{"id":43017,"date":"2020-10-05T19:05:39","date_gmt":"2020-10-05T13:35:39","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=43017"},"modified":"2020-10-05T19:05:39","modified_gmt":"2020-10-05T13:35:39","slug":"mensuration-questions-for-ssc-cgl-tier-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/mensuration-questions-for-ssc-cgl-tier-2-pdf\/","title":{"rendered":"Mensuration Questions for SSC CGL Tier 2 PDF"},"content":{"rendered":"

Mensuration Questions for SSC CGL Tier 2 PDF<\/strong><\/span><\/h2>\n

Download SSC CGL Tier 2 Mensuration Questions PDF. Top 10 SSC CGL Tier 2 Mensuration questions based on asked questions in previous exam papers very important for the SSC exam.<\/p>\n

Download Mensuration Questions for SSC CGL Tier 2 PDF<\/a><\/p>\n

Get 10 SSC CGL Tier-2 Mocks for Rs. 149<\/a><\/p>\n

Take SSC CGL Tier-2 Mock Tests<\/a><\/p>\n

Download SSC CGL Tier-2 Previous Papers PDF<\/a><\/p>\n

Question 1:\u00a0<\/b>Length and breadth of a rectangle are increased by 40% and 70% respectively. What will be the percentage increase in the area of rectangle?<\/p>\n

a)\u00a0118<\/p>\n

b)\u00a0110<\/p>\n

c)\u00a0138<\/p>\n

d)\u00a0128<\/p>\n

Question 2:\u00a0<\/b>The perimeter of a square is 40 cm, find its area?<\/p>\n

a)\u00a0100 sq cm<\/p>\n

b)\u00a025 sq cm<\/p>\n

c)\u00a050 sq cm<\/p>\n

d)\u00a0160 sq cm<\/p>\n

Question 3:\u00a0<\/b>The circumference of a circle is 88 cm, \ufb01nd its area?<\/p>\n

a)\u00a0616 sq cm<\/p>\n

b)\u00a0308 sq cm<\/p>\n

c)\u00a0154 sq cm<\/p>\n

d)\u00a077 sq cm<\/p>\n

Question 4:\u00a0<\/b>The total surface area of a hemisphere is 166.32 sq cm, find its curved surface area?<\/p>\n

a)\u00a0221.76 sq cm<\/p>\n

b)\u00a036.96 sq cm<\/p>\n

c)\u00a0110.88 sq cm<\/p>\n

d)\u00a055.44 sq cm<\/p>\n

Question 5:\u00a0<\/b>If G is the centroid of triangle ABC and area of triangle ABC = 48cm2<\/sup>, then the area of triangle BGC is<\/p>\n

a)\u00a08 cm2<\/sup><\/p>\n

b)\u00a016 cm2<\/sup><\/p>\n

c)\u00a024 cm2<\/sup><\/p>\n

d)\u00a032 cm2<\/sup><\/p>\n

Take SSC CGL Tier-2 Mock Tests<\/a><\/p>\n

Download SSC CGL Tier-2 Previous Papers PDF<\/a><\/p>\n

Question 6:\u00a0<\/b>In $\\triangle$ PQR, the line drawn from the vertex P intersects QR at a point S. If QR = 4.5 cm and SR = 1.5 cm then the ratios of the area of triangle PQS and triangle PSR is<\/p>\n

a)\u00a04 : 1<\/p>\n

b)\u00a03 : 1<\/p>\n

c)\u00a03 : 2<\/p>\n

d)\u00a02 : 1<\/p>\n

Question 7:\u00a0<\/b>If G is the centroid and AD, BE, CF are three medians of triangle ABC with area 72 sq cm , then the area of triangle BDG is :<\/p>\n

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

b)\u00a016 sq cm<\/p>\n

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

d)\u00a08 sq cm<\/p>\n

Question 8:\u00a0<\/b>The three medians AD, BE and CF of triangle ABC intersect at point G. If the area of triangle ABC is 60 sq.cm. then the area of the quadrilateral BDGF is :<\/p>\n

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

b)\u00a015 sq.cm<\/p>\n

c)\u00a020 sq.cm<\/p>\n

d)\u00a030 sq.cm<\/p>\n

Question 9:\u00a0<\/b>In triangle ABC, AD, BE and CF are the medians intersecting at point G and area of triangle ABC is 156 cm2. What is the area $(in cm^2)$ of triangle FGE?<\/p>\n

a)\u00a013<\/p>\n

b)\u00a026<\/p>\n

c)\u00a039<\/p>\n

d)\u00a052<\/p>\n

Question 10:\u00a0<\/b>The tangents drawn at points A and B of a circle with centre O, meet at P. If \u2220AOB = 120\u00b0 and AP = 6 cm, then what is the area of triangle (in cm$^2)$ APB?<\/p>\n

