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In the given figure, if AB || CD || FE then find x and AE. - Geometry Mathematics 2

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प्रश्न

In the given figure, if AB || CD || FE then find x and AE. 

योग

उत्तर १

Construction: Join AF intersecting CD at X.

In ΔABF, DX || AB 

`"FD"/"DB"="FX"/"XA"`  ...(1) (By Basic proportionality theorem)

In ΔAEF, XC || FE 

`"FX"/"XA"="EC"/"CA"` ... (2) (By Basic proportionality theorem)

From (1) and (2), we get 

`"FD"/"DB"="EC"/"CA"`

`4/8 = x/12`

x = 6 

Now, AE = AC + CE

= 12 + 6

= 18

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उत्तर २

line AB || line CD || line FE   ...(given)

∴ `("BD")/("DF") = ("AC")/("CE")`   ...(Property of three parallel lines and their transversals)

∴ `8/4 = 12/x`

∴ x = `(12 xx 4)/8`

∴ x = 6 units

Now, AE = AC + CE [A - C - E]

= 12 + x

= 12 + 6

= 18 units

∴ x = 6 units and AE = 18 units

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Notes

Students can refer to the provided solutions based on their preferred marks.

Property of an Angle Bisector of a Triangle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Similarity - Practice Set 1.2 [पृष्ठ १४]

APPEARS IN

बालभारती Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board
अध्याय 1 Similarity
Practice Set 1.2 | Q 7 | पृष्ठ १४

संबंधित प्रश्न

Given below is the triangle and length of line segments. Identify in the given figure, ray PM is the bisector of ∠QPR.


Given below is the triangle and length of line segments. Identify in the given figure, ray PM is the bisector of ∠QPR.


Measures of some angles in the figure are given. Prove that `"AP"/"PB" = "AQ"/"QC"`.


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In ∆LMN, ray MT bisects ∠LMN If LM = 6, MN = 10, TN = 8, then Find LT. 


In the given fig, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find `"AX"/"XY"`.


Seg NQ is the bisector of ∠ N
of Δ MNP. If MN= 5, PN =7,
MQ = 2.5 then find QP.


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If ΔABC ∼ ΔDEF such that ∠A = 92° and ∠B = 40°, then ∠F = ?



In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR. 

Complete the proof by filling in the boxes.

solution:

In ∆PMQ,

Ray MX is the bisector of ∠PMQ.

∴ `("MP")/("MQ") = square/square` .............(I) [Theorem of angle bisector]

Similarly, in ∆PMR, Ray MY is the bisector of ∠PMR.

∴ `("MP")/("MR") = square/square` .............(II) [Theorem of angle bisector]

But `("MP")/("MQ") = ("MP")/("MR")`  .............(III) [As M is the midpoint of QR.] 

Hence MQ = MR

∴ `("PX")/square = square/("YR")`  .............[From (I), (II) and (III)]

∴ XY || QR   .............[Converse of basic proportionality theorem]


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