Most Important Questions Class 10 Maths Ch 6 Triangles with Solutions

Triangles are a major part of the class 10 math syllabus. The chapter covers crucial topics such as the similarity of triangles, Pythagoras' theorem, and various geometric properties that students need to ace not just for their board exams but also for competitive exams like JEE.

Seeing the importance of the Class 10 Triangles Extra Questions, Educart has come up with Class 10 Triangles extra questions to make sure that our students practice a lot and ace their exams.

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Chapter 6 Triangles: Important Questions

1. The areas of two similar triangles are 64 cm² and 121 cm². If the length of a side of the larger triangle is 55 cm, find the length of the corresponding side of the smaller triangle. Show your work.

Soln. 

Area of first triangle = 64cm²

Altitude of first triangle = 55 cm

Area of second triangle = 121cm²

Let altitude of second triangle = x

As per formula

(Area of first triangle/Area of second triangle) = (Altitude of first triangle/Altitude of second triangle)²

(64cm²/121cm²) = ( 55cm /x)²

(64cm²/121cm²) = ( 55cm² /x²)

x² = ( 55cm² x 121cm² /64cm²)

x² = ( 55cm² x 121cm² /64cm²)

x² = ( 6655cm²/64cm²)

x² = (103.9²)

x = (103.9)

Altitude of second triangle = 103.9 cm

2. Looking at the above figure, Hari said that the shortest distance between Town E and Town F is 15 km.

Is the statement true or false? Justify your answer using the relevant properties.

Soln. 

Shortest distance between Town E and Town F is 15km.

As per theorem

The sum of two sides is greater than the third side.

EH + HG > EF + FG

12 + 7> 15 + 4

19> 19

But, here sum of two sides is equal to the third side

Hence, Hari’s statement is false

3. Are the two quadrilaterals shown below similar? Give a reason for your answer.

Soln. 

Two quadrilaterals are similar if:

1. If their corresponding angles are equal, and

Angles in both quadrilaterals: 50°, 100°, 110°, and 100°.

2. If their corresponding sides are proportional.

Sides of smaller quadrilateral: 2.5cm, 1.5cm, 1.5cm, and 2.5cm

Sides of bigger quadrilateral: (2.5)²cm, (1.5)²cm, (1.5)²cm, and (2.5)²cm

Side ratios= 

(2.5)²cm/2.5cm = 2.5cm

(1.5)²cm/1.5cm = 1.5cm

corresponding sides are not proportional.

Hence, the quadrilaterals aren’t similar

4. An insect sitting at corner P of a room flies along the dotted line PS and reaches corner S. Whereas, an ant sitting at corner P, reaches corner S by crawling along the path PR, followed by RS. Both the paths are shown below.

(Note: The figure is not to scale.)

i) Find the length of the path taken by the ant. 

ii) Find the length of the path of the insect's flight.

Show your steps.

Soln. 

i) Find the length of the path taken by the ant. 

PQ=5m (length of the room),

QR=4 m (width of the room).

PR=√PQ²+QR²

PR=√5²+4²m

PR=√25+16m

PR=√41 m

RS=3 m

Ant’s path= PR + RS

Ant’s path= √41 m + 3 m

Ant’s path= 6.40 m + 3 m

Length of the path taken by the ant= 9.40m

ii) Find the length of the path of the insect's flight.

PQ=5m (length),

QR=4 m (width),

RS=3 m (height).

PS=√PQ²+QR²+RS²​

PS=√5²+4²+3²​

PS=√25+16+9

PS=√50

PS=7.07m

Length of the path of the insect's flight= PS=7.07m

5. Panchami is standing on the ground and flying a kite at a vertical height of 22 m from the ground. The length of the taut string to which the kite is connected is 29 m. Panchami is holding the string roller 1 m above the ground.

i) Draw a figure representing the above scenario.

ii) Find the horizontal distance between the kite and Panchami. Show your work.

Soln. 

i) Draw a figure representing the above scenario.

ii) Find the horizontal distance between the kite and Panchami. Show your work.

Let the horizontal distance AD be x.

