{"id":26147,"date":"2019-03-15T18:32:47","date_gmt":"2019-03-15T13:02:47","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=26147"},"modified":"2019-09-11T15:44:32","modified_gmt":"2019-09-11T10:14:32","slug":"mensuration-questions-for-ssc-cgl-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/mensuration-questions-for-ssc-cgl-pdf\/","title":{"rendered":"Mensuration Questions for SSC CGL PDF"},"content":{"rendered":"

Mensuration Questions for SSC CGL PDF:<\/strong><\/span><\/h2>\n

Download SSC CGL Mensuration questions with answers PDF based on previous papers very useful for SSC CGL exams. 25 Very important Mensuration objective questions for SSC exams.<\/p>\n

Download Mensuration Questions for SSC CGL PDF<\/a><\/p>\n

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Question 1:\u00a0<\/b>A copper wire is bent in the form of an equilateral triangle, and has an area $121\\sqrt{3}$ cm2 <\/sup>. If the same wire is bent into the form of a circle, the area(in cm2<\/sup>) enclosed by the wire in(Take $\\pi = \\frac{22}{7}$)<\/p>\n

a)\u00a0364.5<\/p>\n

b)\u00a0693.5<\/p>\n

c)\u00a0346.5<\/p>\n

d)\u00a0639.5<\/p>\n

Question 2:\u00a0<\/b>An equilateral triangle of side 6cm has its corners cut off to form a regular hexagon. Area (in cm2<\/sup>) of this regular hexagon will be ?<\/p>\n

a)\u00a0$3\\sqrt{3}$<\/p>\n

b)\u00a0$3\\sqrt{6}$<\/p>\n

c)\u00a0$6\\sqrt{3}$<\/p>\n

d)\u00a0$\\frac{5\\sqrt{3}}{2}$<\/p>\n

Question 3:\u00a0<\/b>The area of an equilateral triangle is $25\\sqrt{3}cm^2$. What is the perimeter of the triangle?<\/p>\n

a)\u00a010 cm<\/p>\n

b)\u00a020 cm<\/p>\n

c)\u00a030 cm<\/p>\n

d)\u00a040 cm<\/p>\n

Question 4:\u00a0<\/b>If the length and breadth of a rectangle increases by 20% and 15% each, find the percentage increase in its area.<\/p>\n

a)\u00a035%<\/p>\n

b)\u00a042%<\/p>\n

c)\u00a036%<\/p>\n

d)\u00a038%<\/p>\n

Question 5:\u00a0<\/b>A circle is circumscribed around an equilateral triangle of side 3 cm. What is the area of the circle (in $cm^2$)?<\/p>\n

a)\u00a0$\\sqrt3 \\pi$<\/p>\n

b)\u00a0$3 \\pi$<\/p>\n

c)\u00a0$3\\sqrt3 \\pi$<\/p>\n

d)\u00a0$9\\pi$<\/p>\n

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Question 6:\u00a0<\/b>The length of canvas, 75 cm wide required to build a conical tent of height 14m and the floor area 346.5 m2 is<\/p>\n

a)\u00a0665 m<\/p>\n

b)\u00a0860 m<\/p>\n

c)\u00a0490 m<\/p>\n

d)\u00a0770 m<\/p>\n

Question 7:\u00a0<\/b>If two medians BE and CF of a triangle ABC, intersect each other at G and if BG = CG, \u2220BGC = 60\u00b0 and BC = 8 cm then area of the triangle ABC is<\/p>\n

a)\u00a0$96\\sqrt{3} cm^{2}$<\/p>\n

b)\u00a0$64\\sqrt{3} cm^{2}$<\/p>\n

c)\u00a0$48\\sqrt{3} cm^{2}$<\/p>\n

d)\u00a0$48 cm^{2}$<\/p>\n

Question 8:\u00a0<\/b>5 persons will live in a tent. If each person requires 16m2 of floor area and 100m3 space for air then the height of the cone of smallest size to accommodate these persons would be<\/p>\n

