Yang bisa mohon bantu jawab yaaa insya allah dapet pahala☺️Dikarenakan pr nya susah yang bisa jangan lipa langsung jawab ok

a. nilai y :
[tex] {(y + 4)}^{2} = {16}^{2} + {y}^{2} \\ {y}^{2} + 8y + 16 = {16}^{2} + {y}^{2} \\ 8y + 16 = {16}^{2} \\ 8y + 16 = 256 \\ 8y = 256 - 16 \\ 8y = 240 \\ y = \frac{240}{8} \\ y = 30[/tex]
b. Luas ∆PQR :
[tex] \frac{1}{2} \times a \times t \\ = \frac{1}{2} \times 30 \times 16 \\ = 15 \times 16 \\ = 240 {cm}^{2} [/tex]