{"id":40746,"date":"2020-02-25T15:10:18","date_gmt":"2020-02-25T09:40:18","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=40746"},"modified":"2020-02-25T15:25:56","modified_gmt":"2020-02-25T09:55:56","slug":"mensuration-questions-for-ssc-chsl-set-3-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/mensuration-questions-for-ssc-chsl-set-3-pdf\/","title":{"rendered":"Mensuration Questions for SSC CHSL Set-3 PDF"},"content":{"rendered":"

Mensuration Questions for SSC CHSL Set-3 PDF<\/strong><\/span><\/h1>\n

Download SSC CHSL Mensuration questions with answers Set-3 PDF based on previous papers very useful for SSC CHSL Exams. Top-10 Very Important Maths Questions for SSC Exam<\/p>\n

Download Mensuration Questions for SSC CHSL Set-3 PDF <\/a><\/p>\n

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Take a free mock test for SSC CHSL<\/a><\/p>\n

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

Question 1: <\/b>The radius of the circumcircle of a right angled triangle is 15 cm and the radius of its inscribed circle is 6 cm. Find the sides of the triangle.<\/p>\n

a) 30, 40, 41<\/p>\n

b) 18, 24, 30<\/p>\n

c) 30, 24, 25<\/p>\n

d) 24, 36, 20<\/p>\n

Question 2: <\/b>ABCD is a rhombus. AB is produced to F and BA is produced to E such that AB = AE = BF. Then :<\/p>\n

a) ED > CF<\/p>\n

b) ED \u22a5 CF<\/p>\n

c) $ED^{2}+CF^{2}=EF^{2}$<\/p>\n

d) ED \u2551 CF<\/p>\n

Question 3: <\/b>Two circles of same radius 5 cm, intersect each other at A and B. If AB = 8 cm, then the distance between the centres is :<\/p>\n

a) 6 cm<\/p>\n

b) 8 cm<\/p>\n

c) 10 cm<\/p>\n

d) 4 cm<\/p>\n

Question 4: <\/b>C and C are two concentric circles with centres at 0. Their radii are 12 cm. and 3 cm. respectively. B and C are the points of contact of two tangents drawn to C2 from a point A lying on the circle C1. Then the area of the quadrilateral ABOC is<\/p>\n

a) $\\frac{9\\sqrt{15}}{2}$ sq.cm<\/p>\n

b) $12\\sqrt{15}$ sq.cm<\/p>\n

c) $9\\sqrt{15}$ sq.cm<\/p>\n

d) $6\\sqrt{15}$ sq.cm<\/p>\n

Question 5: <\/b>If a metallic cone of radius 30 cm and height 45 cm is melted and recast into metallic spheres of radius 5 cm, find the number of spheres.<\/p>\n

a) 81<\/p>\n

b) 41<\/p>\n

c) 80<\/p>\n

d) 40<\/p>\n

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Question 6: <\/b>A cyclic quadrilateral ABCD is such that AB = BC, AD = DC, AC \u22a5 BD, \u2220CAD = \u03b8. Then the angle \u2220ABC =<\/p>\n

a) $\\theta$<\/p>\n

b) $\\frac{\\theta}{2}$<\/p>\n

c) $2\\theta$<\/p>\n

d) $3\\theta$<\/p>\n

Question 7: <\/b>A rectangular tin sheet is 12 cm long and 5 cm broad. It is rolled along its length to form a cylinder by making the opposite edges Just to touch each other. Then the volume of the cylinder is<\/p>\n

a) $\\frac{60}{\\pi}$ $cm^{3}$<\/p>\n

b) $\\frac{180}{\\pi}$ $cm^{3}$<\/p>\n

c) $\\frac{120}{\\pi}$ $cm^{3}$<\/p>\n

d) $\\frac{100}{\\pi}$ $cm^{3}$<\/p>\n

Question 8: <\/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) 8 cm<\/p>\n

b) 3 cm<\/p>\n

c) 5 cm<\/p>\n

d) 6 cm<\/p>\n

Question 9: <\/b>The base of a right prism is a quadrilateral ABCD. Given that AB = 9 cm, BC = 14 cm, CD = 13 cm, DA = 12 cm and \u0394DAB = 90\u00b0. If the volume of the prism be 2070 cm3, then the area of the lateral surface is<\/p>\n

a) 720 cm2<\/p>\n

b) 810 cm2<\/p>\n

c) 1260 cm2<\/p>\n

d) 2070 cm2<\/p>\n

Question 10: <\/b>If the altitude of an equilateral triangle is 12\u221a3 cm, then its area would be :<\/p>\n

a) $12 cm^{2}$<\/p>\n

b) $144\\sqrt{3} cm^{2}$<\/p>\n

c) $72 cm^{2}$<\/p>\n

d) $36 \\sqrt{3} cm^{2}$<\/p>\n

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FREE SSC STUDY MATERIAL<\/a><\/p>\n

