In ∆PTQ
Apply Pythagorean theorem
[tex]\\ \tt\hookrightarrow PT^2=PQ^2+QT^2[/tex]
[tex]\\ \tt\hookrightarrow PT^2=15.4^2+9.3^2=323.16[/tex]
[tex]\\ \tt\hookrightarrow PT=17.9[/tex]
In ∆TRQ
[tex]\\ \tt\hookrightarrow tan44.4=\dfrac{QT}{TR}=\dfrac{9.3}{TR}[/tex]
[tex]\\ \tt\hookrightarrow 0.9=\dfrac{9.3}{TR}[/tex]
[tex]\\ \tt\hookrightarrow TR=\dfrac{9.3}{0.9}=10.3[/tex]
- PR=PT+TR=10.3+17.9=28.2cm
Now
Area:-
[tex]\\ \tt\hookrightarrow 1/2bh[/tex]
[tex]\\ \tt\hookrightarrow 1/2(28.2)(9.3)[/tex]
[tex]\\ \tt\hookrightarrow 14.1(9.3)[/tex]
[tex]\\ \tt\hookrightarrow 131.13cm^2[/tex]
