Question

in trapezium ABCD side AB||PQ|| DC, AP=15 ,PD=12,QC=14 find BQ

Answer 1

Given : ABCD is a trapezium
AB || PQ || DC
AP =15 cm , PD = 12 cm and QC =14 cm
In figure join AC,
Since, AB | | DC & PQ || AB
PQ || DC
[Since lines parallel to the same lines are also parallel to each other]
In ∆ ADC
PO || DC
[ PQ | | DC]

By using B.P.T ,
AP/ PD = AO/CO …………(1)

In ∆ABC
QO | | AB ,
[PQ | | AB]

By using B.P.T ,
CQ/BQ = CO/AO
BQ/CQ = AO/CO …………(2)
[Reciprocal the term]

From eq. 1 & 2 ,
BQ/ CQ =AP/ PD ,
BQ/ 14 =15/ 12
12 BQ = 14 × 15
BQ = (14×15)/12 = (14 × 5)/4
BQ = 70/4 = 35/2
BQ = 17.5 cm

Hence, the value of BQ = 17.5 cm

HOPE THIS WILL HELP YOU….

Answer 2

Step-by-step explanation:

is Gurjar 2028 and ab is equals to C G is equals to B is equals to B is equals to 6 is equals to 25 CM calculate the total area of the shaded region