मराठी
महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

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

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प्रश्न

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

बेरीज
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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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Similarity - Practice Set 1.2 [पृष्ठ १४]

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In ∆ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.


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.

In △PMQ, ray MX is bisector of ∠PMQ.

∴ `square/square = square/square` .......... (I) theorem of angle bisector.

In △PMR, ray MY is bisector of ∠PMQ.

∴ `square/square = square/square` .......... (II) theorem of angle bisector.

But `(MP)/(MQ) = (MP)/(MR)` .......... M is the midpoint QR, hence MQ = MR.

∴ `(PX)/(XQ) = (PY)/(YR)`

∴ XY || QR .......... converse of basic proportionality theorem.


In ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that `"AP"/"PD" = "PC"/"BP"`.


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seg QS are angle bisector.
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Prove that : Δ ABC ∼ Δ PQR


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Proof :

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∴ `square/("BC") = ("AD")/("DC")`   ...(I) (`square`)

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∴ `(square)/("EB") = ("AD")/("DC")`   ...(II) (`square`)

∴ `("AB")/square = square/("EB")`   ...[from (I) and (II)]


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