a)\u00a016 metre<\/p>\n

b)\u00a010.25 metre<\/p>\n

c)\u00a020 metre<\/p>\n

d)\u00a018.75 metre<\/p>\n

Question 9:\u00a0<\/b>If the altitude of an equilateral triangle is 12\u221a3 cm, then its area would be :<\/p>\n

a)\u00a0$12 cm^{2}$<\/p>\n

b)\u00a0$144\\sqrt{3} cm^{2}$<\/p>\n

c)\u00a0$72 cm^{2}$<\/p>\n

d)\u00a0$36 \\sqrt{3} cm^{2}$<\/p>\n

Question 10:\u00a0<\/b>The perimeters of a circle, a square and an equilateral triangle are same and their areas are C, S and T respectively. Which of the following statement is true ?<\/p>\n

a)\u00a0C = S =T<\/p>\n

b)\u00a0C > S > T<\/p>\n

c)\u00a0C < S < T<\/p>\n

d)\u00a0S < C < T<\/p>\n

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Question 11:\u00a0<\/b>A wire of length 44 cm is first bent to form a circle and then rebent to form a square. The difference of the two enclosed areas is<\/p>\n

a)\u00a044 cm2<\/p>\n

b)\u00a033 cm2<\/p>\n

c)\u00a055 cm2<\/p>\n

d)\u00a066 cm2<\/p>\n

Question 12:\u00a0<\/b>The area of the iron sheet required to prepare a cone 24 cm high with base radius 7 cm is (Take $\\pi = \\frac{22}{7}$)<\/p>\n

a)\u00a0408 cm2<\/p>\n

b)\u00a0708 cm2<\/p>\n

c)\u00a0804 cm2<\/p>\n

d)\u00a0704 cm2<\/p>\n

Question 13:\u00a0<\/b>On decreasing each side of an equilateral triangle by 2 cm, there is a decrease of 4\u221a3 cm 2 in its area. The length of each side of the triangle is<\/p>\n

a)\u00a08 cm<\/p>\n

b)\u00a03 cm<\/p>\n

c)\u00a05 cm<\/p>\n

d)\u00a06 cm<\/p>\n

Question 14:\u00a0<\/b>The height of an equilateral triangle is 15 cm. The area of the triangle is<\/p>\n

a)\u00a050\u221a3 sq. cm.<\/p>\n

b)\u00a070\u221a3 sq. cm.<\/p>\n

c)\u00a075\u221a3 sq. cm.<\/p>\n

d)\u00a0150\u221a3 sq. cm.<\/p>\n

Question 15:\u00a0<\/b>The radius of a circle is a side of a square. The ratio of the areas of the circle and the square is<\/p>\n

a)\u00a01 : \u03c0<\/p>\n

b)\u00a0\u03c0 : 1<\/p>\n

c)\u00a0\u03c0: 2<\/p>\n

d)\u00a02 : \u03c0<\/p>\n

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Question 16:\u00a0<\/b>The height of an equilateral triangle is 15 cm. The area of the triangle is<\/p>\n

a)\u00a050\u221a3 sq. cm.<\/p>\n

b)\u00a070\u221a3 sq. cm.<\/p>\n

c)\u00a075\u221a3 sq. cm.<\/p>\n

d)\u00a0150\u221a3 sq. cm.<\/p>\n

Question 17:\u00a0<\/b>The diagonal of a square is equal to the side of an equilateral triangle. If the area of the square is 15\u221a3 sq cm, what is the area of the equilateral triangle?<\/p>\n

a)\u00a045\/\u221a2 sq cm<\/p>\n

b)\u00a045\u221a2 sq cm<\/p>\n

c)\u00a045 sq cm<\/p>\n

d)\u00a045\/2 sq cm<\/p>\n

Question 18:\u00a0<\/b>The area of an equilateral triangle is 9\u221a3 sq cm, \ufb01nd height of the triangle?<\/p>\n

