Question

3. In adjoining figure 1.14

seg PS ^ seg RQ seg QT ^ seg PR.

If RQ = 6, PS = 6 and PR = 12,

then find QT.​

Answer 1

Answer:

Given:

PS ⊥ RQ

QT ⊥ PR

RQ = 6, PS = 6 and PR = 12

With base PR and height QT,

A(△PQR)=½×PR×QT

With base QR and height PS,

A(△PQR)=½×QR×PS

∴A(△PQR)/A(△PQR)=½×PR×QT/½×QR×PS

⇒1=PR×QT/QR×PS

⇒PR×QT=QR×PS

⇒QT=QR×PS/PR

=6×6/12

=3

Hence, the measure of side QT is 3 units.

Answer 2

GIVEN:

seg PS ⊥ seg RQ

Seg QT ⊥ seg PR

Also ,

seg RQ = 6

seg PS = 6

seg PR = 12

FIND:

seg QT

Explanation :

Area of triangle = $\frac{1}{2}\:×\:base\:×\:height$

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by taking RQ as base and PS as corresponding height .

Area of triangle PQR = $\frac{1}{2}\:×\:base\:×\:height$

Area of triangle PQR = $\frac{1}{2}\:×\:6\:×\:6$

Area of triangle PQR = $1\:×\:3\:×\:6$

Area of triangle PQR = $18\:\:units$

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Now taking PR as base and QT as height

Area of triangle PQR = $\frac{1}{2}\:×\:PR\:×\:QT$

⟶ 18 = $\frac{1}{2}\:×\:12\:×\:QT$

$⟶\:\frac{18\:×\:2}{1\:×\:12}\:\: = \:QT$

$⟶\:\frac{36}{12}\:\: = \:QT$

$3\:=\:QT$

$QT\:=\:3\:units$