In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, Find BQ.
Practice Set 1.2 | Q 5 | Page 14
In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, Find BQ.
In △ ABD, PX || AB
`"PD"/"AP" = "XD"/"XB"` ....(By Basic proportionality theorem)(1)
In △BDC, XQ || DC
`"XD"/"XB" = "QC"/"BQ"` ....(By Basic proportionality theorem)(2)
From (1) and (2), we get
∴ `"PD"/"AP" = "QC"/"BQ"`
∴ `12/15 = 14/"BQ"`
∴ 12 × BQ = 15 × 14
∴ BQ = `(15 × 14)/12`
∴ BQ = `(5 × 7)/2`
∴ BQ = `35/2`
∴ BQ = 17.5 units
Solution:- Given: In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14. We need to find BQ.
To find: BQ
Solution:
In △ABD, PX || AB
From Basic Proportionality Theorem, we get:
"PD"/"AP" = "XD"/"XB" …(1)
In △BDC, XQ || DC
From Basic Proportionality Theorem, we get:
"XD"/"XB" = "QC"/"BQ" …(2)
From (1) and (2), we get:
"PD"/"AP" = "QC"/"BQ"
Substituting the given values, we get:
12/15 = 14/"BQ"
On solving the above equation, we get:
BQ = (15 × 14)/12 = (5 × 7)/2 = 35/2 = 17.5 units
Therefore, BQ = 17.5 units.
Chapter 1. Similarity- Practice Set 1.2 – Page 13
Click Here for All Textbook Soutions of Chapter 1: Similarity