a)\u00a06 cm<\/p>\n

b)\u00a0$6\\sqrt{3}$ cm<\/p>\n

c)\u00a0$3\\sqrt{3}$ cm<\/p>\n

d)\u00a09 cm<\/p>\n

Question 19:\u00a0<\/b>The diagonal of a square equals the side of an equilateral triangle. If the area of the square is 12 sq cm, what is the area of the equilateral triangle?<\/p>\n

a)\u00a012\u221a3 sq cm<\/p>\n

b)\u00a06\u221a3 sq cm<\/p>\n

c)\u00a012\u221a2 sq cm<\/p>\n

d)\u00a024 sq cm<\/p>\n

Question 20:\u00a0<\/b>From an equilateral triangle of side 4 cm, a triangle formed by joining the mid points of the original triangle is cut out. Find the remaining area (in sq. cm)<\/p>\n

a)\u00a0$3 \\sqrt{3}$ sq. cm<\/p>\n

b)\u00a0$3$ sq. cm<\/p>\n

c)\u00a0$4\\sqrt{3}$ sq. cm<\/p>\n

d)\u00a0$2 \\sqrt{3}$ sq. cm<\/p>\n

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Question 21:\u00a0<\/b>The area of the equilateral triangle is $\\frac{81\\sqrt{3}}{4}$ cm$^2$. Find the height of this triangle.<\/p>\n

a)\u00a0$6\\sqrt{3}$ cm<\/p>\n

b)\u00a0$4.5\\sqrt{3}$ cm<\/p>\n

c)\u00a0$4\\sqrt{3}$ cm<\/p>\n

d)\u00a0$8\\sqrt{3}$ cm<\/p>\n

Question 22:\u00a0<\/b>The perimeters of a square and an equilateral triangle are equal. If the diagonal of the square is 9 cm, what is the area of the equilateral triangle?<\/p>\n

a)\u00a018\u221a3 sq cm<\/p>\n

b)\u00a09\u221a3 sq cm<\/p>\n

c)\u00a018\u221a2 sq cm<\/p>\n

d)\u00a09\u221a2 sq cm<\/p>\n

Question 23:\u00a0<\/b>The diagonal of a square equals the side of an equilateral triangle. If the area of the square is 6\u221a3 sq cm, what is the area of the equilateral triangle?<\/p>\n

a)\u00a09\u221a3 sq cm<\/p>\n

b)\u00a09 sq cm<\/p>\n

c)\u00a09\u221a2 sq cm<\/p>\n

d)\u00a012 sq cm<\/p>\n

Question 24:\u00a0<\/b>If the length of the side of an equilateral triangle is 6 cm, what is its area?<\/p>\n

a)\u00a018\u221a3 sq cm<\/p>\n

b)\u00a036\u221a3 sq cm<\/p>\n

c)\u00a027\u221a3 sq cm<\/p>\n

d)\u00a09\u221a3 sq cm<\/p>\n

Question 25:\u00a0<\/b>The base of the prism is in shape of an equilateral triangle of side 11 cm. The height of the prism is 4$\\sqrt{3}$ cm. What is the volume of the prism?<\/p>\n

a)\u00a0447 cu. cm.<\/p>\n

b)\u00a0363 cu. cm.<\/p>\n

c)\u00a0219 cu. cm.<\/p>\n

d)\u00a0394 cu. cm.<\/p>\n

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Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Area of equilateral triangle is $\\frac{\\sqrt{3}}{4} a^{2}$ where a is side of triangle
\nwhich is equals to $121{\\sqrt{3}}$
\nor a = 22 and whole length of wire will be 66
\nfrom here when it is bend to make a circle, circumference will be $2\\pi r$ = 66
\nr = 10.5
\nhence area of circle will be $\\pi r^{2}$ = 346.5<\/p>\n

