-15) = (x+2) (x+3) (x-5)

8 0

2 years ago

Read 2 more answers

A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)

Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

$36^6$

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

$5\cdot 36^5$

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

$\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}$

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

$P(A)=P(B)=\dfrac{5}{36}$

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

$axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8$