Question

12. The diagonals of a rhombus ABCD intersect at O. If Z ADC = 120° and OD = 6 cm, OC = 8cm, find (i) OAD (ii) side AB (iii) perimeter of the thombus ABCD.​

Answer 1

Answer:

$\sf Find \angle OAD\\ \\ \sf \sf \implies\angle OAD=\angle OCD (isosceles \:triangle) \\ \\ \sf \angle \: OAD= \frac{180 - 120}{2} \\ \\ \sf \implies \sf \: \angle \: OAD = \frac{60}{2} = \red {30 {}^{0} } \\ \\ \sf \implies \sf \: hence \: \angle \: OAD = 30 {}^{0}$

$\sf \:\sf \implies BD^2=AB^2+AD^2-2(AB)(AD) cos(\angle DAB)\\ \\ \sf \implies \sf 12^2=x^2+x^2-2(x)(x) cos(60)\\ \\ \sf \implies \sf 144=2x^2-2x^2(0.5)\\ \\ \sf 144=x^2 \\ \\ \sf \implies \sf \: x = \sqrt{144} = \bold{ 12} \\ \\ \sf \implies \boxed{ \sf \red{AB = 12cm}}$

$\sf \:\sf \implies parimeter \: of \: tringle \: = 4 \times side \\ \\ \sf \implies \: parimeter \: of \: tringle \: = 4 \times \: 12 = \boxed{ \pink{48}}$

Hence, This is Answer

parimeter of triangle is 48

AB=12cm.

[tex]\angle OAD=30°[\tex]

Answer 2

Hii Mate..!

Step-by-step explanation:

ABCD is a rhombus.

AC and BD are diagonals of rhombus.

We know that, diagonals of a rhombus bisect each other perpendicularly .

∴ AC⊥BD

i.e.AO⊥BO

∴ ∠AOB=90 °