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Draw Seg Ab = 6.8 Cm and Draw Perpendicular Bisector of It.

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प्रश्न

Draw seg AB = 6.8 cm and draw perpendicular bisector of it. 

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उत्तर

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Property of an Angle Bisector of a Triangle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2018-2019 (July) Set 1

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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.


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


In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then find QP. 


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


Find QP using given information in the figure.


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


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


In ΔABC, ∠ACB = 90°. seg CD ⊥ side AB and seg CE is angle bisector of ∠ACB.

Prove that: `(AD)/(BD) = (AE^2)/(BE^2)`.


In the following figure, ray PT is the bisector of QPR Find the value of x and perimeter of QPR.


From the information given in the figure, determine whether MP is the bisector of ∠KMN.



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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