Advertisements
Advertisements
Question
In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC
If `"AD"/"DB" = 3/4` and AC = 15 cm find AE
Advertisements
Solution
Let AE be x
∴ EC = 15 – x
In ∆ABC we have DE || BC
By Basic proportionality theorem, we have
`"AD"/"DB" = "AE"/"EC"`
`3/4 = x/(15 - x)`
4x = 3(15 – x)
4x = 45 – 3x
7x = 45
⇒ x = `45/7`
= 6.43
The value of x = 6.43
APPEARS IN
RELATED QUESTIONS
In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC
If AD = 8x – 7, DB = 5x – 3, AE = 4x – 3 and EC = 3x – 1, find the value of x
In ΔABC, D and E are points on the sides AB and AC respectively. For the following case show that DE || BC
AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.
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
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.
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.
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
Two circles intersect at A and B. From a point, P on one of the circles lines PAC and PBD are drawn intersecting the second circle at C and D. Prove that CD is parallel to the tangent at P.
