MyQuestionIcon

1

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

Area of a Hexagon

All the three...

Question

All the three quadrilaterals ADEC, ABIH and BCGF are squares and $∠ A B C = 90^{0}$ . If the area of $A D E C = x^{2}$ and area of AHIB $= y^{2} (x^{2} > y^{2}),$ then find the area of BCGF.

A

(x + y) (x + y)

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

B

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

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

C

(x + y) (x - y)

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

D

(x - y) (x - y)

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

Open in App

Solution

The correct option is C $(x + y) (x - y)$
$T h e Δ A B C h a s ∠ B = 90^{o} . ∴ A B C i s a r i g h t Δ w i t h A C a s h y p o t e n u s e . ∵ a l l t h e t h r e e q u a d r i l a t e r a l s a r e s q u a r e s, ∴ w e h a v e {A C}^{2} = a r . A D E C = x^{2}, {A B}^{2} = a r . A B I H = y^{2}, {B C}^{2} = a r . A = a r . B C G F . A p p l y i n g P y t h a g o r a s T h e o r e m, w e g e t {B C}^{2} = {A C}^{2} - {A B}^{2} = x^{2} - y^{2} (x + y) (x - y) = a r . B C G F . T h i s i s a p o s i t i v e q u a n t i t y s i n c e x^{2} > y^{2} . A n s - a r . B C G F = (x + y) (x - y) .$

Suggest Corrections

thumbs-up

0

Similar questions

Q. All the three quadrilaterals ADEC, ABIH and BCGF are squares and

∠ A B C = 90^{\circ}

. If the area of ADEC

= x^{2}

and the area of AHIB

= y^{2} (x^{2} > y^{2})

, then the area of BCGF is:

x^{y} = y^{x},

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

is equal to:

b) If

\frac{x}{y} = \frac{6}{5}

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

x^{y} = y^{x},

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

is equal to:

b) If

\frac{x}{y} = \frac{6}{5}

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

Q. Assertion :Let a solution

y = y (x)

of the differential equation

x \sqrt{x^{2} - 1} d y - y \sqrt{y^{2} - 1} d x = 0

y (2) = 2 / \sqrt{3}

y (x) = sec ({sec}^{- 1} x - \frac{π}{6})

y (x)

\frac{1}{y} = \frac{2 \sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^{2}}}

Q.
Let a solution

y = y (x)

of the differential equation

x \sqrt{x^{2} - 1} d y - y \sqrt{y^{2} - 1} d x = 0

y (2) = \frac{2}{\sqrt{3}}

y (x) = sec (s e c^{- 1} x - \frac{π}{6})

y (x)

\frac{1}{y} = \frac{2 \sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^{2}}}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

How Big Is the Number?

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Area of a Hexagon

Standard VIII Mathematics

Join BYJU'S Learning Program

CrossIcon