NCERT Exemplar Solution for CBSE Class 10 Mathematics: Triangles (Part-IB)
NCERT Exemplar Problems and Solutions (Part-IB) for CBSE Class 10 Mathematics chapter 6, Triangles, is available here. This part contains only the multiple choice type questions and a detailed solution is provided for each question. These questions will prove to be very helpful while preparing for CBSE Class 10 Board Exam 2017-2018.
Here you get the CBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar Problems and Solutions (Part-IB). This part of the chapter includes solutions of Question Number 7 to 12 from Exercise 6.1 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Triangles. This exercise comprises only the Multiple Choice Questions (MCQs) framed from various important topics in the chapter. Each question is provided with a detailed solution.
NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Triangles:
Multiple Choice Questions (Q. No. 7-12)
Question. 7 In ΔABC and ΔDEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE then, the two triangles are
(a) congruent but not similar
(b) similar but not congruent
(c) neither congruent nor similar
(d) congruent as well as similar
Here, two triangles DABC and DDEF do not satisfy any rule of congruency, (SAS, ASA, SSS), so both are not congruent.
Question. 9 If ΔABC ~ ΔDFE, ∠A = 30°, ∠C = 50°, AB = 5cm, AC = 8cm and DF = 7.5cm. Then, which of the following is true?
(a) DE =12cm, ∠F =50°
(b) DE =12cm, ∠F =100°
(c) EF =12cm, ∠D =100°
(d) EF =12cm, ∠D =30°
And, angle formed by DE and FD is ∠D.
So, ∠B = ∠D
⟹ DABC ~ DEDF (By SAS criterion)
As the ratio of area of two similar triangles is equal to the ratio of square of their corresponding sides.
Therefore, we get:
Question. 12 If S is a point on side PQ of a ΔPQR such that PS = QS = RS, then
(a) PR . QR = RS2
(b) QS2 + RS2 =QR2
(c) PP2 + QP2 = PQ2
(d) PS2 + RS2 = PR2
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