$A B C D$ is a rectangle. Taking $A D$ as a diameter, a semicircle $A X D$ is drawn which intersects the diagonal $B D$ at $X$ . If $A B = 12 c m$ , $A D = 9 c m$ , T he values of $B D$ and BX are

B D = 21 c m a n d B X = 5.4 c m .

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

B D = 9.6 c m a n d B X = 15 c m .

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

B D = 15 c m a n d B X = 9.6 c m .

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

B D = 25 c m a n d B X = 5.5 c m .

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

Open in App

Solution

The correct option is B $B D = 15 c m a n d B X = 9.6 c m .$
$G i v e n - A B C D i s a r e c t a n g l e a n d A X D i s a s e m i c i r c l e w i t h i n A B C D . T h e d i a g o n a l B D i n t e r s e c t s A X D a t X . A B = 12 c m, B D = 9 c m . T o f i n d o u t - t h e v a l u e s o f B D = ? B X = ? . S o l u t i o n - A c t u a l l y, B D i s a t a n g e n t t o t h e c i r c l e w i t h A D a s d i a m e t e r a n d D X i s a s e c a n t t o t h e s a m e c i r c l e . A B i n t e r s e c t s t h e g i v e n c i r c l e a t X . A l s o ∠ D A B = 90^{o} s i n c e A B C D i s a r e c t a n g l e . S o, b y P y t a g o r a s t h e o r e m, w e h a v e B D = \sqrt{{A D}^{2} + {A B}^{2}} c m = \sqrt{9^{2} + 12^{2}} c m = 15 c m . A g a i n, b y a p p l y i n g t a n g e n t - s e c a n t r u l e, w e h a v e B D \times B X = {A B}^{2} ⟹ B X = \frac{{A B}^{2}}{B D} = \frac{12^{2}}{15} c m = 9.6 c m . ∴ A n s - O p t i o n C .$

Suggest Corrections

Similar questions

Diagonal AC and BD of quadrilateral ABCD intersects each other at point O.

Prove that AB + BC + CD + AD < 2 (AC + BD).

⊥

⊥

□

⊥

⊥

Join BYJU'S Learning Program