Find the area of the garden planted with flowers.

Question

A garden is shaped in the form of a regular heptagon (seven-sided), $MNSRQPO$. A circle with centre $T$ and radius $25\ \text{m}$ circumscribes the heptagon as shown in the diagram below. The area of $\triangle MSQ$ is left for a children's playground, and the rest of the garden is planted with flowers. Find the area of the garden planted with flowers.

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I am confused about this question. Can you please give me the hint. Do I need to find first area of a circle and then subtract the area of a triangle from it?

Answer 1

You are supposed to find the area of the triangle and subtract it from the area of heptagon.

Area of heptagon is pretty straightforward i.e. $\pi r^2$

And for area of triangle $MSQ$, consider line $TM $ and $TS$ whose length are nothing but the circle's radius. And the triangle formed will be $MTS$ and the angle formed at $T$ will be equal to 2 parts of the $360$ degree divided into 7 equal parts, because between $TM$ and $TS$ 2 sides of 7 sided quadrilateral is included, so the angle at $T$ will be $\frac{360}{7} * 2$. And after that, you should be able to find the length of the side $MS$ of triangle $MSQ$ by making a perpendicular bisector at angle $T$ formed by triangle $MTS$.

Likewise you can find all the sides of the isosceles triangle and then its area.

Hope you get my hint. Try reading it one more time, if it doesn't make sense.

Answer 2

The side $a$ of a regular $n$-sided polygon, given radius of its circumscribed circle $r$ is $$a=2r\sin\left(\frac{π}{n}\right)\tag 1$$ Also, the area of polygon is $$A=\frac{1}{2}nr^2\sin\left(\frac{2π}{n}\right)\tag 2$$ Refer this post.

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Now, to find the area of $\triangle MSQ$, we need the base $MQ$ and height $SA$. $A$ is the point of intersection of $MQ$ with the $x$-axis.

Knowing that any two consecutive vertices of heptagon subtend an angle of $\frac{2π}{7}$ at the center of the circumscribed circle (here the origin $T$), we get $\angle MTA=\frac{3π}{7}$.

$$\begin{align}\text{Area}(\triangle MSQ)&=\frac{1}{2}MQ\cdot SA\\ &=\frac{1}{2}\left(2\cdot 25\sin\frac{3π}{7}\right)\left(25+25\cos\frac{3π}{7}\right)\\ &=625\sin\frac{3π}{7}\left(1+\cos\frac{3π}{7}\right)\tag 3\end{align}$$

From $(2)$ and $(3)$ the area of the garden planted with flowers is $$1875\sin\left(\frac{2π}{7}\right)-625\sin\left(\frac{π}{7}\right)\ \text{m}^2$$

Answer 3

To find the area of Triangle MSQ, we need the measure of two sides and one included angle.

MS = SQ, so if we just find the measure of one, we'll have two sides.

The angle MSQ can be found using circle theorems.

Angle MTS = 2(360/7) because it is formed by two central angles. Angle MQS = 1/2 of Angle MTS because an angle subtended by a chord at the centre is twice the angle subtended at the circumference. Angle SMQ = Angle MQS = 360/7 (because the triangle MSQ is isosceles - base angles are equal) Angle MSQ = 180 - 2(360/7)

To find MS, first try finding MN, then use the isosceles triangle MNS to find MS.

Worked solution attached.