A geometry problem by A Former Brilliant Member

Geometry Level 2

The lengths of sides of a right triangle form an arithmetic progression with a common difference of 4. If the area of the triangle is 96 square units, what is the sum of this progression?


The answer is 48.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Let x x be the smallest side, then the other sides are x + 4 x+4 and x + 8 x+8 .

In A.P.

a 1 = x a_1 = x

a 2 = a 1 + 4 = x + 4 a_2 = a_1 + 4 = x + 4

a 3 = a 1 + 8 = x + 8 a_3 = a_1 + 8 = x + 8

Since x + 8 x+8 is the largest, it is the hypotenuse.

A A = = 1 2 \frac{1}{2} ( x ) ( x + 4 ) (x)(x+4)

96 96 = = 1 2 \frac{1}{2} ( x 2 + 4 x ) (x^2+4x)

192 192 = = x 2 + 4 x x^2+4x

solving for x x , we have

x x = = 12 12

The terms of the A.P. are 12 , 16 , a n d 20 12,16, and 20 , so the sum is 48 \boxed{48}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

The area of the triangle being 96 (or else the common difference being 4!) is superfluous to requirements. The only right-angled triangles with sides in Arithmetic Progression would be those where the sides are in the ratio 3:4:5. Multiplying by 4 (to get a common difference of 4) gives 12, 16, 20 as the sides. The problem would, in my opinion, be more interesting without mentioning the common difference being 4...

Paul Hindess - 4 years, 5 months ago

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