2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

<\/figure>\n

The hexagon is composed of 6 equilateral triangles, each with a side of 2.
\nArea of an equilateral triangle = $\\frac{\\sqrt3}{4}s^2$
\n=> Area of regular hexagon = 6 $\\times$ area of small equilateral triangles
\n= $6\\times\\frac{\\sqrt3}{4}\\times(2)^2$
\n= $6\\sqrt3$ $cm^2$
\n=> Ans – (C)<\/p>\n

3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let the side of the triangle be \u2018a\u2019 cm
\nArea of the triangle = $\\frac{\\sqrt{3}a^2}{4}$
\nIt is given that,
\n$\\frac{\\sqrt{3}a^2}{4}=25\\sqrt{3}$
\nOn solving we get $a=10$ cm
\nThus the perimeter of the triangle will be $3*10=30$ cm<\/p>\n

4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let L and B be the initial values of length and breadth respectively.
\nTherefore, the final values of length and breadth are 1.2L and 1.15B respectively.
\nInitial area of the rectangle = LB
\nFinal area of the rectangle = 1.2*1.15 LB = 1.38 LB
\nTherefore, required percentage = ((1.38-1)LB\/LB)*100 = 38%.
\nSo, the correct option to choose is D.<\/p>\n

5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

The circle is the circumcircle of the equilateral triangle.
\nCircumradius = $\\frac{2}{3}*\\frac{\\sqrt3}{2}*3=\\sqrt3$ cm
\nArea of the circle = $\\pi*(\\sqrt3)^2=3 \\pi$ $cm^2$
\nHence Option B.<\/p>\n

6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Base area = 346.5 $m^2$ = $\\pi r^2$
\n=> $r^2 = \\frac{346.5 * 7}{22}$
\n=> $r = \\sqrt{110.25} = 10.5 m$
\nHeight of tent = 14 m
\nNow, slant height$(l)$ of cone = $\\sqrt{r^2 + h^2}$
\n=> $l = \\sqrt{10.5^2 + 14^2}$
\n=> $l = \\sqrt{306.25} = 17.5m$
\nLet length of cloth be $x$
\nSurface area of cone = $\\pi rl$
\n=> $0.75 * x = \\frac{22}{7} * 10.5 * 17.5$
\n=> $x = \\frac{22 * 10.5 * 17.5}{7 * 0.75}$
\n=> $x = 770m$<\/p>\n

7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

<\/p>\n

In \u0394 BGC<\/p>\n

\u2220BGC = 60\u00b0 and BG = GC
\n$\\therefore$ \u0394BGC is an isosceles triangle
\n\u2220GBC = \u2220GCB
\nAs, we know sum of angles of a triangle = 180\u00b0
\n$\\therefore$ \u2220GBC + \u2220GCB + \u2220BGC = 180\u00b0
\n=> 2\u2220GBC = 180\u00b0 – 60\u00b0
\n=> \u2220GBC = 120\u00b0\/2 = 60\u00b0
\n\u2220GBC = \u2220GCB = \u2220BGC = 60\u00b0
\n$\\therefore$ \u0394GBC is equilateral triangle
\nArea of equilateral triangle = $\\frac{\\sqrt{3}}{4} * side^2$
\nArea of $\\triangle$GBC = $\\frac{\\sqrt{3}}{4} * 8^2$
\n= 16$\\sqrt{3}$
\nMedian of triangle divides the triangle into two parts of equal area.
\n3 medians of a triangle divide the triangle into six parts of equal area.
\n$\\therefore$ Area of \u0394GBC = 2 \u00d7 area of \u0394GDC
\nArea of \u0394ABC = 6 \u00d7 area of \u0394GDC
\n= 3 \u00d7 2 \u00d7 \u0394GDC
\n= 3 \u00d7 area of \u0394GBC
\n= 3 \u00d7 16$\\sqrt{3}$
\n= 48$\\sqrt{3} cm^2$<\/p>\n

