Download Mensuration Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>What is the area of the shaded portion, if the side of the square measure 21 ms.? a)\u00a086.5 sq. cms.<\/p>\n b)<\/p>\n c)\u00a094.5 sq. cms.<\/p>\n d)\u00a081.5 sq. cms.<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Area of shaded portion will be the area of square minus the area of the circle.<\/p>\n i.e<\/p>\n 21^2 – (pi * 21^2 )\/4<\/p>\n =94.5<\/p>\n Question 2:\u00a0<\/b>The area of the square of side 8 cm is equal to a rectangle. Which of the following statements\/is\/are definitely true about the rectangle (Sides of the rectangle are integers)?<\/p>\n a)\u00a0The length of the rectangle is 16 times its breadth<\/p>\n b)\u00a0The length of the rectangle is 32 times its breadth<\/p>\n c)\u00a0The breadth of the rectangle is $\\frac{1}{6}$ of its length<\/p>\n d)\u00a0The breadth of the rectangle is $\\frac{1}{9}$ of its length<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Area of rectangle = $8^{2}$. Question 3:\u00a0<\/b>In the following figure, ABC is an equilateral triangle which is inscribed inside a circle and whose radius is r. Which of the following is the area of the triangle ? a)\u00a0$(r-DE)^\\frac{1}{2}(r+DE)^2$<\/p>\n b)\u00a0$(r-DE)^2$ $(r+DE)^2$<\/p>\n c)\u00a0$(r-DE)^\\frac{1}{2}$ $(r+DE)^\\frac{3}{2}$<\/p>\n d)\u00a0$(r+DE)^2$ $(r-DE)^\\frac{3}{2}$<\/p>\n e)\u00a0None of these<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Hence Option C is the correct answer.<\/span><\/span><\/span><\/span><\/span><\/p>\n \u00a0<\/span><\/p>\n Question 4:\u00a0<\/b>The radius of a circle is 20% more than the height of a right angled triangle. The base of triangle is 36 cm. If the area of triangle and circle be equal, what will be the area of circle ?<\/p>\n a)\u00a072 $cm^2$<\/p>\n b)\u00a0144 $cm^2$<\/p>\n c)\u00a0216 $cm^2$<\/p>\n d)\u00a0128 $cm^2$<\/p>\n e)\u00a0Cannot be determined<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the height of the triangle be ‘h’. Area of the triangle = (1\/2) * 36 * h = 18h<\/p>\n Radius of the circle = 1.2h<\/p>\n Area of the circle = $\\pi * 1.44 h^2$<\/p>\n Since the areas are equal, 18h = (22\/7) * 1.44 * $h^2$<\/p>\n => h = 18*7\/1.44\/22 =\u00a03.97727273<\/p>\n So, the area of the circle = Area of the triangle = 18 *\u00a03.97727273 = 72 $cm^2$ approx.<\/p>\n Question 5:\u00a0<\/b>The area of a rectangular field is 460 square metres.If the length is 15 per cent more than the breadth, what is breadth of the rectangular field ?<\/p>\n a)\u00a015 metres<\/p>\n b)\u00a026 metres<\/span><\/p>\n c)\u00a034.5 metres<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 5)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n let the length and breadth if rectangle be L and B respectively.<\/p>\n It is given that L = 1.15B<\/p>\n Area is given as 460 sqmtr<\/p>\n L x B = 460<\/p>\n So B = 20 m<\/p>\n Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>What will be the cost of gardening 1 metre broad boundary around a rectangular plot having perimeter of 340 meters at the rate of Rs 10 per square metre?