Triangles

Important Questions

Here are some important questions for Class 10 Mathematics Chapter 6, Triangles, carefully selected to help students prepare for the CBSE Class 10 Mathematics Examination in 2024-25. These questions cover various types of problems and are designed to assist students in understanding Triangles better. By practicing these diverse question types, students can clarify any doubts they may have and improve their problem-solving skills, leading to better performance in the chapter on Triangles.

Table of Contents

In Chapter 6 of Class 10 Mathematics,Triangles, We will learn about the significance of corresponding angles and corresponding sides in establishing the similarity between triangles. Along the way, we will discover the beauty of geometric proofs and explore how to apply them to establish the similarities between triangles.

A triangle is a closed, three-sided polygon. Its basic properties include:

- The sum of the interior angles of a triangle is always 180 degrees.
- The sum of any two sides of a triangle is always greater than the length of the third side (Triangle Inequality Theorem).
- The longest side of a triangle is opposite the largest angle, and the shortest side is opposite the smallest angle.

(b). 8.25 cm

(c). 5.28 cm

(d). 8.52 cm

**Ans.** (b)

**Explanation:**

Since PS is the internal bisector of ∠P and it meets QR at S image.

(b). 26 m

(c). 62 m

(d). 13 m

**Ans.** (b)

**Explanation:**

Applying Pythagoras’ Theorem

AC^2 = AB^2 +BC^2 \\[4.5 bp]
⇒ AC^2 =122 +52

= 144+25=169=132

⇒ AC=13m

**Ans.** AE = 2 cm

**Explanation:**

Given, DE || BC, AD = 2.4 cm, DB = 3.6 cm and AC = 5 cm

In ABC, DE || BC

\therefore \dfrac{AD}{DB}=\dfrac{AE}{CE} \space \space \text{[by basic proportionality theorem]} \\[4.5 bp]
\Rightarrow \dfrac{AD}{DB}=\dfrac{AE}{AC-AE} \\[4.5 bp]
\Rightarrow \dfrac{2.4}{3.6}=\dfrac{AE}{5-AE} \\[4.5 bp]
\Rightarrow \dfrac{2}{3}=\dfrac{AE}{5-AE} \\[4.5 bp]
\Rightarrow 3AE + 2AE = 10 \\[4.5 bp]
\Rightarrow 3AE = 10 - 2AE \\[4.5 bp]
\Rightarrow AE = \dfrac{10}{5} \\[4.5 bp]
\Rightarrow AE = 2 cm

**Ans.** Proved

**Explanation:**

In ∆BMC and ∆EMD, we have

∠1 = ∠2 [Vertically opposite angles] MC = MD [M being the mid-point of CD] ∠BCM = ∠EDM [Alternate angles] Thus ∆BMC ≅ ∆EMD [by ASA] \Rightarrow BC = DE (cpct)

Again, BC = AD [Opposite sides of the parallelogram ABCD] \therefore BC = AD = DE

So, AE = AD + DE = 2BC …(i)

Again, in ∆AEL and ∆CBL, ∠5 = ∠6 [Vertically opposite angles]

∠3 = ∠4 [Alternative angles]

So, ∆AEL ~ ∆CBL

\therefore \dfrac{EL}{BL}=\dfrac{AE}{BC}=\dfrac{2BC}{BC}=2 [From (i)]Thus, EL = 2BL.

**Ans.** AC = 4.9 cm

**Explanation:**

Given, AD is the bisector of ∠A.

Thus, ∠BAD = ∠CAD

Now BC = BD + DC

\Rightarrow 6 = 3.2 + DC \\[4.5 bp]
\Rightarrow DC = 2.8 cm

Since AD is the angle bisector of ∠A,

Chapter No. | Chapter Name |
---|---|

Chapter 1 | Real Numbers |

Chapter 2 | Polynomials |

Chapter 3 | Pair of Linear Equations in Two Variable |

Chapter 4 | Quadratic Equations |

Chapter 5 | Arithmetic Progressions |

Chapter 6 | Triangles |

Chapter 7 | Coordinate Geometry |

Chapter 8 | Introduction to Trigonometry |

Chapter 9 | Some Applications of Trigonometry |

Chapter 10 | Circles |

Chapter 11 | Areas Related to Circle |

Chapter 12 | Surface Areas and Volumes |

Chapter 13 | Statistics |

Chapter 14 | Probability |

If you want to improve your understanding of the concepts in the Triangle chapter, you can visit oswal.io. They offer a comprehensive collection of questions designed to help you practice and reinforce what you’ve learned. By engaging with these questions, you can strengthen your knowledge of triangles and become more proficient at solving problems related to this fundamental geometric shape.

Ans: The Side-Side-Side (SSS) Similarity Criterion states that if the corresponding sides of any two triangles are in the same ratio, then their corresponding angles will be equal, and the triangles will be considered similar triangles.

Ans: The Basic Proportionality Theorem states that if a line is drawn parallel to one side of a triangle, intersecting the other two sides at distinct points, then the segments formed on those sides are divided in the same ratio. In other words, if we have a triangle ABC with a line PQ parallel to BC, intersecting AB at P and AC at Q, then the theorem asserts that AP/PB = AQ/QC.

Ans: The six main types of triangles in mathematics are:

**Scalene Triangle:**A scalene triangle has all three sides of different lengths, and all three angles are also different**Isosceles Triangle:**An isosceles triangle has two sides of equal length, and consequently, two angles are also equal.**Equilateral Triangle:**An equilateral triangle has all three sides of equal length, making all three angles equal to 60 degrees.**Acute Triangle:**An acute triangle has all three angles less than 90 degrees.**Obtuse Triangle:**An obtuse triangle has one angle that is greater than 90 degrees.**Right Triangle:**A right triangle has one angle precisely equal to 90 degrees.

Ans: Congruence in triangles refers to the condition where two triangles have the same shape and size, and all their corresponding angles and sides are equal. When two triangles are congruent, they can be superimposed on each other, and their corresponding parts will coincide perfectly. This concept is denoted by the symbol “≅” and is used to establish the equality of triangles in various geometric proofs and constructions.

Ans: The Pythagorean Theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

Chapter Wise Important Questions for CBSE Board Class 10 Maths |
---|

Real Numbers |

Polynomials |

Pair of Linear Equations in Two Variables |

Quadratic Equations |

Arithmetic Progressions |

Triangles |

Coordinate Geometry |

Introduction to Trigonometry |

Some Applications of Trigonometry |

Circles |

Areas Related to Circles |

Surface Areas and Volumes |

Statistics |

Probability |

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