{"id":190431,"date":"2021-11-24T16:26:58","date_gmt":"2021-11-24T10:56:58","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=190431"},"modified":"2021-11-24T16:26:58","modified_gmt":"2021-11-24T10:56:58","slug":"angles-questions-for-nmat-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/angles-questions-for-nmat-pdf\/","title":{"rendered":"Angles Questions for NMAT – Download PDF"},"content":{"rendered":"

Angles Questions for NMAT – Download [PDF]<\/h1>\n

Download Angles Questions for NMAT PDF – NMAT Fill in the blanks questions pdf by Cracku. Top 10 very important Angles Questions for NMAT based on asked questions in previous exam papers.<\/p>\n

Download Angles Questions for NMAT<\/a><\/p>\n

Get 5 NMAT Mocks for Rs. 499<\/a><\/p>\n

Take NMAT mock test<\/p>\n

Question 1:\u00a0<\/b>Which of the following can’t be the ratio of angles of an acute angled triangle?<\/p>\n

a)\u00a01:9:9<\/p>\n

b)\u00a02:3:4<\/p>\n

c)\u00a03:7:8<\/p>\n

d)\u00a01:1:1<\/p>\n

e)\u00a01:2:3<\/p>\n

Question 2:\u00a0<\/b>Two right angled triangles ABC and DCB are drawn on the same side of BC. If BC = 30, AB = 10 and CD = 15, and AC and BD intersect at P, find the distance of P from BC.<\/p>\n

a)\u00a08 cm<\/p>\n

b)\u00a06 cm<\/p>\n

c)\u00a010 cm<\/p>\n

d)\u00a05 cm<\/p>\n

e)\u00a07 cm<\/p>\n

Question 3:\u00a0<\/b>Two sides of a plot measure 32 m and 24 m and the angle between them is a perfect right angle. The other two sides measure 25 m each and the other three angles are not right angles. What is the area of the plot?
\n[CAT 2001]<\/p>\n

a)\u00a0768 $m^2$<\/p>\n

b)\u00a0534 $m^2$<\/p>\n

c)\u00a0696.5 $m^2$<\/p>\n

d)\u00a0684 $m^2$<\/p>\n

Question 4:\u00a0<\/b>Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer then how many such triangles exist?<\/p>\n

[CAT 2008]<\/p>\n

a)\u00a05<\/p>\n

b)\u00a021<\/p>\n

c)\u00a010<\/p>\n

d)\u00a015<\/p>\n

e)\u00a014<\/p>\n

Question 5:\u00a0<\/b>Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer then how many such triangles exist?<\/p>\n

a)\u00a05<\/p>\n

b)\u00a021<\/p>\n

c)\u00a010<\/p>\n

d)\u00a015<\/p>\n

e)\u00a014<\/p>\n

<\/a><\/div>\n

Question 6:\u00a0<\/b>2 polygons with (n-1) and (n+2) sides respectively have their exterior angles in such a way that the difference between the exterior angles is 6 degrees. What is the value of n?<\/p>\n

a)\u00a011<\/p>\n

b)\u00a015<\/p>\n

c)\u00a012<\/p>\n

d)\u00a013<\/p>\n

Question 7:\u00a0<\/b>What is the number of distinct triangles with integral valued sides and perimeter 14?[CAT 2000]<\/p>\n

a)\u00a06<\/p>\n

b)\u00a05<\/p>\n

c)\u00a04<\/p>\n

d)\u00a03<\/p>\n

Question 8:\u00a0<\/b>The sum of all interior angles of an isosagon ?<\/p>\n

a)\u00a03240$^{\\circ}$<\/p>\n

b)\u00a0360$^{\\circ}$<\/p>\n

c)\u00a0720$^{\\circ}$<\/p>\n

d)\u00a0900$^{\\circ}$<\/p>\n

Question 9:\u00a0<\/b>If two complementary angles are in the ratio 1: 4, find the supplement of larger angle ?<\/p>\n

a)\u00a0157.5$^{\\circ}$<\/p>\n

b)\u00a0108$^{\\circ}$<\/p>\n

c)\u00a0137.5$^{\\circ}$<\/p>\n

d)\u00a0120$^{\\circ}$<\/p>\n

Question 10:\u00a0<\/b>A pair of parallel lines are cut by a transversal and the sum of a pair of alternate interior angles A and B is 100 degrees. What is the sum of the pair of corresponding angles that has A as one of its angles?<\/p>\n

a)\u00a0180 degrees<\/p>\n

b)\u00a0100 degrees<\/p>\n

c)\u00a0260 degrees<\/p>\n

d)\u00a0None of the above<\/p>\n

Join 7K MBA Aspirants Telegram Group<\/a><\/p>\n

Download Highly Rated CAT preparation App<\/a><\/p>\n

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

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

In an acute angled triangle, the sum of the smallest two angles is greater than the third angle. If the ratio is 1:2:3, the angles are 30, 60 and 90 which is a right angled triangle.<\/p>\n

