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Find the fourth side of a quadrilateral inscribed in a circle having one of its sides equal to 20 m as its diameter, and the other two sides adjacent to the diameter, and the other two sides adjacent to the diameter are 8 m and 12 m respectively.

Guest Dec 28, 2014

#2**+10 **

I am not sure at the moment if Chris and I have interpreted the question the same.

(This diagram is drawn to scale)

Here is my interpretation.

$$\\

Let length BC = x

Now I will use the cosine rule:

$$\boxed{cosA=\frac{b^2+c^2-a^2}{2bc}}\\\\

so\\\\

cos\theta=\frac{8^2+16^2-x^2}{2\times8\times16}\quad and \quad cos\theta=\frac{\sqrt{336}^2+12^2-x^2}{2\times \sqrt{336}\times12}\\\\

Hence\\\\

\frac{8^2+16^2-x^2}{2\times8\times16}=\frac{\sqrt{336}^2+12^2-x^2}{2\times \sqrt{336}\times12}\\\\

\frac{320-x^2}{128}=\frac{480-x^2}{ 12\times 4\sqrt{21}}\\\\

\frac{320-x^2}{128}=\frac{480-x^2}{ 12\times 4\sqrt{21}}\\\\

\frac{320-x^2}{8}=\frac{480-x^2}{ 3\sqrt{21}}\\\\

3\sqrt{21}(320-x^2)=8(480-x^2)\\\\

3\sqrt{21}\times320-3\sqrt{21}\;x^2\;=\;3840-8x^2\\\\

960\sqrt{21}-3840\;=\;(-8+3\sqrt{21})\;x^2\\\\$$

$$\\x^2=\frac{960\sqrt{21}-3840}{-8+3\sqrt{21}}\\\\

x=+\sqrt{\frac{960\sqrt{21}-3840}{-8+3\sqrt{21}}}\\\\$$

$${\sqrt{{\frac{\left({\mathtt{960}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{21}}}}{\mathtt{\,-\,}}{\mathtt{3\,840}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{21}}}}\right)}}}} = {\mathtt{9.864\: \!242\: \!223\: \!858\: \!689}}$$

so

The forth side is approx 9.86metres long

Since our answers are the same I guess we interpreted the question the same after all

Melody Dec 29, 2014

#1**+10 **

I'm assuming that you're saying that one side of the quadrilateral is the same as the circle's diameter, and we have two adjacent sides to this that are 8 and 12m, respectively. Then the circle's radius = 10 m

To find the remaining side, we need to first find the central angles intercepting the 8 and 12 m sides.....

We can use the Law of Cosines in both cases....for the 12m side, we have

12^2 = 10^2 + 10^2 - 2(10)(10)cosΘ ......simplifying, we have

cosΘ= 7/25 = cos^{-1}(7/25) = about 73.739795291688°

For the 8m side, we have

8^2 = 10^2 + 10^2 - 2(10(10)cosΘ ...simplifying again, we have

cosΘ = 17/25 = cos^{-1}(17/25) = about 47.156356956404°

So, since the diameter would span 180° of arc, the remaing side must span (180 - 73.739795291688 - 47.156356956404)° = 59.103847751908° of arc

So...the remaining side... (s).... is given by

s^2 = 10^2 + 10^2 - 2(10)(10)cos(59.103847751908)

s = √[10^2 + 10^2 - 2(10)(10)cos(59.103847751908)] = about 9.86 m

This makes sense....the greater side intercepts the greatest arc, the next greatest side intercepts the next greatest arc, etc.

CPhill Dec 28, 2014

#2**+10 **

Best Answer

I am not sure at the moment if Chris and I have interpreted the question the same.

(This diagram is drawn to scale)

Here is my interpretation.

$$\\

Let length BC = x

Now I will use the cosine rule:

$$\boxed{cosA=\frac{b^2+c^2-a^2}{2bc}}\\\\

so\\\\

cos\theta=\frac{8^2+16^2-x^2}{2\times8\times16}\quad and \quad cos\theta=\frac{\sqrt{336}^2+12^2-x^2}{2\times \sqrt{336}\times12}\\\\

Hence\\\\

\frac{8^2+16^2-x^2}{2\times8\times16}=\frac{\sqrt{336}^2+12^2-x^2}{2\times \sqrt{336}\times12}\\\\

\frac{320-x^2}{128}=\frac{480-x^2}{ 12\times 4\sqrt{21}}\\\\

\frac{320-x^2}{128}=\frac{480-x^2}{ 12\times 4\sqrt{21}}\\\\

\frac{320-x^2}{8}=\frac{480-x^2}{ 3\sqrt{21}}\\\\

3\sqrt{21}(320-x^2)=8(480-x^2)\\\\

3\sqrt{21}\times320-3\sqrt{21}\;x^2\;=\;3840-8x^2\\\\

960\sqrt{21}-3840\;=\;(-8+3\sqrt{21})\;x^2\\\\$$

$$\\x^2=\frac{960\sqrt{21}-3840}{-8+3\sqrt{21}}\\\\

x=+\sqrt{\frac{960\sqrt{21}-3840}{-8+3\sqrt{21}}}\\\\$$

$${\sqrt{{\frac{\left({\mathtt{960}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{21}}}}{\mathtt{\,-\,}}{\mathtt{3\,840}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{21}}}}\right)}}}} = {\mathtt{9.864\: \!242\: \!223\: \!858\: \!689}}$$

so

The forth side is approx 9.86metres long

Since our answers are the same I guess we interpreted the question the same after all

Melody Dec 29, 2014