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

1) Answer (B)<\/strong><\/p>\n

Circumradius = 15 cm and inradius = 6 cm<\/span><\/p>\n

Let sides be a,b,c<\/p>\n

Since, it’s a right triangle, => Circumradius lies on the mid point of hypotenuse<\/p>\n

=> length of hypotenuse(c) = 15*2 = 30 cm<\/p>\n

The sum of the other two sides of a right triangle is equal to twice the sum of inradius and circumradius.<\/p>\n

=> $(a+b) = 2(15+6)$<\/p>\n

=> a+b = 42 ———-Eqn(1)<\/p>\n

Now, sum of all sides = 42+30 = 72 cm<\/p>\n

Product of the two sides(other than hypotenuse) is equal to the product of inradius and sum of all three sides.<\/p>\n

=> ab = 6 * 72<\/p>\n

=> ab = 432 ———Eqn(2)<\/p>\n

From eqn(1) and (2)<\/p>\n

=> a = 18 or b = 24 (or vice versa)<\/p>\n

=> Sides are = 18,24,30<\/p>\n

2) Answer (B)<\/strong><\/p>\n

<\/p>\n

ABCD is a rhombus and AB = AE = BF<\/p>\n

=> AD = AE and BC = BF<\/p>\n

=> $\\angle$AED = $\\angle$ADE and $\\angle$BCF = $\\angle$BFC<\/p>\n

Let $\\angle$DAB = $2\\theta$ and $\\angle$ABC = $2\\alpha$<\/p>\n

Using exterior angle property, we get :<\/p>\n

=> $\\angle$AED = ADE = $\\theta$<\/p>\n

and $\\angle$BCF = $\\angle$BFC = $\\alpha$<\/p>\n

Also, in ABCD<\/p>\n

=> $\\angle$DAB + $\\angle$ABC = 180<\/p>\n

=> $2\\theta + 2\\alpha = 90$<\/p>\n

=> $\\theta + \\alpha = 90$<\/p>\n

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

=> $\\angle G$ + $\\theta + \\alpha$ = 180<\/p>\n

=> $\\angle G$ = 180 – 90 = 90<\/p>\n

=> $ED \\perp CF$<\/p>\n

3) Answer (A)<\/strong><\/p>\n

<\/p>\n

AB is chord to each of the circle and AB = 8 cm<\/p>\n

Radius of each circle = 5 cm<\/p>\n

A line drawn from the centre of the circle perpendicular to the chord bisects it in two parts.<\/p>\n

=> AC = 8\/2 = 4 cm<\/p>\n

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

=> $OC = \\sqrt{(OA)^2 – (AC)^2}$<\/p>\n

=> $OC = \\sqrt{25-16} = \\sqrt{9}$<\/p>\n

=> OC = 3 cm<\/p>\n

=> OO’ = 2*3 = 6 cm<\/p>\n

4) Answer (C)<\/strong><\/p>\n

<\/figure>\n

AB = AC = tangents from the same point<\/p>\n

OB = OC = 3 and OA = 12<\/p>\n

$\\angle$ABO = 90<\/p>\n

=> AB = $\\sqrt{12^2 – 3^2} = 3\\sqrt{15}$<\/p>\n

Now, area of $\\triangle$OAB = $\\frac{1}{2}$ OB * AB<\/p>\n

= $\\frac{1}{2} * 3 * 3\\sqrt{15} = \\frac{9\\sqrt{15}}{2}$<\/p>\n

$\\therefore$ area of OABC = $9\\sqrt{15}$ sq. cm<\/p>\n

5) Answer (A)<\/strong><\/p>\n

Volume of metallic cone = $\\frac{1}{3} \\pi r^2 h$<\/p>\n

= $\\frac{1}{3} \\pi * 30^2 * 45 = 13500 \\pi cm^3$<\/p>\n

Volume of sphere = $\\frac{4}{3} \\pi R^3$<\/p>\n

= $\\frac{4}{3} \\pi * 5^3 = \\frac{500 \\pi}{3} cm^3$<\/p>\n

=> Required no. of spheres = $\\frac{13500 \\pi}{\\frac{500 \\pi}{3}}$<\/p>\n

= 81<\/p>\n

6) Answer (C)<\/strong><\/p>\n

<\/p>\n

AB = BC and AD = DC , $\\angle CAD = \\theta$<\/p>\n

In isosceles $\\triangle$ADC<\/p>\n

=> $\\angle$CAD + $\\angle$ACD + $\\angle$CDA = 180<\/span><\/span><\/span><\/p>\n

