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# Here's a bunch of old problems I found, they might be fun to try

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1:  The sum of the perimeters of two equilateral triangles is 45 inches, and the area of the larger one is 16 times the area of the smaller one. What is the area, in square inches, of the larger triangle? Express your answer in the simplest radical form.

(Hint: The answer is a square root)

2:  Two right triangles share a side as follows, What is the area of triangle ABE?

3:  The medians AD, BE, and CF of triangle ABC intersect at the centroid G. The line through G that is parallel to BC intersects AB and AC at M and N, respectively. If the area of triangle ABC is 810, then find the area of triangle AMN.

(Hint: It's not 8)

4:  The number of inches in the perimeter of a square is equal to the number of square inches in its area. Find the length, in inches, of a side of the square.

Have fun!

Mar 15, 2018

### 4+0 Answers

#1
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1.Let the side  of the smaller triangle   = S

Since the area  of the larger triangle is 16 times as much....each dimension must be 4 times that of the smaller triangle...so the length of its side must be  4S

And the sum of ther perimeters = 45....so

3S  + 3(4S)  = 45

15S  = 45

S  = 3 in

So....the side of the larger trinagle  = 4S  = 4(3)  =12 in

And its area  =  (1/2) 12^2 sin(60)  = 72√3 / 2   =    36√3  in^2   Mar 15, 2018
#2
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4:  The number of inches in the perimeter of a square is equal to the number of square inches in its area. Find the length, in inches, of a side of the square.

Let S  = the side of the square

P  = 4S

A  = S^2

This impies that

4S  = S^2

S^2  - 4S  = 0     factor

S ( S - 4)  = 0

Setting both factors to 0  and solving for S  produces  S  = 0  and S  =4

Reject the first....accept the second

S  = 4 in   Mar 15, 2018
#3
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2.

Position the lower left corner, A , at (0,0)

The equation of AD  is   y  = x

And  let  B  be  positioned at (4,0)

And the equation of  BC    is   y = (-5/4) ( x  - 4)  =  (-5/4)x + 5

Setting  these two equations equal and solving for x we have

x  = (-5/4)x + 5

(9/4) x  = 5

x  = 20/9  =   y  =  height of triangle ABE

So....the area of ABE  = (1/2) (base) (height)  = (1/2) (4) (20/9)  =

40/9  units^2   Mar 15, 2018
#4
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....or you could just do it this way...Building on CPhil's answer

4s=s^2     divide by s

4=s

Mar 15, 2018