Question

It's given that:-

x+y+z = 12
x²+y²+z² = 53
and y = 5

Then show that x-z = (+-)√7

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Answer 1

Question:-

It's given that:- x + y + z = 12 , x² + y² + z² = 53 and y = 5, then show that x - z = ±√7

Required Answer:-

Given:-

• x + y + z = 12 ,
• x² + y² + z² = 53
• y = 5

To Do:-

• Show that:- x - z = ±√7

Solution:-

• ★First we have to find the value of x + z and xy to find the value of x - y through the formula of (x - y)²★

• x + y + z = 12 ----------------------(1)
• x² + y² + z² = 53 ----------------------(2)

Putting y = 5 in equation 1:-

• x + 5 + z = 12
• => x + z = 12 - 5
• => x + z = 7

Again putting the value of y = 5 in equation 2:-

• x² + 5² + z² = 53
• => x² + z² = 53 - 25
• => x² + z² = 28

We know that:-

• ✥ (a + b)² = a² + b² + 2ab

We have:-

• x + z = 7
• x² + z² = 28

Applying the formula:-

• (x + z)² = x² + z² + 2xz

Substituting the values:-

• => (7)² = 28 + 2xz
• => 2xz + 28 = 49
• => 2xz = 49 - 28
• => 2xz = 21
• => xz = 21/2

Now, let's find the value of x - z

We know that:-

• ✥ (a - b)² = (a + b)² - 4ab

We have:-

• x + z = 7
• xz = 21/2

Applying the formula:-

• (x - z)² = (x + z)² - 4xz
• => (x - z)² = (7)² - 4 x 21/2
• => (x - z)² = 49 - 42
• => (x - z)² = 7
• => x - z = ±√7 [Showed!]

Answer 2

$ABC \: is \: an \: isosceles \: triangle \: inscribed \: in \: a \: circle. \: If AB = AC = 12 \sqrt{5} \: cm \: and \: BC = 24 cm, find \: the \: radius \: of \: the \: circle.$

I need correct answer .. ( no need to solve)