8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Floor area required for 5 persons = 16*5 = 80 $m^2$
\nAir space required for 5 persons = 100*5 = 500 $m^3$
\nCone volume = $\\frac{1}{3} \\pi r^2h$
\nLet height be $h$
\nFor smallest size, volume = 500 = $\\frac{1}{3} * 80 * h$
\n=> $h = \\frac{150}{8}$ = 18.75 m<\/p>\n

9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let each side of the equilateral triangle be $a$
\nThe altitude of an equilateral triangle bisects the opposite side.
\n=> $(\\frac{a}{2})^2 + (12\\sqrt{3})^2 = a^2$
\n=> $a^2 – \\frac{a^2}{4} = 432$
\n=> $a^2 = \\frac{1728}{3}$
\nArea of equilateral triangle = $\\frac{\\sqrt{3}}{4} * a^2$
\n= $\\frac{\\sqrt{3}}{4} * \\frac{1728}{3}$
\n= $144\\sqrt{3}$<\/p>\n

10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the side of equilateral triangle be ‘a’ units
\nthe radius of circle be ‘r’ units.
\nand the side of square be ‘b’ units
\nThen,
\nPerimeter of square = 4b
\nPerimeter of equilateral triangle = 3a
\nCircumference of circle = $2*\\pi*r$
\nThen acc to ques,
\n=> $4b = 3a = 2*\\pi*r$
\n=> b = $\\frac{\\pi*r}{2}$
\n=> a = $\\frac{2}{3} * \\pi*r$
\nNow,
\narea of circle (C) = $\\pi r^2$
\narea of equilateral triangle (T) = $\\frac{\\sqrt{3}}{4} * a^2$
\n=> area of equilateral triangle (T) =$\\frac{\\pi^2 * r^2}{3*\\sqrt{3}}$
\narea of square (S) = b*b
\n=> area of square (S) = $\\frac{\\pi^2 r^2}{4}$
\nHence it is clearly visible that C > S > T.<\/p>\n

11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let radius of circle be $r$ and side of square be $a$
\n=> Circumference of circle = $2 \\pi r = 44$
\n=> $r = 7 cm$
\nSince, the circular wire is bent to form square, => both of their perimeters are equal
\n=> Perimeter of square = $4a = 44$
\n=>$a = 11 cm$
\nNow, area of circular wire = $\\pi r^2$
\n= $\\frac{22}{7} * 7^2 = 154 cm^2$
\nArea of square = $a^2$
\n= $11^2 = 121 cm^2$
\n=> Required difference = 154-121 = $33 cm^2$<\/p>\n

12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Surface area of cone= $\\frac{22}{7}\\times r\\times(r+L)$
\nwhere L=slant height
\nL= $\\sqrt((r)^{2} + (h)^{2})$
\nHere when r=7 and h=24,
\nL= $\\sqrt((7)^{2} + (24)^{2})$ = $\\sqrt625$ = 25cm
\nArea, A= ${\\frac{22}{7}}\\times7\\times(7+25)$
\nA= 704 $(cm)^{2}$<\/span><\/p>\n

 <\/p>\n

13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Area of an equilateral triangle= $(\\frac{\\sqrt3}{4})\\times(a)^{2}$
\nwhere a is side.
\nWhen a reduced to (a-2)
\nArea difference = 4${\\sqrt3}$
\n$(\\frac{\\sqrt3}{4})$($(a)^{2}-(a-2)^{2})$= 4${\\sqrt3}$
\n4a-4=16
\na=5 (C)<\/p>\n

14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

<\/p>\n

AD = 15 cm and ABC is equilateral triangle
\nIn $\\triangle$ADC
\n=> $tan \\angle ACD = \\frac{AD}{DC}$
\n=> $tan 60 = \\frac{15}{DC}$
\n=> DC = $\\frac{15}{\\sqrt{3}} = 5\\sqrt{3}$
\n=> BC = 2*DC = $10\\sqrt{3}$
\nArea of $\\triangle$ ABC = $\\frac{\\sqrt{3}}{4} * side^2$
\n= $\\frac{\\sqrt{3}}{4} * (10\\sqrt{3})^2$
\n= $75\\sqrt{3} cm^2$<\/p>\n

