Question

The figure above shows that the shaded triangular region with a hypotenuse of 5 cm has been removed from a rectangular tile with dimensions x cm by y cm. Of the following which best approximates the area in square cm of the tile before the piece was removed

Answer 1

Check the picture below.

Mind you that we have four 30-60-90 triangles.

Answer 2

The area of given rectangle before the piece was removed is 65 cm² approximately.

What is area of rectangle?

The area of a rectangle (A) is the product of its length 'a' and width or breadth 'b'. So, Area of Rectangle = (a × b) square units.

In Δ DEF,

∠EDF + ∠DEF + ∠DFE = 180° (angle sum property)

30° + 90° + ∠DFE = 180°

∠DFE = 60°

$cos F = \frac{EF}{FD}(\frac{base}{hypotenuse}\\ \\ cos(60) = \frac{5}{FD}\\ \\FD = 5 * 2= 10$

$cos D = \frac{ED}{FD}(\frac{base}{hypotenuse}\\ \\ cos(30) = \frac{ED}{10}\\ \\ED = \frac{\sqrt{3} }{2} * 10= 5\sqrt{3}$

In Δ AED,

∠AED + ∠EAD + ∠EDA = 180° ( angle sum property)

∠AED + 90° + 30°= 180°

∠AED = 60°

$sinE = \frac{AD}{DE} (\frac{perpendicular}{hypotenuse}\\ \\sin(60) = \frac{AD}{5\sqrt{3} } \\\\AD= \frac{\sqrt{3} }{2}*5\sqrt{3} \\\\AD = 7.5$

In Δ DFC,

∠FDC + ∠DCF + ∠DFC = 180° (angle sum property)

30° + 90° + ∠DFC = 180°

∠DFC = 60°

$sinF = \frac{DC}{DF} \\\\sin(60) = \frac{DC}{10}\\ \\\frac{\sqrt{3} }{2} *10 = DC\\\\DC = 5\sqrt{3}$

Area of rectangle = length * breadth = xy = AD * DC = $7.5 * 5\sqrt{3} = 64.95$

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