Perimeter of trapezium

Geometry Level 2

If the altitude and perimeter of the trapezium are $3$ and $A+B \sqrt{2}+C \sqrt{3},$ respectively, find $A+B+C$ .

The answer is 35.

2 solutions

Chew-Seong Cheong
May 29, 2016

Let the trapezium be $PQRS$ , where $PQ = 10$ . The right slanting side $QR = \sqrt{3^2+4^2} = 5$ . The left slanting side $SP = \sqrt{3^2+3^2} = 3\sqrt{2}$ . Therefore, the perimeter $p$ is:

$\begin{aligned} p & = PQ+QR+RS+SP \\ & = 10+5+(4+10+3)+3\sqrt{2} \\ & = 32 + 3\sqrt{2} + (\color{#D61F06}{0})\sqrt{3} \end{aligned}$

$\implies A+B+C = 32+3+0 = \boxed{35}$

Fully explained solution

Prince Loomba - 5 years ago

Perhaps specify that the drawing is not at the right scale is helpful to not be confused with the other (little) side of length 3.

F_Adrien F_Adrien - 2 years, 7 months ago

Thanks for the update

Tabitha Downing - 4 months, 2 weeks ago

Prince Loomba
May 29, 2016

Use pythagorean theorem 2 times and get A=32 and B =3 and C=0

Perimeter of trapezium

The answer is 35.

2 solutions

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