Advertisements
Advertisements
Question
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.
Advertisements
Solution
Let the side of the rhombus be “x”. Since PQRB is a
Rhombus PQ || BC
By basic proportionality theorem
`"AP"/"AB" = "PQ"/"BC" ⇒ (12 - x)/"BC" = x/6`
12x = 6(12 – x)
12x = 72 – 6x
12x + 6x = 72
18x = 72 ⇒ x = `72/18` = 4
Side of a rhombus = 4 cm
PQ = RB = 4 cm
APPEARS IN
RELATED QUESTIONS
If PQ || BC and PR || CD prove that `"AR"/"AD" = "AQ"/"AB"`
If PQ || BC and PR || CD prove that `"QB"/"AQ" = "DR"/"AR"`
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 = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm.
∠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
Construct a ∆ABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is 4 cm
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.