15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the radius be $x$. Since it is equal to the side of square
\n=> Side of square = $x$
\nRatio of area of circle to that of square
\n= $\\frac{\\pi x^2}{x^2}$
\n= $\\frac{\\pi}{1}$
\n=> Required ratio = $\\pi : 1$<\/p>\n

Free SSC Daily Practice Set<\/a><\/p>\n

16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

<\/p>\n

AD = 15 cm and ABC is equilateral triangle
\nIn $\\triangle$ADC
\n=> $tan \\angle ACD = \\frac{AD}{DC}$
\n=> $tan 60 = \\frac{15}{DC}$
\n=> DC = $\\frac{15}{\\sqrt{3}} = 5\\sqrt{3}$
\n=> BC = 2*DC = $10\\sqrt{3}$
\nArea of $\\triangle$ ABC = $\\frac{\\sqrt{3}}{4} * side^2$
\n= $\\frac{\\sqrt{3}}{4} * (10\\sqrt{3})^2$
\n= $75\\sqrt{3} cm^2$<\/p>\n

17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let the side of square = $s$ cm and diagonal = $d$ cm
\n=> Area of square = $(s)^2 = 15\\sqrt{3}$ ———-(i)
\nIn right triangle of the square, => $(s)^2 + (s)^2 = (d)^2$
\nSubstituting value of $s^2$ from equation (i)
\n=> $(d)^2 = 15\\sqrt{3} + 15\\sqrt{3} = 30\\sqrt{3}$ ———-(ii)
\nSide of equilateral triangle = Diagonal of square = $d$ cm
\n$\\therefore$ Area of equilateral triangle = $\\frac{\\sqrt{3}}{4} d^2$
\nSubstituting value of $d^2$ from (ii), we get :
\n= $\\frac{\\sqrt{3}}{4} \\times 30\\sqrt{3} = \\frac{90}{4} = \\frac{45}{2} cm^2$
\n=> Ans – (D)<\/p>\n

18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let the side of equilateral triangle = $a$ cm
\nArea of equilateral triangle = $\\frac{\\sqrt{3}}{4} a^2 = 9 \\sqrt{3}$
\n=> $a^2 = 9 \\times 4 = 36$
\n=> $a = \\sqrt{36} = 6$ cm
\nThe height $h$ of an equilateral triangle bisect the opposite side.
\n=> $(h)^2 = (6)^2 – (3)^2$
\n=> $(h)^2 = 36 – 9 = 27$
\n=> $h = \\sqrt{27} = 3\\sqrt{3}$ cm
\n=> Ans – (C)<\/p>\n

19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the side of square = $s$ cm and diagonal = $d$ cm
\n=> Area of square = $(s)^2 = 12$ ———-(i)
\nIn right triangle of the square, => $(s)^2 + (s)^2 = (d)^2$
\nSubstituting value of $s^2$ from equation (i)
\n=> $(d)^2 = 12 + 12 = 24$ ———-(ii)
\nSide of equilateral triangle = Diagonal of square = $d$ cm
\n$\\therefore$ Area of equilateral triangle = $\\frac{\\sqrt{3}}{4} d^2$
\nSubstituting value of $d^2$ from (ii), we get :
\n= $\\frac{\\sqrt{3}}{4} \\times 24 = 6\\sqrt{3} cm^2$
\n=> Ans – (B)<\/p>\n

20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

<\/figure>\n

 <\/p>\n

The area of the equilateral triangle = $\\frac{\\sqrt{3}}{4} * 4^2$ = $4\\sqrt{3}$
\nThe area of the triangle formed by joining the midpoints is one fourth the original triangle. Which is $\\sqrt{3} $
\nHence, the remaining area = $3 \\sqrt{3}$<\/p>\n

