A circle is inscribed in an isosceles trapezium ABCD with parallel sides AB and CD. If AB … — JAMB UTME

A circle is inscribed in an isosceles trapezium $ABCD$ with parallel sides $AB$ and $CD$ . If $AB = 18\text{ cm}$ and $CD = 8\text{ cm}$ , what is the area of the region inside the trapezium that lies outside the circle?

A
$(156 - 144\pi)\text{ cm}^2$
B
$(312 - 36\pi)\text{ cm}^2$
$(156 - 36\pi)\text{ cm}^2$ Answer
D
$(169 - 36\pi)\text{ cm}^2$

Answer

(156 - 36\pi)\text{ cm}^2

For an isosceles trapezium with an inscribed circle, the sum of opposite sides must be equal (

AB + CD = AD + BC = 26\text{ cm}

), giving slant side length

13\text{ cm}

. Using Pythagoras, the perpendicular distance (height) is

\sqrt{13^2 - 5^2} = 12\text{ cm}

. The area of the trapezium is

\frac{1}{2}(18 + 8)(12) = 156\text{ cm}^2

. The inscribed circle has radius equal to half the height (

6\text{ cm}

), so its area is

\pi \times 6^2 = 36\pi\text{ cm}^2

. Subtracting the circle area from the trapezium area yields

(156 - 36\pi)\text{ cm}^2

Step-by-Step Solution

Apply the property of a tangential quadrilateral to find the non-parallel sides

For a quadrilateral with an inscribed circle, the sum of opposite sides is equal:

AB + CD = AD + BC = 18 + 8 = 26\text{ cm}

. Since trapezium

ABCD

is isosceles,

AD = BC = 13\text{ cm}

Tangential quadrilaterals have equal sums of opposite side lengths.

Calculate the height

h

of the trapezium using the Pythagorean theorem

Dropping vertical altitudes from top vertices

C

and

D

creates right triangles at the base with horizontal leg

\frac{18 - 8}{2} = 5\text{ cm}

. Thus,

h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = 12\text{ cm}

The height of the trapezium forms the perpendicular leg of the right-angled side triangle.

Find the area of the trapezium and the inscribed circle

Area of trapezium

= \frac{1}{2}(AB + CD) \times h = \frac{1}{2}(18 + 8) \times 12 = 156\text{ cm}^2

. The diameter of the inscribed circle equals the height

h = 12\text{ cm}

, so its radius is

r = 6\text{ cm}

. Area of circle

= \pi r^2 = 36\pi\text{ cm}^2

The diameter of a circle inscribed between parallel bases equals the vertical height between those bases.

Subtract the area of the circle from the area of the trapezium

Remaining Area

= 156 - 36\pi\text{ cm}^2

The region inside the trapezium but outside the circle is the difference between their areas.

Key Concept

Perimeter and Area of Composite Figures and Inscribed Shapes

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