In rhombus ABCD, diagonals AC and BD intersect each other at point O.
If cosine of $∠ C A B$ is $0.6$ and $O B = 8 c m$ , find the lengths of the side and the diagonals of the rhombus.

side = 10 cm. diagonal AC = 12 cm and diagonal BD = 16 cm

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side = 15 cm. diagonal AC = 6 cm and diagonal BD = 16 cm

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side = 9 cm. diagonal AC = 14 cm and diagonal BD = 16 cm

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side = 11 cm. diagonal AC = 11 cm and diagonal BD = 16 cm

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Solution

The correct option is A side = 10 cm. diagonal AC = 12 cm and diagonal BD = 16 cm
$A B C D i s a r h o m b u s . ∴ I t s s i d e s = A B = B C = C D = D A . a l s o A C ⊥ B D ⟹ ∠ O B D = 1 r i g h t a n g l e ⟹ Δ A O B i s a r i g h t o n e w i t h A B a s h y p o t e n u s e . N o w cos θ = 0.6 ⟹ s i n θ = \sqrt{1 - {c o s}^{2} θ} = \sqrt{1 - {(0.6)}^{2}} c m = 0.8 c m S o sin θ = \frac{O B}{A B} ⟹ A B = \frac{O B}{sin θ} = \frac{8}{0.8} c m = 10 c m . A g a i n \frac{O A}{A B} = cos θ = 0.6 ⟹ O A = A B \times 0.6 c m = 10 \times 0.6 c m = 6 c m . B u t w e k n o w t h a t t h e d i a g o n a l s o f a r h o m b u s a r e r i g h t b i s e c t o r o f e a c h o t h e r . ∴ A C = 2 A O = 2 \times 6 c m = 12 c m a n d B D = 2 O B = 2 \times 8 c m = 16 c m . ∴ T h e s i d e o f t h e r h o m b u s = 10 c m . M a j o r d i a g o n a l = 16 c m . M i n o r d i a g o n a l = 12 c m . A n s - O p t i o n A .$

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Similar questions

= 10

= 16

ABCD is a rhombus whose diagonals intersect at 0. If AB = 10 cm, diagonal BD = 16 cm, find the length of diagonal AC.

Construct a rhombus ABCD, if ;

(i) AB = 4 cm and $∠ B = 120^{\circ}$ .

(ii) BC = 4.7 cm and $∠ B = 75^{\circ}$

(iii) CD = 5 cm and diagonal BD = 8.5 cm.

(iv) BC = 4.8 cm, and diagonal AC = 7 cm.

(v) diagonal AC = 6 cm and diagonal BD = 5.8 cm.

(vi) diagonal AC = 4.9 cm and diagonal BD = 6 cm.

(vii) diagonal AC = 6.6 cm and diagonal BD = 5.3 cm.

A B C D

O

A B = 10 c m

B D = 16 c m

A C

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