Question

A SEGMENT BC OF LENGTH 4m IS DIVIDED BY POINT D INTO TWO PARTS SUCH THAT BD^2=DC × BC .FIND THE LENGTH OF DC​

Answer 1

Answer:

$DC=6-2\sqrt5\:m$

Step-by-step explanation:

Given:

BC = 4 m

Let BD = x

Then DC=4-x

$BD^2=DC*BC$

$\implies\:x^2=(4-x)*4$

$\implies\:x^2=16-4x$

$\implies\:x^2+4x-16=0$

To find x, we solve this equation

$\implies\:x^2+4x-16=0$

$x^2+4x=16$

$x^2+4x+4=16+4$

$(x+2)^2=20$

$x+2=$±$2\sqrt5$

$x=-2$±$2\sqrt5$

But x cannot be negative

$x=-2+2\sqrt5$

$x=2\sqrt5-2$

Now,

$DC=4-(2\sqrt5-2)$

$DC=4-2\sqrt5+2$

$DC=6-2\sqrt5\:m$

Answer 2

Answer:

Answer:

DC=6-2\sqrt5\:mDC=6−2

5

m

Step-by-step explanation:

Given:

BC = 4 m

Let BD = x

Then DC=4-x

BD^2=DC*BCBD

2

=DC∗BC

\implies\:x^2=(4-x)*4⟹x

2

=(4−x)∗4

\implies\:x^2=16-4x⟹x

2

=16−4x

\implies\:x^2+4x-16=0⟹x

2

+4x−16=0

To find x, we solve this equation

\implies\:x^2+4x-16=0⟹x

2

+4x−16=0

x^2+4x=16x

2

+4x=16

x^2+4x+4=16+4x

2

+4x+4=16+4

(x+2)^2=20(x+2)

2

=20

x+2=x+2= ±2\sqrt52

5

x=-2x=−2 ±2\sqrt52

5

But x cannot be negative

x=-2+2\sqrt5x=−2+2

5

x=2\sqrt5-2x=2

5

−2

Now,

DC=4-(2\sqrt5-2)DC=4−(2

5

−2)

DC=4-2\sqrt5+2DC=4−2

5 +2

DC=6-2\sqrt5\:mDC=6−2

5 m