<\/p>\n a)\u00a0Rs 3,400\/-<\/p>\n b)\u00a0Rs 1,700\/-<\/p>\n c)\u00a0Rs 3,440\/-<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n here the cost of boundary making = 10 per sqmt<\/p>\n Total area to be boundaried = 2 L + 2B + (1\u00d74)…(1)<\/p>\n It is given that perimeter of rectangle is 340 m<\/p>\n So 2(L + B ) = 340….(2)<\/p>\n From equation 1 and 2<\/p>\n Total area to be boundaried = 344 sq mtr<\/p>\n Cost of boundaring 344 sq mtr area = 344 \u00d7 10 = Rs 3440<\/p>\n Question 7:\u00a0<\/b>The area of a square is thrice the area of a rectangle If the area of the square is 225 sq cm and the length of the rectangle is 15 cm what is the difference between the breadth of the rectangle and the side of the square ?<\/p>\n a)\u00a08 cm<\/p>\n b)\u00a010 cm<\/p>\n c)\u00a012 cm<\/p>\n d)\u00a06 cm<\/p>\n e)\u00a0None of these<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n area of square = $(side)^2$ = 225<\/p>\n Side of square = 15 cm<\/p>\n It is given that area of square is thrice area of rectangle ,so<\/p>\n 225 = 3 x L x B where L and B are length and breadth of rectangle.<\/p>\n Given that length of rectangle is 15 cm so B = 5 cm<\/p>\n Side of square – Breadth of rectangle = 15 -5 =10 cm<\/p>\n Question 8:\u00a0<\/b>The diameter of a circle is 7.7 cm What is the circumference of the circle ?<\/p>\n a)\u00a026.4 cm<\/p>\n b)\u00a024.2 cm<\/p>\n c)\u00a028.4 cm<\/p>\n d)\u00a022.2 cm<\/p>\n e)\u00a0None of these<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n diameter = 7.7 cm<\/p>\n Radius = Diameter\/2 = 7.7\/2<\/p>\n Circumference = 2 \u03c0 R = 2 x 3.14 x 3.85 = 24.178 cm ~24.2 cm<\/p>\n Question 9:\u00a0<\/b>A rectangular field has its length and breadth in the ratio of 6:5 respectively A man riding a bicycle completes one lap of this field along its perimeter at the speed of 19.8 km\/h in 2 min What is the area of the field ?<\/p>\n a)\u00a019200 sq m<\/p>\n b)\u00a027000 sq m<\/p>\n c)\u00a032500 sq m<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n here the speed of cycle is given = 19.8 km\/hr = 19.8 \u00d7 $\\frac{5}{18} $ m\/s = 5.5 m\/s<\/p>\n Time = 2 min = 120 sec<\/p>\n Distance = speed \u00d7 time<\/p>\n Distance = 5.5 \u00d7 120 = 660 m<\/p>\n As length : breadth = 6:5<\/p>\n Length = 6y<\/p>\n Breadth = 5y<\/p>\n Perimeter = 660 m<\/p>\n 22y = 660<\/p>\n y = 30 m<\/p>\n So length = 180 m<\/p>\n Breadth = 150 m<\/p>\n Area = 180\u00d7150 = 27000 sq mtr<\/p>\n Question 10:\u00a0<\/b>What would be the area of a circle whose circumference is 35.2 cm?<\/p>\n a)\u00a067.22 sq cm<\/p>\n b)\u00a075.54 sq cm<\/p>\n c)\u00a098.56 sq cm<\/p>\n d)\u00a086.75 sq cm<\/p>\n e)\u00a0None of these<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n circumference of circle = 2\u03c0R = 35.2<\/p>\n R is the radius if circle<\/p>\n R= 5.6 cm<\/p>\n Area of circle = \u03c0$(R)^2$ = 3.14 \u00d7 $(5.6)^2$ = 98.56 sq cm<\/p>\n Question 11:\u00a0<\/b>The diameter of a circle is 7.7 cm What is the circumference of the circle ?