# Area 9

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**(a) In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides.**

**Find the area of the remaining portion of the triangle.**

**(b) The area of a circle inscribed in an equilateral triangle is 154 cm ^{2}.**

**Find the perimeter of the triangle and the area of the triangle not included in the circle.**

**(Use pi = 22 / 7 and √3 = 1.73). Give you answer correct to one decimal place.**

**Solution:** (a) Area of equilateral triangle ABC with side 24 cm

= √3/4 x a^{2} = √3/4 x 24^{2} = 144 √3 cm^{2} ----------------------(1)

Let r be the radius of the inscribed circle. Then

Area of triangle ABC = Area of triangle OBC + Area of triangle OCA + area of triangle OAB

= 1/2 x r x BC + 1/2 x r CA + 1/2 x r x AB

= 1/2 x r x (BC + CA + AB)

= 1/2 x r x (24 + 24 + 24)

= 1/2 x r x 72 = 36 r cm^{2} ------------------------------------------------------------------------ (2)

From (1) and (2), we get 36r = 144 √3

Implies, r = 4 √3

Therefore, Area of the inscribed circle = Π r^{2}

= 22/7 x (4 √3)^{2} = 150.85 cm^{2}

Therefore, Area of the remaining portion of the triangle

= area of triangle ABC – area of inscribed circle

= 144 √3 – 150.85 = 144 x 1.73 – 150.85

= 249.408 – 150.85 = 98.551 cm^{2}

(b) Let the radius of the inscribed circle be r. Then, the area of this circle = Π r^{2} = 154 (given)

That is 22/7 x r^{2} = 154

Implies r^{2} = 154 x 7/22 = 49

r = √49 = 7 cm.

In a triangle, the center of the inscribed circle is the point of intersection of the angular bisectors and in an equilateral triangle,

these bisectors are also the altitudes and medians whose point of intersection divides the medians in the ratio 2 : 1.

Let altitude AD = h

Therefore, angle ADB = 90^{0} and OD = 1/3 AD; i.e. r = h/3

Implies, h = 3r = 3 x 7 = 21 cm

Let each side of the triangle be a, then

In right triangle ADB, AB^{2} = AD^{2} + BD^{2}

Implies, a^{2} = h^{2} + (a/2)^{2}

Implies, 4 a^{2} = 4 h^{2} + a^{2}

Implies, 3 a^{2} = 4h^{2}

Implies, a = 2h / √3

Implies, a = 2 / √3 x 21 = 2 / √3 x 21 x √3 / √3

= 14 x √3 cm

Therefore, Perimeter of the triangle = 3a = 3 x 14 √3 = 42√3

= 42 x 1.73 = 72.7 cm

Area of the triangle = √3/4 x a^{2}

= √3/4 x (14 √3)^{2}

= 147 √3 cm^{2}

= 147 x 1.73 cm^{2} = 254.31 cm^{2}

Hence, Area of the triangle not included in the circle

= area of the triangle – area of the inscribed circle

= (254.31 - 154) = 100.31 cm^{2} = 100.3 cm^{2} (correct to one decimal place)