Question

The diagonals of a rhombus ABCD intersect at O. AO=3cm, BO=4cm,then find the lengthof BC

Answer 1

$\huge{\pink{\underline{\ulcorner{\star\: Solution \: \star}\urcorner}}}$

Given : ABCD is a Rhombus whose diagonal's bisect each other at O.

AO = 3 cm , BO = 4 cm

To find : BC a side of Rhombus

Solution :

Diagonal's of Rhombus bisect each other at 90°.

$\therefore \: \angle{AOB} \: = \: 90\degree$

Apply Pythagoras in ∆ AOB

AB² = AO² + BO²

AB = √{3²+4²}

AB = √{9 + 16}

AB = √{25}

AB = 5 cm

$Statement \colon\: The\: sides\: of\: rhombus\: are\: always \:equal$

AB = BC = CD = DA

5 cm = BC = CD = DA

$\red{\underline{Answer\: \colon }}$

$\bold{\red{\boxed{ BC\: = \: 5 \: cm}}}$

Answer 2

HERE IS YOUR ANSWER
Given : ABCD is a Rhombus whose diagonal's bisect each other at O.

AO = 3 cm , BO = 4 cm

To find : BC a side of Rhombus

Solution :

Diagonal's of Rhombus bisect each other at 90°.

\therefore \: \angle{AOB} \: = \: 90\degree∴∠AOB=90° 

Apply Pythagoras in ∆ AOB 

AB² = AO² + BO²

AB = √{3²+4²}

AB = √{9 + 16}

AB = √{25}

AB = 5 cm

Statement the sides RHOMBUS are equal 

AB = BC = CD = DA 

5 cm = BC = CD = DA

ANSWER IS 5cm

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