Question

□ABCD is a trapezium. Side AD || side PQ || side BC. If AP = 15 cm, PB = 12 cm, QC = 14 cm then find DQ.​

Answer 1

$\huge\star\mathfrak\blue{{Answer:-}}$

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