Question

In Fig ABCD is a rhombus. Diagonals AC and BD intersect at O. E and F are mid points of AO and BO respectively. If AC = 16cm and BD = 12cm then EF is:

Answer 1

Answer:

EF=5cm

Step-by-step explanation:

AC=16 cm(given)

BD=12 cm(given)

OA=OC(diagonals of a parallelogram bisect each other)

OD=OB(diagonals of a parallelogram bisect each other)

OE=EA(E is the midpoint of OA)

OF=FB(F is the midpoint of OB)

So,

(EF)^2=(OE)^2+(OF)^2

EF^2=(4)^2+(3)^2

EF^2=16+9

EF=25

EF=5 cm

HOPE IT HELPS YOU

Answer 2

$\huge\sf\pink{Answer}$

☞ Your answer is 5 cm

━━━━━━━━━━━━━

$\huge\sf\blue{Given}$

✭ ABCD is a Rhombus

✭ AC & BD are the diagonals which meat at O

✭ E & F are the midpoints

✭ AC is 16 cm & BD is 12 cm

━━━━━━━━━━━━━

$\huge\sf\gray{To \:Find}$

◈ Length of EF?

━━━━━━━━━━━━━

$\huge\sf\purple{Steps}$

$\large\underline{\underline{\sf Concept}}$

We know that in a Rhombus the diagonals bisect at 90° so then we shall use pythagoras theorem to find the length of EF

━━━━━━━━━━

In a Rhombus we know that Diagonals bisect each other,

➝ $\sf AC = 16 \ cm$

➝ $\sf AC = 2AO$

➝ $\sf \dfrac{AC}{2} = OA$

➝ $\sf \dfrac{16}{2} = OA$

➝ $\sf OA = 8 \ cm$

And further E is the mid point of OA

➝ $\sf OA = 2OE$

➝ $\sf 8 = 2OE$

➝ $\sf \dfrac{8}{2} = OE$

➝ $\sf \green{OE = 4 \ cm}$

Similarly,

➳ $\sf BD = 12 \ cm$

➳ $\sf BD = 2OB$

➳ $\sf 12 = 2OB$

➳ $\sf \dfrac{12}{2} = OB$

➳ $\sf OB = 6 \ cm$

As F is the mid point of OB

➳ $\sf OB = 2OF$

➳ $\sf 6 = 2OF$

➳ $\sf \dfrac{6}{2} = OF$

➳ $\sf \red{OF=3 \ cm}$

Now that we know two sides we shall use Pythagoras theorem to find the third side, that is,

$\underline{\boxed{\sf OE^2 + OF^2 = EF^2}}$

»» $\sf 4^2+3^2 = EF^2$

»» $\sf 16+9 = EF^2$

»» $\sf 25 = EF^2$

»» $\sf \sqrt{25} = EF$

»» $\sf \orange{EF = 5 \ cm}$

$\sf\star\: Diagram \:\star$

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(1,3)\qbezier(3,0)(3,0)(4,3)\qbezier(1,3)(1,3)(4,3)\qbezier(3,0)(0,0)(0,0)\qbezier(0,0)(0,0)(4,3)\qbezier(1,3)(3,0)(3,0)\qbezier(0.9,0.5)(0,0.5)(2.68,0.5)\put(1.3,0.59){ \bf 5 \ cm }\put(1.9,1.7){ \bf O }\put(-0.3,-0.2){ \bf A }\put(3.1,-0.2){ \bf B }\put(4,3){ \bf C }\put(0.7,3){ \bf D }\end{picture}$

━━━━━━━━━━━━━━━━━━