You visited us 1 times! Enjoying our articles? Unlock Full Access!

Distance between Two Points Using Pythagoras Theorem

In ABC,∠ AB...

Question

In $△ A B C, ∠ A B C = 90^{o}$ . If $A C = (x + y)$ and $B C = (x - y)$ , then the length of $A B$ is:

x^{2} - y^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 x y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 \sqrt{x y}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x^{2} + y^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2 \sqrt{x y}$
Given, $∠ A B C = 90^{\circ},$
Also, $A C = (x + y)$ and $B C = (x - y)$
We need to find the length of $A B$ .
Now as $∠ A B C = 90^{\circ},$ at $B$ we can use Pythagoras theorem
$∴$ $A B^{2} + B C^{2} = A C^{2}$
$⟹$ $A B^{2} = A C^{2} - B C^{2}$
$⟹$ $A B^{2} = (x + y)^{2} - (x - y)^{2}$
Using the formula for $(a + b)^{2}$ and $(a - b)^{2}$
$⟹$ $A B^{2} = x^{2} + 2 x y + y^{2} - (x^{2} - 2 x y + y^{2})$
By cancelling the like terms we get,
$A B^{2} = 4 x y$
Taking square root on both the sides we get,
$A B = 2 \sqrt{x y}$ .
Hence, the answer is C,

Suggest Corrections

Similar questions

A (- 1, - 1)

x^{2} + y^{2} = 1

B, C .

P

A B C

A B, A P, A C

A . P

P

x \neq \pm y

\frac{x^{2} - y^{2}}{x^{2} + 2 x y + y^{2}} : \frac{x - y}{x + y}

x^{2} + y^{2} - x y = 3

y - x = 1

\frac{x y}{x^{2} + y^{2}}

u = {tan}^{- 1} (\frac{x^{3} + y^{3}}{x - y})

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = ?

Z = x {tan}^{- 1} (\frac{y}{x}) + x e^{x / y}

x^{2} \frac{\partial^{2} z}{\partial x^{2}} + 2 x y \frac{\partial^{2} z}{\partial x \partial y} + y^{2} \frac{\partial^{2} z}{\partial y^{2}} = ?

Join BYJU'S Learning Program