{"id":37270,"date":"2022-02-09T18:19:12","date_gmt":"2022-02-09T12:49:12","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=37270"},"modified":"2022-02-25T09:39:44","modified_gmt":"2022-02-25T04:09:44","slug":"mcq-questions-for-class-10-maths-chapter-11","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-10-maths-chapter-11\/","title":{"rendered":"MCQ Questions for Class 10 Maths Chapter 11 Constructions"},"content":{"rendered":"
Question 1.<\/p>\n
(A) 8
\n(B) 10
\n(C) 11
\n(D) 12
\nAnswer:
\n(D) 12<\/p>\n
Explanation:
\nMinimum number of the points marked = sum of ratios = 5 + 7 = 12.<\/p>\n
<\/p>\n
Question 2.<\/p>\n
(A) A12<\/sub> Explanation: Question 3.<\/p>\n (A) A5<\/sub> and B6<\/sub> Explanation: Question 4.<\/p>\n (A) 135\u00b0 Explanation: Case – Based MCQs<\/span><\/p>\n Attempt any four sub-parts from each question. Each sub-part carries 1 mark. Question 1.<\/p>\n (A) PD Explanation: <\/sub><\/p>\n Question 2.<\/p>\n (A) 98\u00b0 Explanation: Question 3.<\/p>\n (A) 3 : 4 Explanation: Question 4.<\/p>\n (A) 2:5 Explanation: <\/p>\n Question 5.<\/p>\n (A) 6 m Explanation: Constructions Class 10 MCQ Questions with Answers Question 1. To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn so that \u2220BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is …<\/p>\n
\n(B) A11<\/sub>
\n(C) A10<\/sub>
\n(D) A9<\/sub>
\nAnswer:
\n(B) A11<\/sub><\/p>\n
\nWe have to divide the line segment into 7 + 4 = 11 equal parts and 11th part will bem joined to B, here A12<\/sub> will never appear.<\/p>\nTo divide a line segment AB in the ratio 5 : 6 draw a ray AX such that\u2220BAX is an acute angle, then draw a ray BY parallel to AX, and the points, A1<\/sub>, A2<\/sub>, A3<\/sub>,… and B1<\/sub>, B2<\/sub>, B3<\/sub>, … are located at. Equal distances on ray AX and BY, respectively. Then the points joined are<\/h2>\n
\n(B) A6<\/sub> and B5<\/sub>
\n(C) A4<\/sub> and B5<\/sub>
\n(D) A5<\/sub> and B4<\/sub>
\nAnswer:
\n(A) A5<\/sub> and B6<\/sub><\/p>\n
\nIn the figure; segment AB of given length is divided into 2 parts of ratio 5 : 6 in following steps:
\n(i) Draw a line-segment AB of given length.
\n(ii) Draw an acute angle BAX as shown in figure either upside or down side.
\n(iii) Draw angle \u2220ABY = \u2220BAX on other side of AX, that is, down side.
\n(iv) Divide AX into 5 equal parts by using compass.
\n(v) Divide BX into same distance in 6 equal parts as AX was divided.
\n(vi) Now, join A5 and B6 which meet AB at P. P divides AB in ratio AP : PB = 5 : 6.
\n<\/p>\nTo draw a pair of tangents to a circle which are in\u00acclined to each other at an angle of 60\u00b0, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be:<\/h2>\n
\n(B) 90\u00b0
\n(C) 60\u00b0
\n(D) 120\u00b0
\nAnswer:
\n(D) 120\u00b0<\/p>\n
\n
\nWe know that tangent and radius at contact point are perpendicular to each other. So, \u2220P and \u2220Q in quadrilateral TPOQ formed by tangents and radii will be of 90\u00b0 each.
\nSo, the sum of
\n\u2220T+ \u2220O = 180\u00b0
\nas T = 60\u00b0 [Given]
\n\u2234\u2220O = 180\u00b0 – 60\u00b0
\n= 120\u00b0<\/p>\n
\nI. Read the following text and answer the below questions:
\nA school conducted Annual Sports Day on a triangular playground. On the ground, parallel lines have been drawn with chalk powder at a distance of 1 m. 7 flower pots have been placed at a distance of from each other along DM as shown in the figure.
\n
\nNow answer the following questions:<\/p>\nPD3<\/sub> is parallel to:<\/h2>\n
\n(B) PE
\n(C) ED7<\/sub>
\n(D) None of these.
\nAnswer:
\n(C) ED7<\/sub><\/p>\n
\nWe have,
\nPD(C) ED3<\/sub> || ED7<\/sub><\/p>\nIf \u2220PD3D = 82\u00b0, then the measure of \u2220ED7D is:<\/h2>\n
\n(B) 82\u00b0
\n(C) 90\u00b0
\n(D) 45\u00b0
\nAnswer:
\n(B) 82\u00b0<\/p>\n
\nWe have,
\nPD(C) ED3<\/sub> || ED7<\/sub>
\nThen, \u2220ED7<\/sub>D = \u2220PD3<\/sub>D
\n(Corresponding angles)
\n\u2234\u2220ED7<\/sub>D = 82\u00b0.<\/p>\nThe ratio in which P divides DE, is:<\/h2>\n
\n(B) 7 : 3
\n(C) 3 : 7
\n(D) 2 : 5
\nAnswer:
\n(A) 3 : 4<\/p>\n
\nP divides DE in the ratio 3 : 4.<\/p>\nThe ratio of DE to DP will be:<\/h2>\n
\n(B) 3 : 4
\n(C) 3 : 7
\n(D) 7 : 3
\nAnswer:
\n(D) 7 : 3<\/p>\n
\nDE = 7 m
\n[\u2235 DD1<\/sub> = D1<\/sub>D2<\/sub> = D2<\/sub>D3<\/sub> = D3<\/sub>D4<\/sub>
\n= D4<\/sub>D5<\/sub>= D5<\/sub>D6<\/sub> = D6<\/sub>D7<\/sub>]
\nand DP = 3 m
\n[\u2234 DD1<\/sub> = D1<\/sub>D2<\/sub> = D2<\/sub> D3<\/sub>]
\n\u2234 \\(\\frac{D E}{D P}\\) = \\(\\frac{7 \\mathrm{~m}}{3 \\mathrm{~m}}\\) = \\(\\frac{7}{3}\\)
\nHence, the ratio of DE to DP is 7 : 3.<\/p>\nThe total distance used for putting 7 flower pots is:<\/h2>\n
\n(B) 7 m
\n(C) 5 m
\n(D) 8 m.
\nAnswer:
\n(B) 7 m<\/p>\n
\nSince, 7 flower pots have been placed at a distance of 1 m from each other, then total distance = 7 m.<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n
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