flower bed rectangle plus semi circle paper test

The rectangle's perimeter can potentially measure any of the following:

1+1+13+13
2+2+12+12
3+3+11+11
4+4+10+10
5+5+9+9
6+6+8+8

The trick is to realize the diagonal is 10 and figure out what triangle works with perfect squares. If the diagonal is 10, then you can have a triangle with sides 6-8-10, which is in proportion to 3-4-5.

Once you figure out the sides, you can work with perimeter/circumference of the semi circle which you can get easily because you already have the diameter (in this case 8 ). Perimeter would be 28+ 4 Pi...

I have a doubt why is B alone not suff

It says the diagonal is 10 and QR>RS
So applying the special triangles rule wouldn't the dimensions of the triangle be 6-8-10. And the dimensions of the rectangle would be 8 by 6.
This information is Suff to solve the ques

Am I missing something here?

good reasoning mals24

yep B seems to be correct answer here because 6-8-10 is the only phythagorus (pardon my spelling :0) triplet possible.

You cant assume its a triplet without knowing the lengths of two of the sides. You have two unknowns, all you know is one side is greater than the other from the prompt.

With just the diagonal you have the following equation:

a^2+b^2 = 10^2
a^2+b^2 = 100

The test wants you to assume its a 6-8-10 triangle but you cant because you only have the measure of one side and one angle (the right 90 angle). One of the sides for example could potentially measure 9, rather than 8. If this were true the leftover side would measure the square root of 19.

answer should be C...

a+b=14

a^2+b^2=100

using these two's values should be 6 and 8.....

Nice example

in hurry many of us make this kind of mistake

when ever we see 10 as diagonal we take other two side as 6 and 8.. need to focus more on DI

Thanks

reason, why B isn't sufficient

diag is 10 feet & QR >RS
consider this scenario: QR = 5*sqrt(3) and RS = 5
this also fits for this rectangle.

So, answer is (C) only.