How much bigger is your screen?

Well if you haven't managed to get the answer I would recommend drawing 2 televisions on a bit of scrap paper. You know the width and length of both televisions but drawing it helps you visualise it now try it and see if you can come up with it. Yes Pythagoras' theorem which counts for right angled triangles like these 2! Right lets work out the diagonal length of the first television. Width if 60 and length is 80. $60^2$ + $80^2$ =10000 $\sqrt{10000}$ =100 so the diagonal length of TV1 is 100. Now same for TV2, width is 60 and length is 91. $60^2$ + $80^2$ =11881 $\sqrt{11881}$ =109 so diagonal length for TV2 is 109. 109-100=9 which is your answer, please ask me if you do not understand pythagoras' theorem and I can explain in more detail! PS Draw diagrams next time!!!!

The dimension of the first screen is $\sqrt{60^{2} + 80^{2}} =100$ .The dimension of the second screen is $\sqrt{60^{2} + 91^{2}} =109$ .So the absolute difference is $109-100=9$ .

In the first case,

Width of TV = 60 cm

Length of TV = 80 cm

By Pythagoras theorem, we have

Diagonal = sqrt [{60}^2 + {80}^2]

. = 100 cm

Similarly for the second TV, diagonal will be 109 cm.

The absolute difference between the dimensions (diagonals) of the two will be (109 - 100 =) 9 cm.

First television

w = 80cm, l = 60 cm

by Pythagorean Theorem, its diagonal = 100cm

Second television

w = 91cm, l = 60 cm

by Pythagorean Theorem, its diagonal = 109cm

Since we're looking for the absolute difference,

|100-109| = 9

As the formulae to find diagonal = root over of {length^2 + breadth^2} {root over of [80^2 + 60^2]} = 100 {root over of [91^2 + 60^2]} = 91 As the question is given to find the difference of diagonals in centimetres,we have to subtract 91 cm from 100 cm which is equal to 9 cm So, the answer is = { 9 cm }.

Using pythagorean theorema just minus both diagonal length (dimensions).

when $width \times length$ = $60 \times 80$ the diagonal is $100$

when $width \times length$ = $60 \times 91$ the diagonal is $109$

minus both of them the absolute difference of the dimension = $9$

c=√(60²+80²)

=√(10000)

=100

c'=√(60²+91²)

=√(11881)

=109

109-100=9

the diagonal length of the 1st TV is $\sqrt{60^2 + 80^2} = 100 cm$ and the 2nd one is $\sqrt{60^2 + 91^2} = 109 cm$

And the difference is $109 -100 = 9 cm$

Using pythagoras's theorem, 60^2+80^2=x^2 x=100 60^2+91^2=y^2 y=109 109-100=9

Square root ( 60^2) + (80^2) = 100 = a

Square root ( 60^2) + (91^2) = 109 = b

| a - b | = 9

for first rectangular TV dim1= square root(60^2+80^2)=100
for second rectangular TV dim2= square root(60^2+91^2)=109
difference = dim2-dim1=9

First, find the dimension of first television using Pythagorean Theorem, with a=60 and b=80, the dimension(c) will come out to be 100. Then, find the dimension of the second television using Pythagorean Theorem with a=60 and b=91, the dimension(c) will come out to be 109. The absolute difference is therefore, 9

Vamos utilizar o teorema de Pitágora para encontrar a diagonal da televisão: Sendo assim:

1ª Televisão: a^2=60^2+80^2

a^2=3600+6400

a^2=10000

a=100cm

2ªTelevisão:

a^2= 60^2+91^2

a^2=3600+8281

a^2=11881

a=109

Diferença entre a diagonal das dduas televisões é:

109-100= 9cm

D=diagonal D^2=w^2+l^2 D^2=60^2+80^2 D^2=10000 D=100

D^2=60^2+91^2 D^2=11881 D=109

D1-D2=ans 109-100= 9

According to the Pythagorean Theorem, 60^2+80^2= 10000. The square root of 10000 is 100. 60^2+91^2=11881. The square root is 109.

109-100=9