geometry GRE question please help?

Question

in a square PQRS, T is the mid point of side RS. the lenght of PT= 8(5)^1/2 . i.e 8 root of 5
what is the side of the square.
please explain the answer

Answer 1

let side of square is = x ----[PQ=QR=RS =SP =x]

T is mid point of RS .

Therefore ST =RS/2 =x/2 ,PT =8 root 5 ,PS =x

In right angle PST

(PS)^2+(ST)^2 = (PS)^2

(x)^2 +(x/2)^2 = [8root 5]^2

or x^2 +x^2/4 = 64*5

5x^2/4 =320

x^2 =( 320*4)/5

x^2 =256

x = 16 or -16

-16 rejected ,side always postive.

therefore x = 816

side of the square is 16 unit.ans.

Answer 2

Let, the sides of square are :

PQ = QR = RS = SP = x



RT = TS = x/2

Thus

x^2 + (x/2)^2 = [8*(5)^1/2]^2 = 320

4x^2 + x^2 = 1280

5x^2 = 1280 ==> x^2 = 256

x = 16

Hene,

The Side = 16 >=========================< ANSWER

Answer 3

Let "a" be the side of the square PQRS.since T is the mid point of RS, ST = a/2.

in right triangle PTS, PT is the hypotenuse; SP = a; ST = a/2.

by Pythagorus thm.,

SP^2+ST^2 = PT^2

a^2+(a/2)^2 = (8(5)^1/2)^2

(5a^2)/4 = 320

a^2 = 1280/5

=256

a= 16.

the side of the square is 16

Answer 4

RT = 1/2 RS (midpoint)

RS = PR ( square)



PRT is a rt angled triangle



PT^2 = PR^2 + 1/4 PR^2 = 5/4 PR^2 = 8(5)^(1/2)



PR^2 = 8^2*5 *4/5 = 256

PR = 16



does not deserve best ans, corrected it so that no person is misguided

Answer 5

the reason being that the final concept you're questioning approximately (ie, the facets advert, DC being equivalent forcing the angles to be equivalent) heavily isn't authentic. the least perplexing thank you to verify it is merely to verify an exceedingly slender sliver of a triangle--- say you draw a triangle ABC with AB somewhat small and AC, CB the two style of massive. in fact, cause them to the comparable length so it is an isosceles triangle it somewhat is way less complicated to charm to, and make AB somewhat, somewhat small so it is almost like a drawing of a pin. Now bisect BC with a factor P, and draw AP. it will be sparkling even from a poorly drawn photograph that the perspective BAP is plenty greater suitable than the perspective p.c.. (If the triangle is somewhat slivery, the scale of perspective BAP is going to be very on the threshold of the scale of perspective BAC, and the scale of perspective %is going to be very on the threshold of 0, merely like the skinny perspective PCA.) As for why it could't be desperate, properly, as quickly as you already know the sliver phenomenon properly sufficient, it will be sparkling that version interior the perspective length can take place even exterior of exaggerated situations. you would be waiting to mentally envision shifting C around somewhat in one in all those vogue that the perspective at C remains at 60 levels, and noticing that in case you progression C greater in direction of A it is achievable for ACD to be greater suitable than DCB, while in case you progression C greater in direction of B it is achievable for any incorrect way around to take place. So it could't be desperate from the given strategies--- regardless of if the angles interior the given photograph look equivalent, or in fact *are* equivalent. i think of the regulations at the back of those kinds of issues are that in the event that they do no longer explicitly inform you a fact, you're no longer allowed to deduce it, regardless of if it variety of feels present, or _is_ present, interior the photograph. (It somewhat facilitates to permit your strategies to wander from the only static photograph of the area they gave you, to verify different, comparable photos that meet all the given strategies.) good luck on the GRE, it is not any exciting in any respect.