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

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

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प्रश्न

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

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उत्तर

In ∆ PQR,

PQ = 7, PR = 3, QM = 3.5, and MR = 1.5 ...(Given)

`"PQ"/"PR" = 7/3` ...(i)

`"QM"/"MR" = 3.5/1.5 = (3.5 × 10)/(1.5 × 10) = 35/15 = 7/3` ...(ii)

From (i) and (ii)

∴ `"PQ"/"PR" = "QM"/"MR"` 

∴ by converse of angle bisector theorem,

∴ Ray PM is the bisector of ∠QPR.

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पाठ 1: Similarity - Practice Set 1.2 [पृष्ठ १३]

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बालभारती Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
पाठ 1 Similarity
Practice Set 1.2 | Q 1.1 | पृष्ठ १३

संबंधित प्रश्‍न

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


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


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 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"`.


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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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`("AB")/("BC") = ("AE")/("EB")`

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

∴ `("AB")/square = square/("EB")`   [from (i) and (ii)]


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

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


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