Advertisements
Advertisements
प्रश्न
Check whether AD is bisector of ∠A of ∆ABC of the following
AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm.
उत्तर
In ∆ABC, AB = 4 cm, AC = 6 cm, BD = 1.6 cm, CD = 2.4 cm
`"BD"/"DC" = 1.6/2.4 = 16/24 = 2/3`
`"AB"/"AC" = 4/6 = 2/3`
∴ `"BD"/"DC" = "AB"/"AC"`
By angle bisector theorem; AD is the internal bisector of ∠A
APPEARS IN
संबंधित 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
If PQ || BC and PR || CD prove that `"AR"/"AD" = "AQ"/"AB"`
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
Check whether AD is bisector of ∠A of ∆ABC of the following
AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm
∠QPR = 90°, PS is its bisector. If ST ⊥ PR, prove that ST × (PQ + PR) = PQ × PR
Draw a triangle ABC of base BC = 5.6 cm, ∠A = 40° and the bisector of ∠A meets BC at D such that CD = 4 cm
Draw ∆PQR such that PQ = 6.8 cm, vertical angle is 50° and the bisector of the vertical angle meets the base at D where PD = 5.2 cm
ST || QR, PS = 2 cm and SQ = 3 cm. Then the ratio of the area of ∆PQR to the area of ∆PST is
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?