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

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

c)\u00a0$9$<\/p>\n

d)\u00a0$9\\sqrt{3}$<\/p>\n

18000+ Questions – Free SSC Study Material<\/a><\/p>\n

Free SSC Preparation (Videos Youtube)<\/a><\/p>\n

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

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

Let length and breadth of rectangle initially be $10$ units<\/p>\n

=> Area, $A=10\\times10=100$ sq.units<\/p>\n

Length is increased by 40%, => New length = $10+(\\frac{40}{100}\\times10)=14$ units<\/p>\n

Similarly,\u00a0new breadth = $10+(\\frac{70}{100}\\times10)=17$ units<\/p>\n

=> New area,\u00a0$A’=14\\times17=238$ sq.units<\/p>\n

$\\therefore$ % increase in area = $\\frac{(238-100)}{100}\\times100=138\\%$<\/p>\n

=> Ans – (C)<\/p>\n

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

Let side of square = $s$ cm<\/p>\n

Perimeter of square = $4 \\times s = 40$<\/p>\n

=> $s = \\frac{40}{4} = 10$ cm<\/p>\n

$\\therefore$ Area of square = $(s)^2$<\/p>\n

= $(10)^2 = 100 cm^2$<\/p>\n

=> Ans – (A)<\/p>\n

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

Let the radius of circle = $r$ cm<\/p>\n

Circumference of circle = $2 \\pi r = 88$<\/p>\n

=> $2 \\times \\frac{22}{7} \\times r = 88$<\/p>\n

=> $r = \\frac{88 \\times 7}{44} = 14$ cm<\/p>\n

$\\therefore$ Area of circle = $\\pi r^2$<\/p>\n

= $\\frac{22}{7} \\times (14)^2$<\/p>\n

= $22 \\times 2 \\times 14 = 616 cm^2$<\/p>\n

=> Ans – (A)<\/p>\n

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

Let radius of hemisphere = $r$ cm<\/p>\n

Total surface area of hemisphere = $3 \\pi r^2 = 166.32$ ————(i)<\/p>\n

Multiplying equation\u00a0(i) by $\\frac{2}{3}$<\/p>\n

=> $\\frac{2}{3} \\times 3 \\pi r^2 = \\frac{2}{3} \\times 166.32$<\/p>\n

=> Curved Surface area of hemisphere =\u00a0$2 \\pi r^2 = 2 \\times 55.44 = 110.88 cm^2$<\/p>\n

=> Ans – (C)<\/p>\n

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

As we know area of triangle, with centroid as one of the vertices and remaining 2 triangle vertices, is $\\frac{1}{3}$rd of the area of whole triangle.
\nHence area will be $\\frac{48}{3}$ = 16<\/p>\n

Download SSC CGL General Science Notes PDF<\/a><\/p>\n

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

NOTE :- The ratio of area of two triangles on same base is equal to the ratio of two corresponding sides of the two triangles.<\/p>\n

\n
<\/figure>\n

Given : QR = 4.5 cm and SR = 1.5 cm<\/p>\n

=> QS = QR – SR = 4.5-1.5 = 3 cm<\/p>\n

Since, the two triangles PQS and PSR have same base PS<\/p>\n

=> $\\frac{area(\\triangle PQS)}{area(\\triangle PSR)} = \\frac{QS}{SR}$<\/p>\n

= $\\frac{3}{1.5}$ = 2 : 1<\/p>\n

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

The area of triangle formed by any two vertices and centroid is (1\/3) times the area of ABC.
\nAlso the median divides the triangle into two equal areas.
\nSo, area of BDG = (1\/6) times of ABC
\n= (1\/6)*72
\n=12<\/p>\n

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

Given \u2206ABC, G is the centroid and AD,BE, CF are three medians and the area of \u2206AGE = 10
\n<\/span><\/span>As we know the median divides the triangle into 6 triangles of equal area
\nHence area of the quadrilateral BDGF = 2*\u2206AGE = 2*10
\narea of the quadrilateral BDGF = 20<\/p>\n

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

<\/figure>\n

Medians of a triangle divides the triangle into 6 parts of equal areas.<\/p>\n

Also,\u00a0ar($\\triangle$ ABC) =\u00a0156<\/p>\n

=> ar($\\triangle$ AFG) =\u00a0ar($\\triangle$ FBG)\u00a0=\u00a0ar($\\triangle$ BGD)\u00a0=\u00a0ar($\\triangle$ DGC)\u00a0=\u00a0ar($\\triangle$ CGE)\u00a0=\u00a0ar($\\triangle$ EGA) = $\\frac{156}{6}=26$\u00a0$cm^2$<\/p>\n

