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This is one of our many quizzes on Triangle area and perimeter. Click on this to access more such competitive quizzes on Mensuration.

Triangle area and perimeter Quiz 2

mathematics

 Radhe’s plot is in the shape of a triangle. There are 4 poles along the longest side of the plot and 2 and 3 poles along the other two sides respectively. If none of these poles is at any of the corners of the plot, find the number of triangles that are possible with vertices at these poles.

 By putting a combination formula, we can get the total number of triangles that can be formed then by subtracting the number of triangles of the same side we will get the required number.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/8e2e1b61-512e-4038-ac89-3f72710583407800626319783323448.png"/>$ \\ \text{The total number of points }= 9 \\ \text{The total number of triangles which } \text{can be formed from these points } = { }^{9}{{C}_{3}} = 84 \\ \text{Subtracting the triangles of the same side }= { }^{4}{{C}_{3}}{{-}^{3}}{{C}_{3}}= 5 \\ \text{Hence, }84 - 5 = 79$

 A piece of paper in the form of an isosceles triangle ABC as in the given figure, where $AB = AC = a$ and $BC = b$ is rolled along one of the equal sides to form a cone. Then, the cone is opened again and a smaller isosceles triangle ADE of the triangle is cut out such that $AD=AE=\dfrac{a}{3}$ and $DE=\dfrac{b}{3}$. Again, the piece of paper is rolled along AD. What is the ratio of the surface area of the cone formed by rolling ABC to the surface area of the frustum formed by rolling BDEC if $a=8cm$ and $b=9cm (\text{ take } \pi = 3) \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/05d9bb8d-7001-4965-8cf2-2a41f8c43dae2563198560431757382.png"/>

 First find the area of the cone by the formula $A=\pi rl$ then find the area of the frustum by the formula $A=\pi l(r+r')$.

  $\text{In }\Delta ABC, AB = AC = a = 8cm\text{ and }BC = b = 9cm \\ \text{So, from the cone formed by rolling }\Delta ABC,\text{ slant height, }l = a = 8cm\text{ and } 2\pi r=b \\ r=\dfrac{b}{2\pi }=\dfrac{3}{2}cm \\ \text{So, the area of the cone, }A=\pi rl=36c{{m}^{2}} \\ \text{Now, for the frustum, }l=\dfrac{2}{3}a=\dfrac{16}{3}cm \\ 2\pi r'=\dfrac{b}{3} \\ \Rightarrow r'=\dfrac{9}{6\pi }=\dfrac{1}{2}cm \\ \text{Area of frustum, }A'=3\times \dfrac{16}{3}\times \left( \dfrac{3}{2}+\dfrac{1}{2} \right)=32c{{m}^{2}} \\ \text{Ratio }= A : A' = 36 : 32 = 9 : 8$ 

 In the diagram given below, $\Delta ABC$, a semicircle is drawn on one of its sides BC taken as radius of $14cm$, to meet the other two sides AB and AC at D and E respectively. Find the area of the shaded region if $AB=BC=CA \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/f30933a5-aaae-44f4-bb01-0f518c4a97902904866485855699589.png"/>

 First divide the given figure into four equilateral triangles then subtract the area of the quadrilateral formed by the two triangles from the area of the semi-circle.

  $\text{Here, }\Delta ABC\text{ can be divided into four equilateral triangles.} \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0d1a0cbc-3040-4f1b-b212-23b5eb0c44074113679903615487878.png"/>$ \\ \text{Area of quadrilateral }BDEC=\dfrac{3}{4}\times \dfrac{\sqrt{3}}{4}\times {{(28)}^{2}} \\ \text{Area of semicircle }=\dfrac{1}{2}\times \pi \times {{(14)}^{2}} \\ \text{Area of the shaded region }=\dfrac{2}{3}\left( \dfrac{1}{2}\times \pi \times 196-\dfrac{3}{4}\times \dfrac{\sqrt{3}}{4}\times 784 \right)=196\left( \dfrac{\pi }{3}-\dfrac{\sqrt{3}}{2} \right)c{{m}^{2}}$ 

 An equilateral triangle, ABC, of side a units, is shown in the figure below. If the in-centre of the triangle is denoted as I, find the area of the shaded region, $($in sq. units$)$ given that ID and IE are the in-radii of the triangle and the circle shown passes through the points B, D, I and E. $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/bf8f83c4-f301-4787-85d2-e2493f90eb183780046168742201708.png"/>

 First find the area of the quadrilateral $BDIE$ by the two triangles constituting it, then find the area of the circle, then subtract both for the required shaded region.

  $\text{The circum-radius of the triangle }(IB) = \dfrac{a}{\sqrt{3}} \\ \text{In-radius of the triangle }(ID) = \dfrac{a}{2\sqrt{3}} \\ \text{In, } \Delta BDI,\angle BDI={{90}^{0}} BD=\sqrt{I{{B}^{2}}-I{{D}^{2}}}=\dfrac{a}{2} \\ \text{Area of quadrilateral }BDIE = 2\times \Delta BDI=\left( \dfrac{a}{2} \right)\times \dfrac{a}{2\sqrt{3}}=\dfrac{{{a}^{2}}}{4\sqrt{3}} \\ \text{Radius of circle circumscribing }BEID = \dfrac{1}{2}BI=\dfrac{a}{2\sqrt{3}} \\ \text{Area of the circle }= \pi {{\left( \dfrac{a}{2\sqrt{3}} \right)}^{2}}=\dfrac{\pi {{a}^{2}}}{12} \\ \text{Area of shaded region }= \dfrac{\pi {{a}^{2}}}{12}-\dfrac{{{a}^{2}}}{4\sqrt{3}}=\dfrac{{{a}^{2}}}{12}(\pi -\sqrt{3})$ 

 Let A B be a given straight line segment of a certain positive length. An interior point P of A, B is chosen randomly $($but uniformly$)$. Let P be the probability that AP, PB form the sides of a right triangle both of whose acute angles are at least ${{30}^{0}}$. Then which of the following is correct?

 We know that each acute angle is $\ge {{30}^{0}}$, from there we are getting a range of tangent function, then from there we will get a range for $x$, will give the length of the line segment AP.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0d456736-917a-4733-b7ab-77915e2ef576576793808149064858.png"/>$ \\ \text{Let }AB=a, AP=x, PB=y\text{ and the triangle be named as }A’ B’ P’ \\ \text{Since each acute angle is }\ge {{30}^{0}},\text{ then } \dfrac{y}{x}\ge \dfrac{1}{\sqrt{3}}and\dfrac{x}{y}\ge \dfrac{1}{\sqrt{3}} \\ \text{Since }\tan {{30}^{0}}=\dfrac{1}{\sqrt{3}}\text{ and tan is an increasing function on }\left( 0,\dfrac{\pi }{2} \right) \\ \text{So, }\dfrac{y}{\sqrt{3}}\le x\le \sqrt{3}y,\dfrac{a-x}{\sqrt{3}}\le x\le \sqrt{3}(a-x) \\ \dfrac{a}{\sqrt{3}+1}\le x\le \dfrac{\sqrt{3}a}{\sqrt{3}+1}\Rightarrow a\dfrac{(\sqrt{3}-1)}{2}\le x\le a\dfrac{(3-\sqrt{3})}{2} \\ \text{Thus, }P\text{ lies on a line segment of length }a\left[ \dfrac{(3-\sqrt{3})}{2}-\dfrac{(\sqrt{3}-1)}{2} \right]=a(2-\sqrt{3}) \\ \text{Hence, }P\text{ lies between }0.2\text{ and }0.3$