21)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

We know that area of the equilateral triangle = $\\frac{\\sqrt{3}}{4}*s^2$
\nTherefore,
\n$\\frac{81\\sqrt{3}}{4}$ = $\\frac{\\sqrt{3}}{4}s^2$
\n=> s = 9 cm
\nWe know that for an equilateral triangle, h = $\\frac{\\sqrt{3}}{2}$side
\nTherefore the required height is $4.5\\sqrt{3}$ cm<\/p>\n

22)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let the side of square = $s$ cm and diagonal, $d=9$ cm
\n=> $(s)^2+(s)^2=(d)^2$
\n=> $2s^2=9^2=81$
\n=> $s^2=\\frac{81}{2}$
\n=> $s=\\sqrt{\\frac{81}{2}} = \\frac{9}{\\sqrt{2}}$
\nThus, perimeter of square = $4s$
\n= $4 \\times \\frac{9}{\\sqrt{2}} = 18\\sqrt{2}$ cm
\nAlso, perimeter of square = Perimeter of equilateral triangle = $18\\sqrt{2}$ cm
\nLet side of equilateral triangle = $a$ cm
\n=> $3a=18\\sqrt{2}$
\n=> $a=\\frac{18\\sqrt{2}}{3} = 6\\sqrt{2}$ cm
\n$\\therefore$ Area of equilateral triangle = $\\frac{\\sqrt{3}}{4} a^2$
\n= $\\frac{\\sqrt{3}}{4} \\times (6\\sqrt{2})^2$
\n= $18\\sqrt{3}$ $cm^2$
\n=> Ans – (A)<\/p>\n

23)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the side of square = $s$ cm and diagonal = $d$ cm
\n=> Area of square = $(s)^2 = 6\\sqrt{3}$ ———-(i)
\nIn right triangle of the square, => $(s)^2 + (s)^2 = (d)^2$
\nSubstituting value of $s^2$ from equation (i)
\n=> $(d)^2 = 6\\sqrt{3} + 6\\sqrt{3} = 12\\sqrt{3}$ ———-(ii)
\nSide of equilateral triangle = Diagonal of square = $d$ cm
\n$\\therefore$ Area of equilateral triangle = $\\frac{\\sqrt{3}}{4} d^2$
\nSubstituting value of $d^2$ from (ii), we get :
\n= $\\frac{\\sqrt{3}}{4} \\times 12\\sqrt{3} = \\frac{36}{4} = 9 cm^2$
\n=> Ans – (B)<\/p>\n

24)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Side of an equilateral triangle, $a=6$ cm
\nArea = $\\frac{\\sqrt{3}}{4}(a)^2$
\n= $\\frac{\\sqrt{3}}{4} \\times (6)^2 = \\frac{36\\sqrt{3}}{4}$
\n= $9\\sqrt{3}$ $cm^2$
\n=> Ans – (D)<\/p>\n

25)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Volume of the prism = area of base * height = $\\sqrt{3} * 11^2$ * 4$\\sqrt{3}$ \/ 4
\n= 121*3 = 363 cu. cm.
\nHence, option B is the right answer.<\/p>\n

 <\/p>\n

SSC Free Previous Papers App<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Mensuration Questions for SSC CGL PDF: Download SSC CGL Mensuration questions with answers PDF based on previous papers very useful for SSC CGL exams. 25 Very important Mensuration objective questions for SSC exams. Question 1:\u00a0A copper wire is bent in the form of an equilateral triangle, and has an area $121\\sqrt{3}$ cm2 . If the […]<\/p>\n","protected":false},"author":21,"featured_media":26149,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,504],"tags":[462,1519,1719,1663],"class_list":{"0":"post-26147","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-ssc-cgl","10":"tag-ssc-cgl","11":"tag-ssc-cgl-2019","12":"tag-ssc-cgl-mocks","13":"tag-ssc-cgl-previous-papers"},"better_featured_image":{"id":26149,"alt_text":"Mensuration Questions for SSC CGL PDF","caption":"Mensuration Questions for SSC CGL PDF","description":"Mensuration Questions for SSC CGL 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