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CAT 2018 Slot 1 — QA Question 15

Basics of QuadrilateralsEasy

Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of length of EB to that of CG is

Answer & solution

  • A

    2 : 5

  • B

    4 : 9

  • C

    3 : 8

  • 1 : 3

Solution

Easy

Each side of the inner square is the hypotenuse of a right triangle cut from a corner, with legs that add to a full side of the outer square. Equate areas, get a quadratic in the ratio of the two legs, then pick the root consistent with CG>EBCG > EB.

E F G H A B C D
1

Label the corner cuts. Let EB=xEB = x and BF=yBF = y. By the symmetry of the inscribed square the four corner triangles are congruent, so each side of the outer square equals x+yx+y, and each side of the inner square equals the hypotenuse x2+y2\sqrt{x^2+y^2}.

outer side=x+y,inner side=x2+y2\begin{aligned} &\text{outer side} = x+y,\qquad \text{inner side} = \sqrt{x^2+y^2} \end{aligned}
2

Apply the area condition. Inner area is 62.5%=5862.5\% = \tfrac58 of the outer area:

x2+y2=58(x+y)2(area ratio)⇒ 8x2+8y2=5x2+5y2+10xy⇒ 3x2−10xy+3y2=0\begin{aligned} &x^2+y^2 = \tfrac58 (x+y)^2 \quad\text{(area ratio)}\\ &\Rightarrow\ 8x^2+8y^2 = 5x^2+5y^2+10xy\\ &\Rightarrow\ 3x^2 - 10xy + 3y^2 = 0 \end{aligned}
3

Solve for the ratio. Divide by y2y^2 and set r=xyr=\tfrac{x}{y}:

3r2−10r+3=0⇒ r=10±100−366=10±86⇒ r=3  or  r=13\begin{aligned} &3r^2 - 10r + 3 = 0\\ &\Rightarrow\ r = \frac{10\pm\sqrt{100-36}}{6} = \frac{10\pm 8}{6}\\ &\Rightarrow\ r = 3 \ \text{ or }\ r = \tfrac13 \end{aligned}
4

Pick the right root. Here EB=xEB=x and CG=yCG=y (the longer leg). Since CG>EBCG>EB we need $x EB:CG=x:y=1:3\begin{aligned} &EB : CG = x : y = 1 : 3 \end{aligned}

EB:CG=1:3EB : CG = \mathbf{1 : 3}
CAT 2018 Slot 1 QA Q15: Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a squar — Solution | TheCATExam