In the following figure, the large quadrilateral is a square. What is the area of the shaded region in terms of x and y?

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

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

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

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

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

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

\frac{x^{2} + y^{2}}{4}

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

\frac{(x + y)^{2}}{4}

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

Open in App

Solution

The correct option is A $\frac{x^{2} + y^{2}}{2}$

$A B = \sqrt{x^{2} + y^{2}}$
$A C = \sqrt{x^{2} + y^{2}}$
$tan θ_{1} = \frac{y}{x}, cos θ_{1} = \frac{x}{\sqrt{x^{2} + y}}$
$tan θ_{2} = \frac{y}{x}, cos θ_{2} = \frac{y}{\sqrt{x^{2} + y}}$
$cos (180 - {(θ}_{1} + θ_{2})) = - cos {(θ}_{1} + θ_{2})$
$= sin θ_{1} sin θ_{2} - cos θ_{1} cos θ_{2} = y x$
$sin (180 - {(θ}_{1} + θ_{2})) = sin {(θ}_{1} + θ_{2})$
$= sin θ_{1} cos θ_{2} + cos θ_{1} sin θ_{2}$
$= \frac{x^{2} + y^{2}}{x^{2} + y^{2}} = 1$ \\ $θ_{1} + θ_{2} = 90$ \\ $△ A B C = \frac{1}{2} (x^{2} + y^{2}) sin 90$
$△ A B C = \frac{1}{2} (x^{2} + y^{2})$

Suggest Corrections

Similar questions

x^{2} + y^{2} = 4

(x - 2) + y^{2} = 4

x

y

\frac{x^{2}}{y^{2}} + \frac{y^{2}}{x^{2}} - \frac{1}{i} (\frac{x}{y} - \frac{y}{x}) - \frac{9}{4}

x = 3 sec θ - 2

y = 3 tan θ + 2

x, y

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

Join BYJU'S Learning Program