The perimeter of a rhombus is $148 \text{ cm}$ and one of its diagonals is $70 \text{ cm}$. Its area is equal to the area of a trapezium whose parallel sides are $x \text{ cm}$ and $50 \text{ cm}$ and the distance between the parallel sides is $14 \text{ cm}$. What is the value of $x$?

What we know: Perimeter = $148$ cm

A rhombus has 4 equal sides, so:

Side length = $\dfrac{148}{4} = 37$ cm

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What we know:

• One diagonal = $70$ cm
• Side length = $37$ cm

The diagonals of a rhombus cut each other at right angles and split each other in half. This creates right-angled triangles with:

• One leg = $\dfrac{70}{2} = 35$ cm
• Other leg = $\dfrac{d_2}{2}$ (unknown)
• Hypotenuse = $37$ cm

Using Pythagoras theorem:

$35^2 + \left(\dfrac{d_2}{2}\right)^2 = 37^2$

$1225 + \left(\dfrac{d_2}{2}\right)^2 = 1369$

$\left(\dfrac{d_2}{2}\right)^2 = 144$

$\dfrac{d_2}{2} = 12$

So $d_2 = 24$ cm

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Area of rhombus = $\dfrac{1}{2} \times d_1 \times d_2$

Area = $\dfrac{1}{2} \times 70 \times 24 = 840$ cm²

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For the trapezium:

• Parallel sides are $x$ cm and $50$ cm
• Height = $14$ cm

Area of trapezium = $\dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

Area = $\dfrac{1}{2} \times (x + 50) \times 14 = 7(x + 50)$

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Since both areas are equal:

$840 = 7(x + 50)$

$\dfrac{840}{7} = x + 50$

$120 = x + 50$

$x = 70$ cm