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,
Now We know that
So;
Now;
Also;
Now We know that
[By Pythagoras theorem]
Hence,
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.
Answer:
(6, 9 ) and r = 3
Step-by-step explanation:
The equation of a circle in standard form is
(x - h)² + (y - k)² = r²
where (h, k) are the coordinates of the centre and r is the radius
Given
x² + y² - 12x - 18y + 108 = 0
Rearrange the x- terms and the y- terms together and subtract 108 from both sides, that is
x² - 12x + y² - 18y = - 108
To obtain standard form use the method of completing the square
add ( half the coefficient of the x and y terms )² to both sides
x² + 2(- 6)x + 36 + y² + 2(- 9)y + 81 = - 108 + 36 + 81
(x - 6)² + (y - 9)² = 9 ← in standard form
with centre = (6, 9 ) and r = 
= 3
Here's my interpretation: 4 whole km, plus zero tenths of a km, plus 3 hundredths of a km, plus 5 thousandths of a km.
The origin is 0,0 the y axis is the line going up, the x axis is the line going like this — the top right is quadrant 1 the one below it is quadrant 2 the one beside that is quadrant 3 and the one above that is quadrant 4
Answer:
6, 0, 10
b
Step-by-step explanation:
