OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O. i) If the radius of the circle is 10 cm, find the area of the rhombus. ii) If the area of the rhombus is 32√3 cm2, find the radius of the circle. - Noon Academy

OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O. i) If the radius of the circle is 10 cm, find the area of the rhombus. ii) If the area of the rhombus is 32√3 cm2, find the radius of the circle.

Class 10 | Exercise 18C | ICSE | Maths | Selina | Tangents and Intersecting Chords

OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O. i) If the radius of the circle is 10 cm, find the area of the rhombus. ii) If the area of the rhombus is 32√3 cm2, find the radius of the circle.

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(C) - 2

(i) Given,

$radius\text{ }=\text{ }10\text{ }cm$

In rhombus

$OABC$

$OC\text{ }=\text{ }10\text{ }cm$

So,

$OE\text{}=\text{}{\scriptscriptstyle1\!/\!{}_2}\text{}x\text{}OB\text{}=\text{}{\scriptscriptstyle1\!/\!{}_2}\text{}x\text{}10\text{}=\text{}5\text{}cm$

Now, in right

$\vartriangle OCE$

$O{{C}^{2}}~=\text{ }O{{E}^{2}}~+\text{ }E{{C}^{2}}$

${{10}^{2}}~=\text{ }{{5}^{2}}~+\text{ }E{{C}^{2}}$

Or,

$E{{C}^{2}}~=\text{ }100\text{ }-\text{ }25\text{ }=\text{ }75$

$EC\text{ }=\text{ }\surd 75\text{ }=\text{ }5\surd 3$

Hence,

$AC\text{ }=\text{ }2\text{ }x\text{ }EC\text{ }=\text{ }2\text{ }x\text{ }5\surd 3\text{ }=\text{ }10\surd 3$

We know that,

Area of rhombus

$=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }OB\text{ }x\text{ }AC$

$=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }10\text{ }x\text{ }10\surd 3$

So,

$=\text{ }50\surd 3\text{ }c{{m}^{2}}~\approx \text{ }86.6\text{ }c{{m}^{2}}$

(ii) We have the area of rhombus

$=\text{ }32\surd 3\text{ }c{{m}^{2}}$

But area of rhombus

$OABC\text{ }=\text{ }2\text{ }x\text{ }area\text{ }of\text{ }\vartriangle OAB$

Area of rhombus

$OABC\text{ }=\text{ }2\text{ }x\text{ }\left( \surd 3/4 \right)\text{ }{{r}^{2}}$

Where

$r$

is the side of the equilateral triangle

$OAB$

$2\text{ }x\text{ }\left( \surd 3/4 \right)\text{ }{{r}^{2}}~=\text{ }32\surd 3$

$\surd 3/2\text{ }{{r}^{2}}~=\text{ }32\surd 3$

${{r}^{2}}~=\text{ }64$

So,

$r\text{ }=\text{ }8$

Therefore, the radius of the circle is

$8\text{ }cm$

i

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