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प्रश्न
Find QP using given information in the figure.

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

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

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 Δ ABC and Δ PQR,
∠ ABC ≅ ∠ PQR, seg BD and
seg QS are angle bisector.
`If (l(AD))/(l(PS)) = (l(DC))/(l(SR))`
Prove that : Δ ABC ∼ Δ PQR


Seg NQ is the bisector of ∠ N
of Δ MNP. If MN= 5, PN =7,
MQ = 2.5 then find QP.
From the top of a light house, an abserver looking at a boat makes an angle of depression of 600. If the height of the lighthouse is 90 m then find how far is the boat from the lighthouse. (3 = 1.73)

In ΔABC, ray BD bisects ∠ABC.
If A – D – C, A – E – B and seg ED || side BC, then prove that:
`("AB")/("BC") = ("AE")/("EB")`
Proof :
In ΔABC, ray BD bisects ∠ABC.
∴ `("AB")/("BC") = (......)/(......)` ......(i) (By angle bisector theorem)
In ΔABC, seg DE || side BC
∴ `("AE")/("EB") = ("AD")/("DC")` ....(ii) `square`
∴ `("AB")/square = square/("EB")` [from (i) and (ii)]
In the figure, ray YM is the bisector of ∠XYZ, where seg XY ≅ seg YZ, find the relation between XM and MZ.

Draw seg AB = 6.8 cm and draw perpendicular bisector of it.
In the following figure, ray PT is the bisector of ∠QPR Find the value of x and perimeter of ∠QPR.

Draw the circumcircle of ΔPMT in which PM = 5.6 cm, ∠P = 60°, ∠M = 70°.
From the information given in the figure, determine whether MP is the bisector of ∠KMN.

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


In ΔABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing `("AB")/("BC") = ("AE")/("EB")`, complete the following activity:
Proof :
In ΔABC, ray BD bisects ∠B.
∴ `square/("BC") = ("AD")/("DC")` ...(I) (`square`)
ΔABC, DE || BC
∴ `(square)/("EB") = ("AD")/("DC")` ...(II) (`square`)
∴ `("AB")/square = square/("EB")` ...[from (I) and (II)]
