Rectangle perimeter from area and diagonal

Geometry Level 1

Suppose each diagonal of a rectangle is of length 17 while the area is 120. Find the perimeter of the rectangle.


The answer is 46.

Lawrence Chiou by Lawrence Chiou , 28, USA

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4 solutions

Sam Bealing
Apr 7, 2016

Let x , y x,y be the sides of the rectangle and let P be the perimeter. The question gives us:

x 2 + y 2 = 1 7 2 = 289 , x y = 120 x^2+y^2=17^2=289,xy=120

So: P 2 = ( 2 x + 2 y ) 2 = 4 ( x 2 + y 2 + 2 x y ) = 4 ( 289 + 240 ) = 2116 P^2=(2x+2y)^2=4(x^2+y^2+2xy)=4(289+240)=2116 giving P = ( 2116 ) = 46 P=\sqrt{(2116)}=46 .

This occurs for a rectangle with side lengths 8 8 and 15 15 .

4 Helpful 3 Interesting 1 Brilliant 3 Confused

Moderator note:

For geometric questions, one should check that there are indeed solutions that make geometric sense.

For example, there is no rectangle with a diagonal of 1 and an area of 100.

Let the sides be a , b a,b . Note that we are given , a 2 + b 2 = 289 and a b = 120 a^2+b^2=289 \text{ and } ab=120 Thus , a^2+b^2+2ab=(a+b)^2=529 \tag*{} ⟹ a + b = 23 \implies a+b=23 Or, 2 a b = P = 46 2ab=P=46

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Mevlut Esen
May 9, 2016

Let us call rectangle's long side a and short side b. a^2+b^2=17^2.............. (1)(Pythagoras) a×b=120...................... (2)(Area of rect.) Put (1) and (2) on a^2+b^2+2ab=(a+b)^2 addition square (a+b)^2=17^2+2×120 a+b=23 Perimeter of rectangle is: 2×(a+b)=46

0 Helpful 0 Interesting 1 Brilliant 1 Confused
Sebastian Dulong
May 9, 2016

a^2+b^2=289 (Pythagoras) ab=120 (a+b)^2=589 a+b=23 2a+2b=46

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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