Answer:
The dimension of the TV = 11.2 inches × 8.4 inches
Step-by-step explanation:
Let x be the length of the TV
Let y be the width of the TV
x : y = 4:3
[tex]\frac{x}{y} = \frac{4}{3} \\cross-multiplying\\3x = 4y \\x = \frac{4}{3} y- - - - - (1)[/tex]
From the figure attached:
using Pythagoras theorem
x² + y² = 42²
where:
[tex]x = \frac{4}{3} y\\(\frac{4}{3}y)^2 + y^2 = 42^2\\\frac{16}{9}y^2 + y^2 = 1764\\ \frac{16y^2}{9} + \frac{y^2}{1} = 1764\\\frac{16y^2 +9y^2}{9}= 1764\\ cross-multiplying\\16y^2 + 9y^2 = 1764\\25y^2 = 1764\\y^2 = \frac{1764}{24}\\y^2 = 70.56\\y = \sqrt{70.56} \\y = 8.4\ inches\\Finding\ x:\\x = \frac{4}{3} y\\x = \frac{4}{3} \times 8.4\\ x = 1.33 \times 8.4\\x = 11.2\ inches\\\therefore\ the\ dimensions\ are\ 11.2\ inches \times 8.4\ inches[/tex]