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

<\/p>\n

Let PT be the perpendicular drawn from P to BC.
\nLet PT be ‘y’. Let CT be ‘x’. This means BT = 30-x<\/p>\n

Triangles ABC and PTC are similar.
\n=> $\\frac{PT}{AB}=\\frac{CT}{BC} => \\frac{y}{10}=\\frac{x}{30}$ —– (1)<\/p>\n

Triangles DBC and PBT are similar.
\n=> $\\frac{PT}{CD}=\\frac{BT}{BC} => \\frac{y}{15}=\\frac{30-x}{30}$ —– (2)<\/p>\n

Solving equations (1) and (2) we get y = 6 => PT = 6 cm.<\/p>\n

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

Length of the diagonal of the right triangle is 40. The height of the isosceles triangle formed, with 40 as its base is 15.
\nSo, area = $\\frac{1}{2}* 32 * 24 + \\frac{1}{2} * 40 * 15 = 384 + 300 = 684 m^2$<\/p>\n

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

For obtuse-angles triangle, $c^2 > a^2 + b^2$ and c < a+b<\/p>\n

If 15 is the greatest side, 8+x > 15 => x > 7 and 225 > 64 + x^2 => x^2 < 161 => x <= 12<\/p>\n

So, x = 8, 9, 10, 11, 12<\/p>\n

If x is the greatest side, then 8 + 15 > x => x < 23<\/p>\n

x^2 > 225 + 64 = 289 => x > 17<\/p>\n

So, x = 18, 19, 20, 21, 22<\/p>\n

So, the number of possibilities is 10<\/p>\n

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

For obtuse-angles triangle, $c^2 > a^2 + b^2$ and c < a+b
\nIf 15 is the greatest side, 8+x > 15 => x > 7 and $225 > 64 + x^2 => x^2 < 161 => x <= 12$
\nSo, x = 8, 9, 10, 11, 12
\nIf x is the greatest side, then 8 + 15 > x => x < 23
\n$x^2$ > 225 + 64 = 289 => x > 17
\nSo, x = 18, 19, 20, 21, 22
\nSo, the number of possibilities is 10<\/p>\n

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

The exterior angles are given by 360\/(n-1) and 360\/(n+2)
\nSo, 360\/(n-1) – 360\/(n+2) = 6 => 720 + 360 = 6(n^2 -3n – 2) => (n-1)(n+2) = 180.
\n15*12 = 180, so, n-1 =12 and n+2 = 15 and n = 13<\/p>\n

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

Let the sides be x, y and 14-(x+y)
\nx+y > 14-(x+y) => x+y > 7
\nx+14-x-y > y => y < 7
\nSimilarly, x < 7
\nIf x = 1, y = 7 (not possible)
\nSo, if x = 2, y = 6
\nif x = 3, y = 5
\nif x = 4, y = 4, 5
\nThe cases for x = 5 and 6 are already taken care of by y.
\nNumber of possible cases = 4<\/p>\n

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

Isosagon is a regular polygon with 20 sides.<\/p>\n

Sum of interior angles = (n-2)*180; where n is the number of sides.<\/p>\n

Sum of angles =\u00a0(20-2)*180 = 3240<\/p>\n

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

Let the two angles be X and 90-X.
\nHence, X = 4(90-X) or X = 72 $^{\\circ}$
\nHence, the supplement of the larger angle is
\n180$^{\\circ}$ – 4\/5(90$^{\\circ}$)=108$^{\\circ}$<\/p>\n

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

<\/figure>\n

Since A and B are alternate interior angles, they are equal and each is equal to 50 degrees. Since corresponding angles are equal, the sum of A and its corresponding angle, say C is 50+50 = 100 degrees.<\/p>\n

We hope these Angles Questions for NMAT pdf for the NMAT exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Angles Questions for NMAT – Download [PDF] Download Angles Questions for NMAT PDF – NMAT Fill in the blanks questions pdf by Cracku. Top 10 very important Angles Questions for NMAT based on asked questions in previous exam papers. Take NMAT mock test Question 1:\u00a0Which of the following can’t be the ratio of angles of […]<\/p>\n","protected":false},"author":32,"featured_media":190540,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,169,125,4977],"tags":[5195,4979],"class_list":{"0":"post-190431","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-downloads","9":"category-featured","10":"category-nmat","11":"tag-angles","12":"tag-nmat-2021"},"better_featured_image":{"id":190540,"alt_text":"Angles Questions for NMAT","caption":"Angles Questions for NMAT","description":"Angles Questions for 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