The TV Salesman Insists That All 42" TVs Are The Same Size. Is He Right?

For the $4:3$ TV:

$\sqrt {(4x)^{2} + (3x)^{2}} = 42$

$x = \frac {42}{5}$

Thus, the area is:

$(\frac {42 \cdot 4}{5})(\frac {42 \cdot 3}{5}) = \frac {42^{2} \cdot 12}{25} in.^{2}$

For the $16:9$ TV:

$\sqrt {(16y)^{2} + (9y)^{2}} = 42$

$y = \frac {42\sqrt {337}}{337}$

Thus, the area is:

$(\frac {42\sqrt {337} \cdot 16}{337})(\frac {42\sqrt {337} \cdot 9}{337}) = \frac {42^{2}\cdot 337 \cdot 144}{337^{2}} = \frac {42^{2} \cdot 144}{337} in.^{2}$

Subtracting them yields:

$\frac {42^{2} \cdot 12}{25} - \frac {42^{2} \cdot 144}{337}$

That if you take your time carefully you can simplify to

$\frac {783216}{8425} \approx 92.96$

So rounding gives our answer which is:

${\color{#3D99F6}{\boxed{93 in.^{2}}}}$

The ratio of the sides in the first TV is $3:4:5$ and the area can be expressed as the area of a $3\times 4$ multiplied by the square of the scale factor, $\frac{42}{5}$ $3\cdot 4\cdot \dfrac{42^2}{5^2}=846.72$ The ratio of the sides in the second TV is $9:16:\sqrt{337}$ and the area can be expressed as the area of a $9\times 16$ multiplied by the square of the scale factor, $\frac{42}{\sqrt{337}}$ $9\cdot 16\cdot \dfrac{42^2}{337}\approx 753.76$ The difference is then $846.72-753.76\approx 92.96$ which rounds to $\boxed{93}$ .

Lets consider the 42" 4:3 television first .

Let the diagonal be D the width be w and the height be H .

We are given that $\frac{W}{H}=\frac{4}{3}$

from which ${W}=\frac{4H}{3}$

By Pythagoras theorem,

${D^2}={W^2}+{H^2}$

i.e. ${42}^2=(\frac{4H}{3})^2+{H^2}$

which simplifies to ${1764}=\frac{25{H^2}}{9}$

From which ${H}={25.2inches}$

and ${W}=\frac{4\times{25.2}}{3}$

${W}={33.6inches}$

Hence the area equals ${H}\times{W}$

Which equals ${846.72 sq.inches}$

Next we consider the 42" 16:9 .

Let the diagonal be D the width be w and the height be h .

We are given that $\frac{w}{h}=\frac{16}{9}$

from which ${W}=\frac{16h}{9}$

By Pythagoras theorem,

${D^2}={w^2}+{h^2}$

i.e. ${42}^2=(\frac{16h}{9})^2+{h^2}$

which simplifies to ${1764}=\frac{256{h^2}}{81}$

From which ${h}={20.5910inches}$

and ${w}=\frac{16\times{20.5910}}{9}$

${W}={36.6062inches}$

Hence the area equals ${h}\times{w}$

Which equals ${753.7577sq.inches}$

Hence the difference in area equals ${846.72}-{753.7577}$

Which is approximately equal to $\boxed{93sq.inches}$

Let 4x and 3x be the sides of 4:3 TV. Then diagonal will be 5x which is 42". Hence x=8.4". Area of this TV will be 12x sqrd. With x==8.4, it works out to be 847 Sq. inch

Similarly let 16y and 9y be sides of 16:9 TV. Diagonal will be 18.36y which is 42". Hence y=2.29". Area of this TV will be 144y sqrd. With y==2.29, it works out to be 754 Sq. inch

Hence differece is 93 Sq. Inch

a=42^2/(4^2 +3^2) a =

70.56

-->b=42^2/(16^2 +9^2) b =

5.2344214

-->(12 a)-(16 9*b) ans =

92.963323

let for the 4:3 and 16:9 TVs the lengths and width of screens be (4x,3x) and (16x,9x) respectively. Since screen-size (length of diagonal) is 42, by Pythagoras theorem ((4x)^2 + (3x)^2)= 42^2 and ((16x)^2 + (9x)^2)= 42^2. These 2 equations give 2 values of 'x', which can then be used to find the areas: (4x * 3x)=847 and (16x * 9x)=753. Hence difference is 93.

Area of 42" 4:3 TV is 4s x 3s where s can be calculated as (4s)^2 + (3s)^2 = 42^2. After solving we get area of this TV as 846.72 square inches. Similarly, for 16:9 TV, the value of s becomes different and area up to two decimal places is 753.75 square inches. Approximate difference between areas of both these TVs is 93 square inches.

let the sides of first tv be 4x,3x then as diagonal =42'' it implies x=42''/5 then area=846.72 //ly by considering sides of other tv as 16y,9y we get the area is 753.75 so their difference is around 93

solving the area. 4:3, let the sides be 4X and 3X.. to solve for X... (it will be easier to solve X^2) (4X)^2 + (3X)^2 = 42 ^2 ; X^2 = 42^2 / 25 the area of the tv would be, 4X*3X = 12X^2 = 12(42^2/25) = 846.72 sq. in.

16:9 let the sides be 16X and 9X.. solve for X^2 (16X)^2 + (9X)^2 = 42 ^2 ; X^2 = 42^2 / (16^2 +9^2) the area of the tv would be, 16X*9X = 144X^2 = 144(42^2/(16^2 + 9^2)) = 753.76 sq. in.

Difference would be, 846.72 - 753.76 = 92.96 ~ 93 sq. in. (with the 4:3 aspect ratio TV being the larger one)

It's pretty simple......For first television let us say the sides are 4a and 3a. So diagonal 16a^2 + 9a^2 = 42^2.From this we get a=8.4 . Now easily we find area=846.72 . Now for second TV, sides are say 16b and 9b. Similarly 256b^2 +81b^2 =42^2 .Again calculating the area from the value of b we get, area = 753.75. So, difference = 846.72 - 753.75 =(approx) 93.