Question

In the above sided figure, find the value of x. ​

Answer 1

$\setlength{\unitlength}{1.2mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000) \put(21,6){\line( - 1,1){25}} \put(-4,29){\line(1,1){25}} \put(21,54){\line(2, -1){48}} \put(20,6){\line(2,1){49}} \put(25, 26){\circle*{1}} \put(7,14){\bf Q} \put(22,26){\bf O}\put(22,55){\bf F} \put(41,48){\bf S}\put(69,25){\bf E} \put(39,10){\bf R} \put(21,4){\bf D} \put(-7,32){\bf G} \put(7,44){\bf P} \put(19,57){\vector( - 1,-1){10}} \put(19,57){\vector(1,1){1}} \put(46,16){\vector( -2, - 1){18}}\put(46,16){\vector(2,1){18}} \put(51,50){\vector( - 2,1){18}} \put(51,50){\vector(2, - 1){18}} \put(42,8){\sf x} \put(2,51){\sf 4 cm}\put(41,56){\sf 7 cm} \put( -5, 21){\bf 6 cm}\end{picture}$

Given:

• GQ = 6cm
• PF = 4cm
• EF = 7cm

Find:

• Value of DE

Solution:

PF = SF $\sf\bigg(\because Length\: of \: Tangent \: From \: Point \: To \: Circle \: are \: Equal\bigg)$

where,

• PF = 4cm

So,

SF = 4cm

Now,

EF = ES + SF

where,

• EF = 7cm
• SF = 4cm

So,

EF = ES + SF

:=7 = ES + 4

:= 7 - 4 = ES

:= 3cm = ES

:= ES = 3cm

_________________________________

$\setlength{\unitlength}{1.2mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000) \put(21,6){\line( - 1,1){25}} \put(-4,29){\line(1,1){25}} \put(21,54){\line(2, -1){48}} \put(20,6){\line(2,1){49}} \put(25, 26){\circle*{1}} \put(7,14){\bf Q} \put(22,26){\bf O}\put(22,55){\bf F} \put(41,48){\bf S}\put(69,25){\bf E} \put(39,10){\bf R} \put(21,4){\bf D} \put(-7,32){\bf G} \put(7,44){\bf P} \put(19,57){\vector( - 1,-1){10}} \put(19,57){\vector(1,1){1}} \put(46,16){\vector( -2, - 1){18}}\put(46,16){\vector(2,1){18}} \put(51,50){\vector( -2,1){18}} \put(51,50){\vector(2,- 1){0.1}} \put(42,8){\sf x} \put(2,51){\sf 4 cm}\put(41,56){\sf 4 cm} \put( -5, 21){\bf 6 cm}\put(69,32){\vector( -2,1){18}} \put(69,32){\vector(2,- 1){0.1}}\put(65,39){\sf 3 cm}\end{picture}$

_________________________________

Here,

RE = ES

where,

• ES = 3cm

So,

RE = 3cm

$\sf\bigg(\because Length\: of \: Tangent \: From \: Point \: To \: Circle \: are \: Equal\bigg)$

Now,

RD = DQ

where,

• DQ = GQ = 6cm (Tangent From Point D to Circle)

So,

RD = DQ = GQ = 6cm

RD = 6cm

Now,

DE = RD + RE

where,

• DE = x
• RD = 6cm
• RE = 3cm

So,

DE = RD + RE

⇨ x = 6 + 3

⇨ x = 9cm

$\therefore$ DE = x = 9cm

_________________________________

Hence, x = 9cm

Answer 2

Answer:

Could you please explain how GQ=DQ?

Step-by-step explanation: