{"id":27245,"date":"2019-04-08T18:39:19","date_gmt":"2019-04-08T13:09:19","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=27245"},"modified":"2019-04-08T18:39:19","modified_gmt":"2019-04-08T13:09:19","slug":"ssc-cgl-square-root-questions-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/ssc-cgl-square-root-questions-pdf\/","title":{"rendered":"SSC CGL Square Root Questions PDF"},"content":{"rendered":"

SSC CGL Square Root Questions:<\/h1>\n

Download SSC CGL Trignometry questions with answers PDF based on previous papers very useful for SSC CGL exams. 20 Very important Trignometry objective questions for SSC exams.<\/p>\n

Download SSC CGL Square Root Questions<\/a><\/p>\n\n

Question 1:\u00a0<\/b>What is the positive square root of $[19+4\\sqrt21]?$<\/p>\n

a)\u00a0$\\sqrt{7}+2\\sqrt{3}$<\/p>\n

b)\u00a0$\\sqrt{3}+2\\sqrt{7}$<\/p>\n

c)\u00a0$\\sqrt{2}+3\\sqrt{7}$<\/p>\n

d)\u00a0$\\sqrt{7}+3\\sqrt{3}$<\/p>\n

Question 2:\u00a0<\/b>What is the square root of $\\frac{(3-2\\sqrt2)}{(3+2\\sqrt2)}$ ?<\/p>\n

a)\u00a0$3-2\\sqrt2$<\/p>\n

b)\u00a0$3+2\\sqrt2$<\/p>\n

c)\u00a0$1$<\/p>\n

d)\u00a0$17$<\/p>\n

Question 3:\u00a0<\/b>What is the square root of $\\frac{(3-2\\sqrt2)}{(3+2\\sqrt2)}?$<\/p>\n

a)\u00a0$3-2\\sqrt2$<\/p>\n

b)\u00a0$3+2\\sqrt2$<\/p>\n

c)\u00a0$1$<\/p>\n

d)\u00a0$$17$<\/p>\n

Question 4:\u00a0<\/b>Determine the value of m for which $4x+\\frac{\\sqrt{x}}{6}+\\frac{m^2}{4}$ is a perfect square.<\/p>\n

a)\u00a0$\\frac{1}{24}$<\/p>\n

b)\u00a0$\\frac{1}{12}$<\/p>\n

c)\u00a0$12$<\/p>\n

d)\u00a0$24$<\/p>\n

Question 5:\u00a0<\/b>What is the positive square root of $[25+4\\sqrt{39}]$ ?<\/p>\n

a)\u00a0$\\sqrt{13}+2\\sqrt3$<\/p>\n

b)\u00a0$\\sqrt{13}+3\\sqrt2$<\/p>\n

c)\u00a0$\\sqrt{11}+2\\sqrt3$<\/p>\n

d)\u00a0$11+3\\sqrt2$<\/p>\n

SSC CGL Free Mock Test<\/a><\/p>\n\n

Question 6:\u00a0<\/b>Divide 20 into two parts such that the sum of the square of the parts is 232. What is the value of the two parts?<\/p>\n

a)\u00a06, 14<\/p>\n

b)\u00a08, 12<\/p>\n

c)\u00a04, 16<\/p>\n

d)\u00a010, 10<\/p>\n

Question 7:\u00a0<\/b>A triangle is inscribed inside a square such that the base of the triangle coincides with the side of the square. The square is inscribed inside a circle of diameter $24\\sqrt{2}$ cm. Then find the area of the triangle.<\/p>\n

a)\u00a0196 sq.cm<\/p>\n

b)\u00a0169 sq.cm<\/p>\n

c)\u00a0225 sq.cm<\/p>\n

d)\u00a0288 sq.cm<\/p>\n

Question 8:\u00a0<\/b>Find the area of a square which is inscribed in a circle of radius $7\\sqrt{2}$ cm.<\/p>\n

a)\u00a0144 sq.cm<\/p>\n

b)\u00a0196 sq.cm<\/p>\n

c)\u00a0289 sq.cm<\/p>\n

d)\u00a049 sq.cm<\/p>\n

Question 9:\u00a0<\/b>The sum of two positive numbers is 20% of the sum of their squares and 25% of the difference of their squares. If the numbers are x and y the, $\\ \\frac{x+y}{x^{2}}\\ $ is equal to<\/p>\n

a)\u00a0$\\frac{1}{4}$<\/p>\n

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

c)\u00a0$\\frac{1}{3}$<\/p>\n

d)\u00a0$\\frac{2}{9}$<\/p>\n

Question 10:\u00a0<\/b>What is the volume (in cm$^{3}$) of a right pyramid of height $12$ cm and having a square base whose diagonal is $6\\sqrt{2}$ cm?<\/p>\n

