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प्रश्न
Find QP using given information in the figure.
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उत्तर
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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.
In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then find QP.
In the given figure, if AB || CD || FE then find x and AE.
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"`.
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)`.
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)]
