Question

In the given figure, PQRS is a square and TQUV is another square inside it in a corner. If VQ = 4/9QS, find the ratio of the area of the triangle VSU to the area of PQRS

Answer 1

@vishweshshrimali5

Answer 2

Hi sorry for being late in responding :)

Answer 3

First of all can you calculate the area of PQRS ?

Answer 4

its k.i am also late.

Answer 5

81/32

Answer 6

sq=9/4vq
so area of square =81/32

Answer 7

@vishweshshrimali5 is it right?

Answer 8

@neer2890 can you please have a look at this problem. I am busy at other problem.

Answer 9

ok.
we know that
VQ= 4/9 QS
QS=9/4 VQ

now we can use pythagorean theorem to find the side of square.

Answer 10

2x^2=81/16 (VQ)^2
x^2=81/32(VQ)^2
and we know that area of square =x^2

Answer 11

\[so .area. of .\square. PQRS=\frac{ 81 }{ 32 }(VQ)^{2}\]

Answer 12

yes.ov=vq/2=4/9qs/2=2/9qs

Answer 13

so small square area=8/81(qs)^2

Answer 14

hmm.

Answer 15

ok..now i'm not very much good at geometry.,
there may be a shortcut to it here..
But as much i know
To find the area of triangle VSU
SV=SQ-VQ
SV=QS-4/9QS
SV=5/9QS

Answer 16

sv=height,vu=base