Question

Hii ♡

A semi-circle is constructed inside the square with a diameter of one unit.
The yellow line starts at the top corner, and is tangent to the semi circle.What is the length of the yellow line?

Thanks :D

Answer 1

Given :

The diameter of semi-circle is 1 unit.

To Find :

Length of the yellow line

Solution:

First Refer to the attachment and see the labelled points.

As from attachment you can see length QU is equal to 1 unit.

$\textbf{\underline{\underline{From\:Two\:Tangent\:Theorem:}}}$

QU is straight away.

Now, Consider QRO as a right-angled triangle in which hypotenuse OQ is equal to the hypotenuse of another triangle QUO in which O is the centre.

OU and OR both are radius and are exactly same, then for sure QU is equal to QR

Let's assume the distance UT be x

$\textbf{\underline{\underline{Using\:Two\:Tangent\:Theorem:}}}$

$\star$ ST is also equal to x

$\star$ PT is equal to 1 - x

$\textbf{\underline{\underline{Using\: Pythagoras\:Theorem:}}}$

$\diamond\rm(Hypotenuse) ^{2} = \\ \rm (Perpendicular)^{2} + (Base) ^{2}$

$\hookrightarrow \sf \: {QU}^{2} = {PQ}^{2} + {PT}^{2}$

$\hookrightarrow \bf \: (1 + x) ^{2} = {1}^{2} + (1 - x) ^{2}$

$\hookrightarrow \bf \: ( {1})^{2} + 2(x)(1x) + ( {1x})^{2} = 1 + ({1})^{2} - 2(1)( - x) + ( { - x})^{2}$

$\hookrightarrow \bf\: 1 + 2x + {x}^{2} = 1 + 1 - 2x + {x}^{2}$

$\hookrightarrow \bf \: 1 + 2x =2 - 2x$

$\hookrightarrow \bf \: 2x + 2x = 2 - 1$

$\hookrightarrow \bf \: 4x = 1$

$\hookrightarrow \bf \: x = \dfrac{1}{4}$

$\hookrightarrow \bf \: x = 0.25$

We have taken length as 1 + x

Hence,

Length = 1 + x

Length = 1 + 0.25

Length = 1.25 units

Answer 2

${\large\bf{\mid{\overline{\underline{Given:-}}}\mid}}$

• Side of Square = 1 unit .
• A semicircle is inside the square with diameter equal to side of square .
• From diagram we can see that, DC is diameter , and O is centre of circle ..
• let EC length be = x unit .
• Than AE = (1-x) unit ..

$\Large\bold\star\underline{\underline\textbf{Concept\:used}}$

• The tangent segments whose endpoints are the points of tangency and the fixed point outside the circle are equal. In other words, tangent segments drawn to the same circle from the same point (there are two for every circle) are equal.
• Pythagoras Theoram .

___________________________________

$\Large\underline{\underline{\sf{Solution}:}}$

From image , we can see that,

→ EF and ED are tangents of circle from extrenal point E ,

so, above told theoram , both will be Equal .

so,

$\large\boxed{{ \blue{\bf \: EF=ED=x \: units}}}$

Similarly,

→ Tangent BF and BC are from external point B ,

so,

$\large\boxed{ \pink{\bf \: BF=BC = 1units}}$

__________________________________

Now, in ∆BAC, we have ,

→ Angle BAC = 90°( angle of square)

→ side BE = EF + FB = (x+1) units .

→ side AB = 1 unit (side of square)

→ side AE = AD - ED = (1-x) units .

So, Now By Pythagoras Theoram we get,

$\sf \: AB^{2}+AE^{2} =BE^{2} \\ \\ \bf \: \green{putting \: values} \\ \\ \red\leadsto \sf {1}^{2} + {(1 - x)}^{2} = {(x + 1)}^{2} \\ \\ \red\leadsto \sf \: 1 + 1 + \cancel{{x}^{2}} - 2x = \cancel{{x}^{2}} + 2x + 1 \\ \\ \red\leadsto \sf \: 4x = 1 \\ \\ \red\leadsto \red{\large \boxed{\bf \: x = \dfrac{1}{4}}}$

_____________________________________

So, Length of Yellow line will be,

$\bf \: EB = x + FB \\ \\ \bf \: EB = \frac{1}{4} + 1 \\ \\ \bf \: EB = \frac{5}{4} \\ \\ \pink{\large\boxed{\boxed{\bold{EB = \frac{5}{4} \purple{units}}}}}$

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