a)\u00a0864<\/p>\n

b)\u00a0432<\/p>\n

c)\u00a0144<\/p>\n

d)\u00a0288<\/p>\n

FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n

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

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

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

Expression\u00a0:\u00a0$[19+4\\sqrt21]?$<\/p>\n

= $[19+2\\sqrt{4\\times21}]=[19+2\\sqrt{84}]$<\/p>\n

= $7+12+2\\sqrt{7\\times12}$<\/p>\n

= $(7)^2+(12)^2+2\\sqrt{7\\times12}$<\/p>\n

Now, we know that $a^2+b^2+2ab=(a+b)^2$<\/p>\n

= $(\\sqrt{7}+\\sqrt{12})^2$<\/p>\n

Thus, square root is = $\\sqrt{7}+\\sqrt{12}$<\/p>\n

=\u00a0$\\sqrt7+2\\sqrt3$<\/p>\n

=> Ans – (A)<\/p>\n

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

Given\u00a0:\u00a0$x=\\frac{(3-2\\sqrt2)}{(3+2\\sqrt2)}$<\/p>\n

To find\u00a0: $\\sqrt{x}$<\/p>\n

Solution\u00a0: rationalizing the denominator, we get<\/p>\n

=>\u00a0$\\frac{(3-2\\sqrt2)}{(3+2\\sqrt2)}\\times\\frac{(3-2\\sqrt2)}{(3-2\\sqrt2)}$<\/p>\n

=\u00a0$\\frac{(3-2\\sqrt2)^2}{(3)^2-(2\\sqrt2)^2}$<\/p>\n

=\u00a0$\\frac{(3-2\\sqrt2)^2}{9-8}$<\/p>\n

=> $x=(3-2\\sqrt2)^2$<\/p>\n

Taking square root on both sides,<\/p>\n

=> $\\sqrt{x}=3-2\\sqrt2$<\/p>\n

=> Ans – (A)<\/p>\n

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

Expression\u00a0:\u00a0$\\frac{(3-2\\sqrt2)}{(3+2\\sqrt2)}$<\/p>\n

Rationalizing the denominator,<\/p>\n

=\u00a0$\\frac{(3-2\\sqrt2)}{(3+2\\sqrt2)}\\times\\frac{(3-2\\sqrt2)}{(3-2\\sqrt2)}$<\/p>\n

