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प्रश्न
In the given figure, if AB || CD || FE then find x and AE.
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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
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
Notes
Students can refer to the provided solutions based on their preferred marks.
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"`.
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
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.
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 ∆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]
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)]
