The diagram below consists of a semicircle with center Z and squares LIKE, MORE, and MATH. If the area of square LIKE is 100, what is the area of square MATH? Your answer must be exact. Show work

Question

Answer 1

WE ARE GIVEN THAT SQUARE LIKE HAS AN AREA OF 100 SQUARE UNITS.

THIS MEANS EACH SIDE MEASURES 10 LINEAR UNITS.

USING THE PYTHAGOREAN THEOREM, THE DIAGONAL MEASURES

FROM THE CENTER OF THE CIRCLE TO THE CIRCUMFERENCE, THAT

IS FROM Z TO K, IS A DISTANCE THAT CAN BE DETERMINED USING

THE PYTHAGOREAN THEOREM. 5*5 + 10*10 = 25 + 100 = 125, SO

THIS DISTANCE = √125 = 5√5. NOTE THAT THIS IS THE RADIUS OF THIS

CIRCLE.

ZO IS ALSO A RADIUS, SO IT MEASURES 5√5 AS WELL. IT IS ALSO THE

HYPOTENUSE OF THE RIGHT TRIANGLE ZOR. THIS MEANS THAT

ZO*ZO = (√125)*(√125) = 125 = ZR*ZR + OR*OR

ZR=5+RE AND OR=RE BECAUSE MORE IS A SQUARE AND ZR IS HALF OF

THE SIDE LENGTH OF SQUARE LIKE, WHICH WE KNOW HAS A SIDE

LENGTH OF 10.

SO, 125 = (5+RE)*(5+RE) + RE*RE = 25 + RE*RE + 10*RE + RE*RE

AND HENCE 125 = 25 + 2*RE*RE + 10*RE; SUBTRACT 25 FROM BOTH SIDES.

NOW, 2RE^2 + 10*RE - 100 = 0. DIVIDE BOTH SIDES BY 2.

RE^2 + 5*RE -50 = 0; FACTORING: (RE+10)(RE-5)=50, SO RE = -10 AND RE =5.

SINCE RE IS A DISTANCE, WE REJECT THE NEGATIVE VALUE AND

CONCLUDE THAT RE = 5.

ZT IS ALSO A RADIUS, SO ZT = √125 AS WELL. LET B BE THE INTERSECTION

POINT OF LINE TH AND ZR. TRIANGLE ZBT IS ALSO A RIGHT TRIANGLE,

SO ZT*ZT = 125 = BT*BT + ZB*ZB. BT = 5+TH AND ZB = 5+EB.

SO, 125 = (5+TH)*(5+TH) + (5+EB)*(5+EB). NOTE THAT EB=MH, AND MH=TH,

SINCE MATH IS A SQUARE AND ITS INTERIOR ANGLES ARE ALL RIGHT

ANGLES.

HENCE, 125 = 25 +10*TH + TH*TH + 25 + 10*TH + TH*TH

OR 125 = 50 +20*TH+2*TH^2; NEXT SUBTRACT 50 FROM BOTH SIDES,

AND WE GET 2*TH^2 +20TH -75 = 0

THE SOLUTIONS TO THIS QUADRATIC EQUATION ARE:

-5 +/- 2.5√10. REJECT THE NEGATIVE SOLUTION, SINCE WE ARE

WORKING WITH DISTANCES. SO THE ANSWER IS -5+2.5√10

THE AREA OF SQUARE MATH IS THEREFORE:

(-5+2.5√10)*(-5+2.5√10) = 25 + 2.5*2.5*10 -25√10

WHICH IS EQUAL TO (87.5 – 25√10) SQUARE UNITS

Answer 2

Here is a start: Label any and all dimensions!

ZK = 5√5, Verify with Pythagoras

125 = OR² + (5 + RE)², yet OR = RE

125 = RE² + (5 + RE)²

125 = 2RE² + 10RE + 25

Can you solve for RE and continue? Mark all new dimensions!