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This is one of our many quizzes on Polygon properties. Click on this to access more such competitive quizzes on Geometry.

Polygon properties Quiz 2

mathematics

 Let ABCD be a square and E, F, G and H be midpoints of line AB, BC, CD and AD respectively. If area of triangle EFGH is $16\text{ c}{{\text{m}}^{2}} $, then perimeter of triangle AEH is $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c1aed22f-dbe2-42ef-bf3a-43aae57e62758875711642252355346.png"/>

  Use the theorem of a square formed by joining midpoints of another square to calculate the area of a big square. Then, to find the perimeter of the triangle, evaluate sides of both squares and add accordingly.

$E, F, G$ and $H $ are midpoints of line AB, BC, CD and AD respectively which are sides of square ABCD. $ \\ $ Then, the area of a square ABCD is twice the area of a square EFGH. $ \\ $ So, area of square ABCD is $ \\ \Rightarrow 2\times 16\text{ c}{{\text{m}}^{2}} \\ \Rightarrow 32\text{ c}{{\text{m}}^{2}} \\ $  Area of the square with side $a$ is ${{a}^{2}}$. $ \\ $ So, side of square EFGH $ \\ \Rightarrow {{a}^{2}}=16\text{ c}{{\text{m}}^{2}} \\ \Rightarrow a=4\text{ cm} $ $ \\ $ And, side of square ABCD $ \\ \Rightarrow {{a}^{2}}=32\text{ c}{{\text{m}}^{2}} \\ \Rightarrow a=4\sqrt{2}\text{ cm} \\ $  Then, AE is  $ \\ \Rightarrow \dfrac{4\sqrt{2}}{2}\text{ cm} \\ \Rightarrow 2\sqrt{2}\text{ cm} \\ $ Thus, perimeter of triangle is, $ \\ \Rightarrow \left( 4+2\sqrt{2}+2\sqrt{2} \right)\text{ cm} \\ \Rightarrow 4\left( 1+\sqrt{2} \right)\text{ cm} \\ $

  The $6$-sided regular polygon ABCDEF has a perimeter equal to$\text{36 cm}$. Perpendicular distance of each side of the polygon from its centre O is equal to$4\text{ cm}$. Then area of polygon is equal to $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/2b5c7c6c-ac90-4dda-83de-120dc48fe1677665651729010157772.png"/>

 Use construction to divide the hexagon into six congruent triangles. Obtain the area of the single triangle and multiply the area by $6$ to calculate the area of the hexagon.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9ab5d1a6-ca74-4544-aed6-49627a3399ff3668405923386742959.png"/> $ \\ $ Join line segments AD, BE and CF. $ \\ $ Perimeter of a regular polygon is the sum of its sides. $ \\  \Rightarrow 6a=36\text{ cm} \\ \Rightarrow a=6\text{ cm} \\ \text{Area of triangle with height ’h’ and base ‘b’ is }\dfrac{bh}{2}.\\ $  Then, area of triangle OED is, $ \\ \Rightarrow \dfrac{6\times 4}{2}\text{ c}{{\text{m}}^{2}} \\ \Rightarrow 12\text{ c}{{\text{m}}^{2}} \\ $ Diagonal of a polygon divides its area into equal measures. $ \\ $ So, area of polygon ABCDEF is, $ \\ \Rightarrow 6\times 12\text{ c}{{\text{m}}^{2}} \\ \Rightarrow 72\text{ c}{{\text{m}}^{2}} $

 Let ABCEF be pentagon. Area of triangle ABF is $10\text{ c}{{\text{m}}^{2}} $, length of BF is $10\text{ cm}$ and length of AD is$22\text{ cm}$, then area of triangle DBF is $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c70795a0-7386-4a68-b5fc-8581124574048078625313875238364.png"/>

 Use area of triangle and determine the height of triangle AFB. Analyse the figure and evaluate the length GD and obtain the area of DBF.

