Trapezoidal Problem

Geometry Level 2

A right trapezoid has an area of ​​125 cm2, a height of X, and bases measuring 2X and X 2 \frac {X}{2} .

Find the measure of the sloping side.

6 \sqrt{6} 8 \sqrt{8} 6 × 73 6 \times \sqrt{73} 5 × 13 5 \times \sqrt{13}

Victor Paes Plinio by Victor Paes Plinio , 21, Brazil

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3 solutions

Filly Mare
Jun 10, 2014

A r e a = h ( b 1 + b 2 ) 2 ⇒ 125 = X ( 2 X + X 2 ) 2 = X 2 + X 2 4 Area=\frac{h(b_1 + b_2 )}{2} \Rightarrow 125=\frac{X(2X+ \frac{X}{2})}{2}=X^2+\frac{X^2}{4} ⇒ 500 = 5 X 2 ⇒ x = 10 \Rightarrow 500=5X^2 \Rightarrow x=10 b 1 = 20 , b 2 = 5 h = 10 b_1=20, \quad b_2=5 \quad h=10 Gives right angled triangle of sides 10 and 15. By Pythagoras the sloped side is 100 + 225 = 325 = 25 ⋅ 13 = 5 13 \sqrt{100+225}=\sqrt{325}=\sqrt{25 \cdot 13}=\boxed{5\sqrt{13}}

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Hello,

as for right trapezoid,

given that those bases , b1 = 2X , b2= X/2, H = X, A = 125 cm^(2)

by applying the trapezoid formula,

A = 1/2 x h x (b1+b2)

125 = 0.5(X)(2X + 0.5X)

125 = 0.5(X)(2.5X)

125 = 1.25 X^(2)

125 / 1.25 = X^(2)

X = 10

as for bases , b1= 2(10) = 20 cm , b2 = 10 /2 = 5 cm, as well as height , h = 10 cm,

as for the slope, by using phythagoras theorem.....

a=10cm , b= 20 -5 = 15cm....

let the length of the slope = c,

c = (225+100)^(0.5)

c = 18.02776 as the the answer is equivalent to 5 x 13^(0.5)....

thanks...

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Mauricio Jadulos
Apr 6, 2014

(125-x^2/2) = 1/2(3x/2)x....x = 10... by Phytagorean theorem .C= .square root ( x^2 + 9x^2/4) = 5 square root13

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