{"id":40423,"date":"2020-02-05T17:50:35","date_gmt":"2020-02-05T12:20:35","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=40423"},"modified":"2020-02-05T17:50:35","modified_gmt":"2020-02-05T12:20:35","slug":"mensuration-questions-for-ssc-chsl-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/mensuration-questions-for-ssc-chsl-set-2-pdf\/","title":{"rendered":"Mensuration Questions for SSC CHSL Set-2 PDF"},"content":{"rendered":"

Mensuration Questions for SSC CHSL Set-2 PDF<\/strong><\/span><\/h2>\n

Download SSC CHSL Mensuration Questions with answers Set-2\u00a0 PDF based on previous papers very useful for SSC CHSL Exams. Top-10 Very Important Mensuration Questions for SSC Exam<\/p>\n

Download Mensuration Questions for SSC CHSL Set-2 PDF<\/a><\/p>\n

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More SSC CHSL Important Questions and Answers PDF<\/a><\/p>\n

Question 1:\u00a0<\/b>A rectangle with one side of length 4 cm is inscribed in a circle of diameter 5 cm . Find area of the rectangle<\/p>\n

a)\u00a0$21 cm^2$<\/p>\n

b)\u00a0$12 cm^2$<\/p>\n

c)\u00a0$4 cm^2$<\/p>\n

d)\u00a0$3 cm^2$<\/p>\n

Question 2:\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

Question 3:\u00a0<\/b>In APQR, 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 4:\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 5:\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

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Question 6:\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 7:\u00a0<\/b>In the given figure, PM is one-third of PQ and PN is one-third of PS. If the area of PMRN is 17 cm2, then what is the area $(in cm^2)$ of PQRS?<\/p>\n

<\/figure>\n

a)\u00a034<\/p>\n

b)\u00a051<\/p>\n

c)\u00a068<\/p>\n

d)\u00a085<\/p>\n

Question 8:\u00a0<\/b>In \u0394ABC, a line through A cuts the side BC at D such that BD : DC = 4 :5. If the area of \u0394ABD = 60 cm , then the area of \u0394ADC is<\/p>\n

a)\u00a090 cm\u00b2<\/p>\n

b)\u00a050 cm\u00b2<\/p>\n

c)\u00a060 cm\u00b2<\/p>\n

d)\u00a075 cm\u00b2<\/p>\n

Question 9:\u00a0<\/b>If D, E and F are the mid points of BC, CA and AB respectively of the AABC then the ratio of area of the parallelogram DEFB and area of the trapezium CAFD is :<\/p>\n

a)\u00a02 : 3<\/p>\n

b)\u00a03 : 4<\/p>\n

c)\u00a01 : 2<\/p>\n

d)\u00a01 : 3<\/p>\n

Question 10:\u00a0<\/b>ABC is a right angled triangle. B being the right angle. Midpoints of AB,BC and AC are respectively B\u2019,C’ and A\u2019. Area of \u0394A\u2019B\u2019C\u2019 is<\/p>\n

a)\u00a0$\\frac{1}{2}$ x area of AABC<\/p>\n

b)\u00a0$\\frac{2}{3}$ x area of AABC<\/p>\n

c)\u00a0$\\frac{1}{4}$ x area of AABC<\/p>\n

d)\u00a0$\\frac{1}{8}$ x area of AABC<\/p>\n

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

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

<\/figure>\n

Given\u00a0: AB = 4 cm and AC = 5 cm<\/p>\n

To find\u00a0: Area of rectangle ABCD = ?<\/p>\n

Solution : In $\\triangle$ ABC,<\/p>\n

=> $(BC)^2=(AC)^2-(AB)^2$<\/p>\n

=> $(BC)^2=(5)^2-(4)^2$<\/p>\n

=> $(BC)^2=25-16=9$<\/p>\n

=> $BC=\\sqrt{9}=3$ cm<\/p>\n

$\\therefore$ Area of rectangle ABCD = $AB \\times BC$<\/p>\n

= $4 \\times 3=12$ $cm^2$<\/p>\n

=> Ans – (B)<\/p>\n

2)\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

3)\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

<\/p>\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

4)\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

5)\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

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6)\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

