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This is one of our many quizzes on Quadrilaterals and polygons. Click on this to access more such competitive quizzes on Geometry.

Quadrilaterals and polygons Quiz 2

mathematics

 If one of the pairs of parallel sides of square with area $\text{625 c}{{\text{m}}^{2}}$ is increased by $75\%$, then increase in the area will be

 Using the formula for area of square and obtaining the side of square. Determine the new side of the rectangle obtained on increasing original side by $75\%$ and then, determine area of rectangle and then increase in area. 

Area of square is $625 cm^{2}$. $ \\ $ Area of square with side $a$ is given by ${{a}^{2}}$ $ \\ \Rightarrow {{a}^{2}}\text{ c}{{\text{m}}^{2}}=\text{625 c}{{\text{m}}^{2}} \\  \Rightarrow a=\text{25 cm} \\ $ Increase factor is given by, $ \\ \Rightarrow \dfrac{75}{100}\times 25 \\  \Rightarrow \dfrac{75}{4} \Rightarrow 18.75 \\ $ So, length of pair of parallel sides become, $ \\ \Rightarrow 25+18.75\text{ cm} \\  \Rightarrow 43.75\text{ cm} \\ $ Area of rectangle with length $l$ and breadth $b$ is equal to $l\times b$ $ \\ $ Then, area of rectangle with length $43.75$ and breadth $25cm$ is $ \\ \Rightarrow 25\times 43.75\text{ c}{{\text{m}}^{2}} \\  \Rightarrow 1093.75\text{ c}{{\text{m}}^{2}} \\ $ Increase in area is, $ \\ $ Area of rectangle with length $43.75$ and breadth $25cm$ $ - $ Area of square given $ \\ \Rightarrow 1093.75\text{ c}{{\text{m}}^{2}}-625\text{ c}{{\text{m}}^{2}} \\  \Rightarrow 468.25\text{ c}{{\text{m}}^{2}}$

 Determine the value of $x$ if the region covered by trapezium is $\text{4 c}{{\text{m}}^{2}}$. <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1c6dd477-0279-47ee-a913-91dd717cc0217896886528365703747.png"/>

 Recall the formula for area of trapezium and obtain the value of $x$. Remember that length cannot be negative.

Area of trapezium is $4cm^2$, $ \\ $ Area of trapezium with parallel sides $a, b$ and height $h$ is equal to $\dfrac{1}{2}\left( a+b \right)h.$ $ \\ $  From given figure, $ \\ \Rightarrow \dfrac{1}{2}\left[ \left( x+2 \right)+\left( x+4 \right) \right]x=4 \\  \Rightarrow \dfrac{1}{2}\left( 2x+6 \right)x=4 \\  \Rightarrow \left( x+3 \right)x=7 \\  \Rightarrow {{x}^{2}}+3x-4=0 \\  \Rightarrow (x-1)(x+4)=0 \\  \Rightarrow x=-4\text{ or 1} \\ $ As height of trapezium cannot be negative, $x = 1$

 A number of students were told to stand in such a way that the distance between any two adjacent students is the same and to make a close polygon by holding the hands with their adjacent students. If interior angle made by students is equal to thrice of the exterior angle made by them, then the number of students is

 Recall that, sum of exterior angles of n - sided polygon is always equal to $360{}^\circ $ and sum of interior angles of n - sided polygon is equal to $\left( n-2 \right)180{}^\circ .$

Let, the number of students be n forming the n - side polygon. $ \\ $ Sum of exterior angles of n - sided polygon is always equal to $360{}^\circ$ and, $ \\ $ Sum of interior angles of n - sided polygon is equal to $\left( n-2 \right)\text{180}{}^\circ$ $ \\ $ Given, sum of interior angles made by students is thrice of sum of exterior angles made by them, $ \\ \Rightarrow \left( n-2 \right)\text{180}{}^\circ =3\cdot 36\text{0}{}^\circ  \\  \Rightarrow \left( n-2 \right)=3\cdot 2 \\  \Rightarrow n=6+2 \Rightarrow n=8 \\ $ So, number of students is equal to 8.

 If the side of the square is $4\text{ cm}$, and E and F are the midpoints of side AB and CD, then find the ratio of area of square ABCD to the area of triangle AOD. <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/2319af28-2508-4473-9e4a-d7f2f3a023302559815125705316487.png"/>

 Using the formula of area of triangle and square and determine the ratio of their ratios.

Side of the square is $4cm$ $ \\ $ Area of square of side $a$ is equal to ${{a}^{2}}$. $ \\ $ So, area of square with side $4cm$ is, $ \\ \Rightarrow {{\left( 4 \right)}^{2}}\text{ c}{{\text{m}}^{2}} \\  \Rightarrow 16\text{ c}{{\text{m}}^{2}} \\ $ From figure, points E and F are midpoint of lines AB and CD $ \\ \Rightarrow \text{AB=2}\cdot \text{AE} \\  \Rightarrow \text{AE=}\dfrac{1}{2}\text{AB} \\  \Rightarrow \text{AE=}\dfrac{1}{2}(4) \\  \Rightarrow \text{AE=}2\text{ cm}\text{.} \\ $ Thus, height of triangle AOD is $\dfrac{1}{2}\left( 4 \right)cm$ and base of triangle is $4cm$ $\\ $ Area of triangle with base  $b$ and height $h$ is $\dfrac{bh}{2}$ $ \\ $ Then, area of triangle with base $4cm$ and height $2cm$ is, $ \\ \Rightarrow \dfrac{\left( 4 \right)\left( 2 \right)}{2}\text{ c}{{\text{m}}^{2}} \\  \Rightarrow 4\text{ c}{{\text{m}}^{2}} \\ $ Ratio of the area of square ABCD to area of triangle AOD is, $ \\ \Rightarrow \dfrac{16}{4} \Rightarrow 4:1$

 If  ABC is an equilateral triangle with area $\dfrac{441\sqrt{3}}{4}\text{ c}{{\text{m}}^{2}}$ , then ${{x}^{2}}+{{y}^{2}}-{{z}^{2}}$  is equal to <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/353774c8-1c77-4f75-8d2c-cc0b2b1d73948905835800100998650.png"/>

 Recall the formula for the area of the equilateral triangle and evaluate the side of an equilateral triangle. Then evaluate the values of $x, y$ and $z$ and then ${{x}^{2}}+{{y}^{2}}-{{z}^{2}}$.

Area of the equilateral triangle is $\dfrac{49\sqrt{3}}{4}\text{ c}{{\text{m}}^{2}}$. $ \\ $ Area of the equilateral triangle with side $a$ is equal to $\dfrac{\sqrt{3}}{4}{{a}^{2}}.$ $ \\  \Rightarrow \dfrac{49\sqrt{3}}{4}\text{ c}{{\text{m}}^{2}}=\dfrac{\sqrt{3}}{4}{{a}^{2}} \\  \Rightarrow 49\text{ c}{{\text{m}}^{2}}={{a}^{2}} \\  \Rightarrow a=7\text{ cm} \\ $  From figure, $ \\  \Rightarrow x+4=7 \\  \Rightarrow x=3 \\  \Rightarrow y+3=7 \\  \Rightarrow y=4 \\  \Rightarrow z+2=7 \\  \Rightarrow z=5 \\ $ So, ${{x}^{2}}+{{y}^{2}}-{{z}^{2}}$ $ \\ \Rightarrow {{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}}-{{\left( 5 \right)}^{2}} \\  \Rightarrow 9+16-25 \\  \Rightarrow 25-25  \Rightarrow 0$