 Area of the triangle with base $b$ and height $h$ is $\dfrac{bh}{2}$ $ \\ $ Then, for triangle ABF $ \\ \Rightarrow \dfrac{10\times h}{2}=10\text{ c}{{\text{m}}^{2}} \\ \Rightarrow h=2\text{ cm} \\ \text{From figure, }\\ GD = AD – AG\\ \Rightarrow GD=\left( 22-2 \right)\text{ cm} \\ \Rightarrow GD=20\text{ cm} \\ $ Then, area of triangle DBF is, $ \\ \Rightarrow \dfrac{10\times 20}{2}\text{ c}{{\text{m}}^{2}} \\ \Rightarrow 100\text{ c}{{\text{m}}^{2}} $

 If CE is straight line, then measure of angle  $\angle \text{ADE} $ is $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/990652d8-2293-4e6b-99b2-4572a28f4eeb4268764969297526269.png"/>

 Use the formula for calculating the sum of an interior angle of n-sided polygon for polygon ABCD. Use the concept of the sum of linear pair angles and obtain a measure of angle $\angle \text{ADE} $.

 Sum of interior angles of a n-sided polygon is $\left( n-2 \right){{180}^{{}^\circ }}$ $ \\ $ Then, for polygon ABCD, $ \\ \Rightarrow {{105}^{{}^\circ }}+{{75}^{{}^\circ }}+{{85}^{{}^\circ }}+{{x}^{{}^\circ }}={{180}^{{}^\circ }}(4-2) \\ \Rightarrow {{x}^{{}^\circ }}={{95}^{{}^\circ }} \\ $ As, CE is a straight line, then angles $\angle ADC$ and $\angle ADE$ forms a linear pair $ \\ $ Then, $ \\ \Rightarrow {{x}^{{}^\circ }}\text{+}\angle \text{ADE=18}{{\text{0}}^{{}^\circ }} \\ \Rightarrow \angle \text{ADE=8}{{\text{5}}^{{}^\circ }}  $

 A park ABCD with area $1600\text{ }{{\text{m}}^{2}}$was divided into two areas, AED for children and ABCDE for adults and old age people. Perpendicular from E to line AD divides it into ratio of $1:3$, then area allotted for adults and old people is $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9b1eea0c-66ac-4fa3-a50b-af743742fb1b2986676853914574794.png"/>

 Find area of square and find measure of square’s side. Using the ratio method, obtain a measure of FD. Find the area of the triangle and subtract the area from the square's area to obtain the area of ABCDE.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/184b21ad-5c8d-48d3-9a10-1429685c29981687046669652896649.png"/> $ \\ $ Construct perpendicular from point E on line AD. $ \\ $ Area of the square with side $a$ is ${{a}^{2}}$ $ \\ $ Then, for square ABCD, $ \\ \Rightarrow {{a}^{2}}=1600\text{ }{{\text{m}}^{2}} \\ \Rightarrow a=40\text{ m} \\ $ Perpendicular from E to line AD divides it into ratio of $1:3$, $ \\ $ So, $ \\ \Rightarrow \dfrac{AF}{FD}=\dfrac{1}{3} \\ \Rightarrow \dfrac{AD-FD}{FD}=\dfrac{1}{3} \\ \Rightarrow 3AD=4FD \\ \Rightarrow 3\times 40=4FD \\ \Rightarrow FD=30\text{ m} \\ $ In triangle, AED $ \\ \Rightarrow \tan {{45}^{{}^\circ }}=\dfrac{EF}{FD} \\ \Rightarrow 1=\dfrac{EF}{30\text{ m}} \\ \Rightarrow EF=30\text{ m} \\ $ Area of triangle with base $b$ and height $h$ is $\dfrac{bh}{2}$ $ \\ $ So, area of triangle AED, $ \\ \Rightarrow \dfrac{30\times 30}{2}\text{ }{{\text{m}}^{2}} \\ \Rightarrow 450\text{ }{{\text{m}}^{2}} \\ $ So, area for adults and old people is, $ \\ \Rightarrow \left( 1600-450 \right)\text{ }{{\text{m}}^{2}} \\ \Rightarrow 1150\text{ }{{\text{m}}^{2}}$