=> $2\\theta$ + $\\angle$CDA = 180<\/span><\/span><\/span><\/p>\n

=> $\\angle$CDA = 180 – $2\\theta$<\/span><\/span><\/span><\/span><\/p>\n

Sum of opposite angles in a cyclic quadrilateral = 180<\/span><\/span><\/span><\/span><\/p>\n

=> $\\angle$CDA + $\\angle$ABC = 180<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

=> $\\angle$ABC + 180 – $2\\theta$ = 180<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

=> $\\angle$ABC = $2\\theta$<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

7) Answer (B)<\/strong><\/p>\n

<\/p>\n

From the above figure circumference of cylinder= 12cm
\nSo 2\u03c0<\/strong>R= 12<\/p>\n

R= $\\frac{12}{2\\pi}$<\/p>\n

Volume of cylinder= \u03c0<\/strong>$(R)^{2}$h
\nwhere h and R are height and radius
\nSo volume, V= \u03c0<\/strong>$(\\frac{12}{2\\pi})^{2}\\times5$<\/span><\/span><\/span><\/p>\n

V=$\\frac{180}{\\pi}$ $cm^{3}$<\/span><\/span><\/span><\/span><\/p>\n

8) Answer (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

9) Answer (A)<\/strong><\/p>\n

<\/p>\n

In right $\\triangle$DAB, using Pythagoras theorem,<\/p>\n

=> $BD = \\sqrt{(AD)^2 + (AB)^2}$<\/p>\n

=> $BD = \\sqrt{12^2 + 9^2} = \\sqrt{225}$<\/p>\n

=> $BD = 15 cm$<\/p>\n

Now, area of $\\triangle$DAB = $\\frac{1}{2} * 9 * 12 = 54 cm^2$<\/p>\n

Area of $\\triangle$BCD = $\\sqrt{s(s-a)(s-b)(s-c)}$<\/span><\/p>\n

where, $s = \\frac{a+b+c}{2}$<\/span><\/p>\n

=> Area of $\\triangle$BCD = $\\sqrt{21 * 6 * 7 * 8} = 84 cm^2$<\/span><\/span><\/p>\n

=> Area of quad ABCD = area of $\\triangle$DAB + area of $\\triangle$BCD<\/span><\/span><\/span><\/span><\/p>\n

= 54+84 = $138 cm^2$<\/span><\/span><\/span><\/span><\/p>\n

Volume of prism = base area * height<\/span><\/span><\/span><\/span><\/p>\n

=> 2070 = 138 * height<\/span><\/span><\/span><\/span><\/p>\n

=> height = 15 cm<\/span><\/span><\/span><\/span><\/p>\n

Lateral surface area of prism = perimeter of base * height<\/span><\/span><\/span><\/span><\/p>\n

= (12 + 9 + 14 + 13) * 15 = 48 * 15<\/span><\/span><\/span><\/span><\/p>\n

= 720 $cm^2$<\/span><\/span><\/span><\/span><\/p>\n

10) Answer (B)<\/strong><\/p>\n

Let each side of the equilateral triangle be $a$<\/p>\n

The altitude of an equilateral triangle bisects the opposite side.<\/p>\n

=> $(\\frac{a}{2})^2 + (12\\sqrt{3})^2 = a^2$<\/p>\n

=> $a^2 – \\frac{a^2}{4} = 432$<\/p>\n

=> $a^2 = \\frac{1728}{3}$<\/p>\n

Area of equilateral triangle = $\\frac{\\sqrt{3}}{4} * a^2$<\/p>\n

= $\\frac{\\sqrt{3}}{4} * \\frac{1728}{3}$<\/p>\n

= $144\\sqrt{3}$<\/p>\n

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

We hope this Mensuration questions of set-3 pdf for SSC CHSL Exam preparation is so helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Mensuration Questions for SSC CHSL Set-3 PDF Download SSC CHSL Mensuration questions with answers Set-3 PDF based on previous papers very useful for SSC CHSL Exams. Top-10 Very Important Maths 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 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