Question

Match the following columns:

Column I	Column II
(a) In a given ΔABC, DE || BC and `(AD)/(DB) = 3/5`. If AC = 5.6 cm then AE = ..... cm.	(p) 6
(b) If ΔABC ~ ΔDEF such that 2AB = 3DE and BC = 6 cm then EF = ...... cm.	(q) 4
(c) If ΔABC ~ ΔPQR such that  ar(ΔABC) : ar(ΔPQR) = 9 : 16 and ВС = 4.5 cm then QR = ...... cm.	(r) 3
(d) In the given figure, AB || CD and OA = (2x + 4) cm, OB = (9x – 21) cm,  OC = (2x – 1) cm and OD = 3 cm. Then x = ?	(s) 2.1

Answer 1

Column I	Column II
(a) In a given ΔABC, DE || BC and `(AD)/(DB) = 3/5`. If AC = 5.6 cm then AE = ..... cm.	(s) 2.1
(b) If ΔABC ~ ΔDEF such that 2AB = 3DE and BC = 6 cm then EF = ...... cm.	(q) 4
(c) If ΔABC ~ ΔPQR such that  ar(ΔABC) : ar(ΔPQR) = 9 : 16 and ВС = 4.5 cm then QR = ...... cm.	(p) 6
(d) In the given figure, AB || CD and OA = (2x + 4) cm, OB = (9x – 21) cm,  OC = (2x – 1) cm and OD = 3 cm. Then x = ?	(r) 3

Explanation:

(a) Let AE = x cm.

Then, EC = (5.6 – x) cm.

`(AD)/(DB) = (AE)/(EC)`

⇒ `3/5 = x/(5.6 - x)`

∴ 3(5.6 – x) = 5x

⇒ 8x = 3 × 5.6

∴ `x = (3 xx 5.6)/8`

= `(16.8)/8`

= 2.1

(b) `(AB)/(DE) = (BC)/(EF)`

⇒ `3/2 = 6/x`

⇒ 3x = 12

⇒ x = 4

(c) `(ar(ΔABC))/(ar(ΔPQR)) = (BC^2)/(QR^2)`

⇒ `9/16 = (BC^2)/(QR^2)`

⇒ `(3/4)^2 = ((BC)/(QR))^2`

⇒ `(BC)/(QR) = 3/4`

⇒ `QR = 4/3 xx BC`

= `(4/3 xx 4.5) cm` 

= 6 cm

(d) ΔОАВ ~ ΔОCD

⇒ `(OA)/(OC) = (OB)/(OD)`

⇒ `(2x + 4)/(2x - 1) = (9x - 21)/3`

⇒ 6x + 12 = 18x² – 51x + 21

⇒ 18x² – 57x + 9 = 0

⇒ 6x² – 19x + 3 = 0

⇒ (x – 3)(6x – 1) = 0

⇒ x = 3 or x = `1/6`

But, `x = 1/6` makes (2x – 1) < 0.

So, we reject it.

∴ x = 3.