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

Let PQ = SR = $3l$ and PS = QR = $3b$ cm<\/p>\n

=> QM = $2l$ and NS = $2b$<\/p>\n

Let the area of rectangle PQRS = $9lb=x$ $cm^2$ ————–(i)<\/p>\n

=> Area of rectangle PQRS = Area of PMNR + ar($\\triangle$ NSR) +\u00a0ar($\\triangle$ QMR)<\/p>\n

=> $x=17+(\\frac{1}{2}\\times NS\\times SR)+(\\frac{1}{2}\\times QM\\times QR)$<\/p>\n

=> $x=17+(\\frac{1}{2}\\times2b\\times3l)+(\\frac{1}{2}\\times2l\\times3b)$<\/p>\n

=> $x=17+6lb$<\/p>\n

Multiplying equation (i) by $\\frac{2}{3}$, => $6lb=\\frac{2x}{3}$<\/p>\n

=> $x-\\frac{2x}{3}=17$<\/p>\n

=> $x=17\\times3=51$ $cm^2$<\/p>\n

=> Ans – (B)<\/p>\n

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

<\/p>\n

Let BD = 4$x$ and DC = 5$x$<\/p>\n

Let height of the triangle be $h$ which will be same for both triangles ABD and ADC.<\/p>\n

Now, area of $\\triangle$ABD = $\\frac{1}{2} * BD * h$ = 60<\/p>\n

=> $\\frac{1}{2} * 4x * h$ = 60<\/p>\n

=> $xh$ = 30<\/p>\n

Now, area of $\\triangle$ADC = $\\frac{1}{2} * DC * h$<\/p>\n

= $\\frac{1}{2} * 5x * h = \\frac{1}{2} * 5 * 30$<\/p>\n

= 75 $cm^2$<\/p>\n

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

Let , each side of the triangle be 2x.
\nSo, AF=FB=BD=DC=CE=EA= x . EF=FD=DE=x ( as always EF= half of BC)
\narea of the parallelogram DEFB is $ base\\times height$ = $ (x)\\times h_p$
\n& area of the trapezium CAFD is $\\frac{1}{2}\\times base\\times height$ = $\\frac{1}{2}\\times (x+2x)\\times h_t$<\/span>
\nClearly , both the heights are to be measured from midpoint of sides to midpoints of the line joining the midpoints of the side. So , $h_p = h_t$ .
\nratio of area of the parallelogram DEFB and area of the trapezium CAFD is
\n= $\\frac{(x)\\times h_p}{\\frac{1}{2}\\times (x+2x)\\times h_t}$
\n= 2 : 3<\/p>\n

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

The triangle obtained by joining the midpoints will also be a right angled triangle.
\nSince the sides are reduced by a factor of 2, the area will be reduced by a factor of 4. (Since area = 0.5*b*h)
\nOption C is the right answer.<\/p>\n

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SSC CHSL Important Q&A PDF<\/a><\/p>\n

We hope this Mensuration Questions for SSC CHSL Exam preparation is so helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Mensuration Questions for SSC CHSL Set-2 PDF Download SSC CHSL Mensuration Questions with answers Set-2\u00a0 PDF based on previous papers very useful for SSC CHSL Exams. Top-10 Very Important Mensuration Questions for SSC Exam Take a free mock test for SSC CHSL Download SSC CHSL Previous Papers More SSC CHSL Important Questions and Answers PDF […]<\/p>\n","protected":false},"author":42,"featured_media":40429,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,9,378],"tags":[3607,165,358],"class_list":{"0":"post-40423","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-chsl","11":"tag-mensuration-questions-for-ssc-chsl","12":"tag-ssc","13":"tag-ssc-chsl"},"better_featured_image":{"id":40429,"alt_text":"Mensuration Questions for SSC CHSL Set-2 PDF","caption":"Mensuration Questions for SSC CHSL Set-2 PDF\n","description":"Mensuration Questions for SSC CHSL Set-2 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