Advertisements
Advertisements
प्रश्न
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.
उत्तर
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
संबंधित प्रश्न
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 = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm
If PQ || BC and PR || CD prove that `"QB"/"AQ" = "DR"/"AR"`
DE || BC and CD || EE Prove that AD2 = AB × AF
Construct a ∆PQR in which the base PQ = 4.5 cm, ∠R = 35° and the median from R to RG is 6 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.
Construct a ∆ABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is 4 cm
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
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.