= $\\frac{(3-2\\sqrt2)^2}{(3+2\\sqrt2)(3-2\\sqrt2)}$<\/p>\n

= $\\frac{(3-2\\sqrt2)^2}{9-8}=(3-2\\sqrt2)^2$<\/p>\n

Thus, square root is =\u00a0$3-2\\sqrt2$<\/p>\n

=> Ans – (A)<\/p>\n

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

Expression\u00a0:\u00a0$4x+\\frac{\\sqrt{x}}{6}+\\frac{m^2}{4}$<\/p>\n

Let $\\sqrt{x}=y$<\/p>\n

= $4y^2+\\frac{y}{6}+\\frac{m^2}{4}$<\/p>\n

= $(2y)^2+2(2y)(\\frac{1}{24})+(\\frac{m}{2})^2$<\/p>\n

Using, $a^2+2ab+b^2=(a+b)^2$<\/p>\n

=> $\\frac{m}{2}=\\frac{1}{24}$<\/p>\n

=> $m=\\frac{2}{24}=\\frac{1}{12}$<\/p>\n

=> Ans – (B)<\/p>\n

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

Expression\u00a0:\u00a0$[25+4\\sqrt{39}]$<\/p>\n

= $[25+2\\sqrt{4\\times39}]=[25+2\\sqrt{156}]$<\/p>\n

= $13+12+2\\sqrt{13\\times12}$<\/p>\n

= $(13)^2+(12)^2+2\\sqrt{13\\times12}$<\/p>\n

Now, we know that $a^2+b^2+2ab=(a+b)^2$<\/p>\n

= $(\\sqrt{13}+\\sqrt{12})^2$<\/p>\n

Thus, square root is = $\\sqrt{13}+\\sqrt{12}$<\/p>\n

=\u00a0$\\sqrt13+2\\sqrt3$<\/p>\n

=> Ans – (A)<\/p>\n

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

Let the first part = $x$ and second part = $(20 – x)$<\/p>\n

According to ques, => $(x)^2 + (20 – x)^2 = 232$<\/p>\n

=> $x^2 + (x^2 + 400 – 40x) = 232$<\/p>\n

=> $2x^2 – 40x + 400 – 232 = 0$<\/p>\n

=> $x^2 – 20x + 84 = 0$<\/p>\n

=> $x^2 – 6x – 14x + 84 = 0$<\/p>\n

=> $x(x-6) – 14(x-6) = 0$<\/p>\n

=> $(x-6)(x-14) = 0$<\/p>\n

=> $x = 6,14$<\/p>\n

=> Ans – (A)<\/p>\n

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

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

<\/u><\/figure>\n

Diameter of the circle = $24\\sqrt{2}$ cm = Diagonal of the square.
\nWe know that, Diagonal of the square = $\\sqrt{2}a$ cm where a is side of the square.
\n$\\sqrt{2}a$ = $24\\sqrt{2}$ \u21d2 a = 24 cm
\nHere, Side of the square = Base of the triangle = Height of the triangle = 24 cm<\/p>\n

Therefore, Area of the triangle $= \\dfrac{1}{2} \\times \\text{Base} \\times \\text{Height} = \\dfrac{1}{2} \\times 24 \\times 24 = 288$ sq.cm<\/p>\n

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

<\/figure>\n

Given that the radius of the circle = $7\\sqrt{2}$ cm.
\nThen, the diameter of the circle = $14\\sqrt{2}$ cm.
\nHere, Diameter of the circle = Diagonal of the square = $14\\sqrt{2}$ cm.
\nWe know that, Diagonal of the square = $\\sqrt{2}a$ cm where a is side of the square.
\n$\\sqrt{2}a$ = $14\\sqrt{2}$ \u21d2 a = 14 cm
\nHence, Area of the square = 14*14 = 196 sq.cm<\/p>\n

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

Given x+y $= \\frac{20}{100}\\times(x^{2}+y^{2})$<\/p>\n

$\\Rightarrow$ x+y $= \\frac{1}{5}\\times(x^{2}+y^{2})$<\/p>\n

$\\Rightarrow$ $x^{2}+y^{2} =$ 5(x+y) $\\rightarrow$ (1)<\/p>\n

Also Given x+y $= \\frac{25}{100}\\times(x^{2}-y^{2})$<\/p>\n

$\\Rightarrow$ x+y $= \\frac{1}{4}\\times(x^{2}-y^{2})$<\/p>\n

$\\Rightarrow$ $x^{2}-y^{2} =$ 4(x+y) $\\rightarrow$ (2)<\/p>\n

Adding equation(1) and equation(2)<\/p>\n

$\\Rightarrow$ 2x$^{2} =$ 9(x+y)<\/p>\n

$\\therefore$ $\\frac{x+y}{x^{2}} = \\frac{2}{9}$<\/p>\n

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

Height of pyramid = $h=12$ cm and diagonal of base = $d=6\\sqrt2$ cm<\/p>\n

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

=> $s^2+s^2=d^2$<\/p>\n

=> $2s^2=(6\\sqrt2)^2=72$<\/p>\n

=> $s^2=\\frac{72}{2}=36$<\/p>\n

$\\therefore$ Volume of pyramid = $\\frac{1}{3}\\times$ Area of base $\\times$ Height<\/p>\n

= $\\frac{1}{3}\\times36\\times12=144$ $cm^3$<\/p>\n

=> Ans – (C)<\/p>\n\n

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

SSC CGL Square Root Questions: Download SSC CGL Trignometry questions with answers PDF based on previous papers very useful for SSC CGL exams. 20 Very important Trignometry objective questions for SSC exams. Question 1:\u00a0What is the positive square root of $[19+4\\sqrt21]?$ a)\u00a0$\\sqrt{7}+2\\sqrt{3}$ b)\u00a0$\\sqrt{3}+2\\sqrt{7}$ c)\u00a0$\\sqrt{2}+3\\sqrt{7}$ d)\u00a0$\\sqrt{7}+3\\sqrt{3}$ Question 2:\u00a0What is the square root of $\\frac{(3-2\\sqrt2)}{(3+2\\sqrt2)}$ ? a)\u00a0$3-2\\sqrt2$ […]<\/p>\n","protected":false},"author":32,"featured_media":27248,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[9,504,1459,1268],"tags":[462],"class_list":{"0":"post-27245","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-ssc","8":"category-ssc-cgl","9":"category-ssc-gd","10":"category-ssc-stenographer","11":"tag-ssc-cgl"},"better_featured_image":{"id":27248,"alt_text":"ssc cgl square root questions","caption":"ssc cgl square root questions","description":"ssc cgl square root 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