<\/p>\n a)\u00a026.4 cm<\/p>\n b)\u00a024.2 cm<\/p>\n c)\u00a028.4 cm<\/p>\n d)\u00a022.2 cm<\/p>\n e)\u00a0None of these<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n diameter = 7.7 cm<\/p>\n Radius = Diameter\/2 = 7.7\/2<\/p>\n Circumference = 2 \u03c0 R = 2 x 3.14 x 3.85 = 24.178 cm ~24.2 cm<\/p>\n Question 12:\u00a0<\/b>The area of a square is thrice the area of a rectangle If the area of the square is 225 sq cm and the length of the rectangle is 15 cm what is the difference between the breadth of the rectangle and the side of the square ?<\/p>\n a)\u00a08 cm<\/p>\n b)\u00a010 cm<\/p>\n c)\u00a012 cm<\/p>\n d)\u00a06 cm<\/p>\n e)\u00a0None of these<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n area of square = $(side)^2$ = 225<\/p>\n Side of square = 15 cm<\/p>\n It is given that area of square is thrice area of rectangle ,so<\/p>\n 225 = 3 x L x B where L and B are length and breadth of rectangle.<\/p>\n Given that length of rectangle is 15 cm so B = 5 cm<\/p>\n Side of square – Breadth of rectangle = 15 -5 =10 cm<\/p>\n Question 13:\u00a0<\/b>What will be the cost of gardening 1 metre broad boundary around a rectangular plot having perimeter of 340 meters at the rate of Rs 10 per square metre?<\/p>\n a)\u00a0Rs 3,400\/-<\/p>\n b)\u00a0Rs 1,700\/-<\/p>\n c)\u00a0Rs 3,440\/-<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n here the cost of boundary making = 10 per sqmt<\/p>\n Total area to be boundaried = 2 L + 2B + (1\u00d74)…(1)<\/p>\n It is given that perimeter of rectangle is 340 m<\/p>\n So 2(L + B ) = 340….(2)<\/p>\n From equation 1 and 2<\/p>\n Total area to be boundaried = 344 sq mtr<\/p>\n Cost of boundaring 344 sq mtr area = 344 \u00d7 10 = Rs 3440<\/p>\n Question 14:\u00a0<\/b>The area of a rectangular field is 460 square metres.If the length is 15 per cent more than the breadth, what is breadth of the rectangular field ?<\/p>\n a)\u00a015 metres<\/p>\n b)\u00a026 metres<\/span><\/p>\n c)\u00a034.5 metres<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 14)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n let the length and breadth if rectangle be L and B respectively.<\/p>\n It is given that L = 1.15B<\/p>\n Area is given as 460 sqmtr<\/p>\n L x B = 460<\/p>\n So B = 20 m<\/p>\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/12\/16\/black.jpg\")
<\/p>\n\n
\nArea of rectangle = 64 $cm^{2}$.
\nLength$\\times$Breadth = 64.
\nCombination of Length and breadth is (32,2).
\nHence, length of rectangle is 16 times the breadth of rectangle.
\nTherefore, Option A is correct.<\/p>\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/12\/16\/tri.jpg\")
<\/p>\n
\nBD=DC = r
\nAccording to Pythagoras Theorem , BD$^{2}$ = r$^{2}$ = DE$^{2}$+ BE$^{2}$
\n$BE^{2}$= $r$$^{2}$ – $ DE$$^{2}$
\n$BE$=$(r^{2} – DE^{2})$ $^{\\frac{1}{2}}$
\n$r$$^{2}$ – $ DE$$^{2}$<\/span>= $(r+DE) \\times (r-DE)$.
\n$BC=2\\times BE$=
\n$AE= (r+DE)$
\nArea of $\\triangle$ ABC = $\\frac{1}{2} \\times BC \\times AE$ = $\\frac{1}{2} \\times 2\\times (r^{2}$ – $DE^{2})$<\/span> $\\times$ $(r+DE)$
\n<\/span><\/span><\/span><\/span><\/span>$\\frac{1}{2} \\times 2\\times (r^{2}$ – $DE^{2})$<\/span> $\\times$ $(r+DE)$ = $(r-DE) \\times (r+DE) \\times (r+DE)$ = $(r-DE)^\\frac{1}{2}$ $(r+DE)^\\frac{3}{2}$<\/p>\n