=> ar(AFGE) =\u00a0ar($\\triangle$ AFG) + ar($\\triangle$ EAG)<\/p>\n

= $26+26=52$ $cm^2$<\/p>\n

$\\therefore$\u00a0ar($\\triangle$ FGE) = $\\frac{1}{4}\\times$\u00a0ar(AFGE)<\/p>\n

= $\\frac{52}{4}=13$ $cm^2$<\/p>\n

=> Ans – (A)<\/p>\n

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

<\/figure>\n

Given\u00a0: \u2220AOB = 120\u00b0 and AP = 6 cm and $\\angle$ OAP = 90\u00b0<\/p>\n

To find\u00a0: ar($\\triangle$ APB) = ?<\/p>\n

Solution\u00a0: $\\angle$ AOP = $\\frac{1}{2} \\angle$ AOB = $\\frac{120}{2}=60^\\circ$<\/p>\n

In $\\triangle$ AOP,<\/p>\n

=> $tan(\\angle AOP)=\\frac{AP}{OA}$<\/p>\n

=> $tan(60^\\circ)=\\frac{6}{OA}$<\/p>\n

=> $\\sqrt3=\\frac{6}{OA}$<\/p>\n

=> $OA=\\frac{6}{\\sqrt3}=2\\sqrt3$ cm<\/p>\n

Thus, area of $\\triangle$ AOP = $\\frac{1}{2}\\times(OA)\\times(AP)$<\/p>\n

$\\frac{1}{2}\\times(2\\sqrt3)\\times(6)=6\\sqrt3$ $cm^2$ ————-(i)<\/p>\n

Now, in $\\triangle$ AOM<\/p>\n

=> $sin(\\angle AOM)=\\frac{AM}{OA}$<\/p>\n

=> $sin(60^\\circ)=\\frac{AM}{2\\sqrt3}$<\/p>\n

=> $\\frac{\\sqrt3}{2}=\\frac{AM}{2\\sqrt3}$<\/p>\n

=> $AM=3$ cm<\/p>\n

Similarly, $OM = \\sqrt3$ cm<\/p>\n

Thus, area of $\\triangle$ AOM = $\\frac{1}{2}\\times(OM)\\times(AM)$<\/p>\n

$\\frac{1}{2}\\times(\\sqrt3)\\times(3)=1.5\\sqrt3$ $cm^2$ ————-(ii)<\/p>\n

=> $ar(\\triangle AMP)=ar(\\triangle AOP)-ar(\\triangle AOM)$<\/p>\n

=\u00a0$6\\sqrt3-1.5\\sqrt3=4.5\\sqrt3$ $cm^2$<\/p>\n

$\\therefore$ $ar(\\triangle APB)=2ar(\\triangle AMP)$<\/p>\n

= $2\\times4.5\\sqrt3=9\\sqrt3$ $cm^2$<\/p>\n

=> Ans – (D)<\/p>\n

Free SSC Online Coaching<\/a><\/p>\n

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We hope this Mensuration Questions pdf for SSC CGL Tier 2 exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Mensuration Questions for SSC CGL Tier 2 PDF Download SSC CGL Tier 2 Mensuration Questions PDF. Top 10 SSC CGL Tier 2 Mensuration questions based on asked questions in previous exam papers very important for the SSC exam. Take SSC CGL Tier-2 Mock Tests Download SSC CGL Tier-2 Previous Papers PDF Question 1:\u00a0Length and breadth […]<\/p>\n","protected":false},"author":42,"featured_media":43019,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,9,504,378,1493,1741,1441],"tags":[451,462,4111,500,461,1461,1847,1242],"class_list":{"0":"post-43017","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-featured","9":"category-ssc","10":"category-ssc-cgl","11":"category-ssc-chsl","12":"category-ssc-cpo","13":"category-ssc-mts","14":"category-stenographer","15":"tag-mensuration","16":"tag-ssc-cgl","17":"tag-ssc-cgl-tier-2","18":"tag-ssc-cpo","19":"tag-ssc-exams","20":"tag-ssc-gd","21":"tag-ssc-mts","22":"tag-ssc-stenographer"},"better_featured_image":{"id":43019,"alt_text":"Mensuration Questions for SSC CGL Tier 2 PDF","caption":"Mensuration Questions for SSC CGL Tier 2 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