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### CBSE Class 10 Maths Chapter 6 – Triangles Objective Questions

CBSE Class 10 Maths Chapter 6 – Triangles Objective Questions

In this Chapter 6- Triangles of CBSE Class 10 Maths, the topics discussed include Introduction to Triangles, Similarity of Triangles, Areas of Similar Triangles, Pythagoras Theorem and so on. An important part of unit 3- Geometry, chapter 6- Triangles has nine theorems in total. Meanwhile, having a clear understanding of the concepts, theorems and the problem-solving methods in this chapter is mandatory to score well in the Board examination of class 10 maths. Taking this into consideration along with the fact that changed exam pattern will include more MCQs, we have compiled the CBSE Class 10 Maths Chapter 6 -Triangles Objective Questions for the students to solve. More practice will help them master the concepts.

6.1 Areas of Similar Triangles (4 MCQs From The Topic)

6.2 Basic Proportionality Theorem (4 MCQs Listed From Topic)

6.3 Criteria for Similarity of Triangles (4 MCQs From The Topic)

6.4 Pythagoras Theorem (4 MCQs Listed From The Given Topic)

6.5 Similar Triangles (4 MCQs From The Topic)

The CBSE Class 10 Maths Objective Questions containing these topics will help the students to prepare well for the exams.

Solution: We know that the ratio of areas of two similar triangles is equal

area of △ ABC / area of △ DEF = (AB/DE) 2 = (12/14) 2 = 36/49

Answer: (i) △ ADE ~ △ ABC and (ii) (area of △ ADE/ area of △ ABC) = (AD 2 /AB 2 )

Therefore, (area of △ ADE / area of △ ABC) = (AD 2 /AB 2 )

In the figure, PB and QA are perpendicular to segment AB. If OA = 5 cm, PO = 7cm and area (ΔQOA) = 150 cm 2 , find the area of ΔPOB.

Therefore, ∠ AQO = ∠ PBO [Alternate angles]

∠ QOA = ∠ BOP [Vertically opposite angles]

Now, area (POB)/ area (QOA) = (OP) 2 / (OA) 2 = 7 2 / 5 2

Solution : For similar isosceles triangles,

Area (Δ 1 ) / Area (Δ 2 ) = (h 1 ) 2 / (h 2 ) 2

The Angle-Bisector theorem states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides (It may be similar or may not depending on type of triangle it divides)

as per the statement AB/ AC= BD/DC i.e. a/b= c/d

In ABC, Given that DE//BC, D is the midpoint of AB and E is a midpoint of AC.

⇒ AD/DB= AE/EC (Basic Proportionality Theorem)

On substituting value of EC in (1), we get

DE = BC/2 only if D and E are the mid points of AB and AC respectively. So this may not be true always.

Solution: Since the given triangles are similar, the ratios of corresponding sides are equal.

Solution: If two triangles are similar, corresponding sides are proportional.

Therefore, △ABC∼△DEC       [by AA similarity]

Answer: (B)a 2  = b 2  +  c 2  + bc

BC 2  =  CD 2  + BD 2       [By Pythagoras Theorem]

BC 2  =  CD 2  + DA 2  + AB 2  + (2×DA×AB)         (i)

CD 2  + DA 2  = AC 2       (ii)  [By Pythagoras Theorem]

Putting the values from (ii) and (iii) in (i), we get

Since  ∠A is an obtuse angle in  ΔABC, so

Solution: Let AB and CD be two trees such that AB = 20 m, CD = 28 m & BD = 17 m

Draw BE parallel to CD. Then, ED = 8 m.

In the figure △ABC is a right angled triangle with right angle at B. BD is perpendicular to AC. Then which of the following options will hold true?

By AA similarity criterion, △ABD ~ △ACB

ar (ΔABD)/ ar(ΔACB) = (AB/AC) 2 = (AD/AB) 2 = (BD/CB) 2

Solution: AB/DE= AC/DF=BC/EF=2/4=1/2

∴ Perimeter of △DEF = (DE + EF + DF) = 15 cm.

Since, the ratio of adjacent sides and the included angles are equal.

∴△ABC is similar to △EDF by SAS similarity criterion.

Now, ∠C = ∠F = 65 °        [Corresponding angles of a similar triangles are equal]

Solution: Ratio of heights = Ratio of sides = 1: 3.

Triangle is the most interesting and exciting chapters of unit 3- Geometry as it takes the students through the different aspects and concepts related to the Geometrical figure triangle. A triangle is a plane figure with three sides and three angles. This chapter deals with several topics and sub-topics related to triangles. Check out below for some of the sub-topics covered in this chapter.

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