Advertisements
Advertisements
प्रश्न
ABCD is a trapezium in which AB || DC and P, Q are points on AD and BC respectively, such that PQ || DC if PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD
Advertisements
उत्तर
Join AC intersecting PQ at S.
Let AP be x
∴ AD = x + 18
In the ∆ABC, QS || AB
By basic proportionality theorem.
`"AS"/"SC" = "BQ"/"QC"`
`"AS"/"SC" = 35/15` ...(1)
In the ∆ACD; PS || DC
By basic proportionality theorem.
`"AS"/"SC" = "AP"/"PD"`
`"AS"/"SC" = x/18` ...(2)
From (1) and (2) we get
`35/15 = x/18`
15x = 35 × 18 ⇒ x = `(35 xx 18)/15` = 42
AD = AP + PD
= 42 + 18 = 60
The value of AD = 60 cm
APPEARS IN
संबंधित प्रश्न
If PQ || BC and PR || CD prove that `"AR"/"AD" = "AQ"/"AB"`
If PQ || BC and PR || CD prove that `"QB"/"AQ" = "DR"/"AR"`
Rhombus PQRB is inscribed in ΔABC such that ∠B is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus.
In trapezium ABCD, AB || DC, E and F are points on non-parallel sides AD and BC respectively, such that EF || AB. Show that = `"AE"/"ED" = "BF"/"FC"`
DE || BC and CD || EE Prove that AD2 = AB × AF
∠QPR = 90°, PS is its bisector. If ST ⊥ PR, prove that ST × (PQ + PR) = PQ × PR
Construct a ∆PQR in which QR = 5 cm, ∠P = 40° and the median PG from P to QR is 4.4 cm. Find the length of the altitude from P to QR.
Construct a ∆PQR such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 cm
ABC is a triangle in which AB = AC. Points D and E are points on the side AB and AC respectively such that AD = AE. Show that the points B, C, E and D lie on a same circle
An Emu which is 8 feet tall is standing at the foot of a pillar which is 30 feet high. It walks away from the pillar. The shadow of the Emu falls beyond Emu. What is the relation between the length of the shadow and the distance from the Emu to the pillar?