AE²= AD² + DE² 

29²= AD² + 21² 

29² - 21² = AD² 

841 - 441= AD² 

AD² = 400

AD = √400

AD = 20m

Since AD and BC are parallel and equal to each other

AD  = BC= 20m

Horizontal distance between the kite and Panchami is  20m

6. During a mathematics class, a teacher wrote the following three algebraic expressions on the board:

(m² - n²), (2 mn ) and ( m² + n²), where m and n are positive integers with n < m

One of the students, Kaivalya, claimed that the above set of expressions ALWAYS represent the sides of a right-angled triangle.Is Kaivalya's claim correct? Justify your answer.

Soln. 

For any triangle to be a right-angled triangle, the sides must satisfy:

a²+b²=c²

Let the expressions (m² - n²), (2 mn ) and ( m² + n²), represent the sides of a triangle

a= (m² - n²)

b=(2 mn )

c= ( m² + n²)

Substituting the new values

a²+b²=c²

(m² - n²)² + (2 mn )² = ( m² + n²)²

Taking LHS

(m² - n²)² + (2 mn )²

(m²)²  + (2 mn )² - 2. (m² - n²) (2 mn ) + 4mn²

(m⁴−2m²n²+n⁴)+4m²n²

m⁴+2m²n²+n⁴

c²=(m²+n²)²=m⁴+2m²n²+n⁴

a2+b2=m4+2m2n2+n4=c2

Since a²+b²=c², the expressions (m²⁻n²), (2mn), and (m²+n²) always satisfy Pythagoras' theorem.

7. In a mathematics class, a teacher drew the following figure where TQ/QR = ⅓. She then asked, "What is the sufficient condition required to prove that ΔTQP ~ ΔRQS?"

(Note: The figure is not to scale.)

• Darsh said that it is sufficient if it is given that TR/SR = ⅓.
• Bhargav said that it is sufficient if it is given that ∠P = ∠S.
• Tanvi said that it is sufficient if it is given that PQ/RS= ⅓.

Examine whether each of their responses is correct or incorrect. Give reasons.

Soln. 

TQ/QR = ⅓

Conditions:

• All corresponding angles are equal (AAA),
• Two corresponding sides are proportional, and the included angles are equal (SAS),
• All corresponding sides are proportional (SSS).

Darsh's Theory: It is sufficient if it is given that TR/SR = ⅓.

• If TR/SR = ⅓, this provides another pair of proportional sides
• If TQ/QR = ⅓, and  TR/SR = ⅓, we need one more condition, such as an included angle ∠TQR=∠RQS, to prove similarity using the SAS criterion.

Since Darsh did not mention angles, his response is incorrect, as the side ratios alone are insufficient to establish similarity.

Bhargava's Theory: It is sufficient if it is given that ∠P=∠S.

• If ∠P=∠S, this gives us one pair of equal angles.
• Given TQ/QR = ⅓, we have one pair of proportional sides.
• To use the SAS similarity criterion, we need the included angles (∠TQP=∠RQS) to be equal, which is not mentioned in Bhargav's condition.

Therefore, Bhargav's response is incorrect, as ∠P=∠S alone is not sufficient to prove similarity.

Tanvi's Theory: It is sufficient if it is given that PQ/RS = ⅓.

• If PQ/RS = ⅓, this provides another pair of proportional sides
• If TQ/QR = ⅓, and  TPQ/RS = ⅓, we need one more condition, such as an included angle ∠TQR=∠RQS, to prove similarity using the SAS criterion.
• To use the SAS similarity criterion, we need the included angles (∠TQP=∠RQS) to be equal, which is not mentioned in Tanvi's condition.

Since Tanvi's response does not mention angles, her response is incorrect, as the side ratios alone are not sufficient to establish similarity.

For △TQP∼△RQS, we need either AAA, SAS, or SSS conditions, hence, all three responses by Darsh, Bhargav, and Tanvi are incorrect

8. In the figure below, OPQ is a triangle with OP = OQ. RS is an arc of a circle with centre

(Note: The figure is not to scale.)

Triangle OSR is similar to triangle OPQ.

Is the above statement true or false? Justify your reason.

Soln. 

Given: 

Δ OPQ is an isosceles triangle with OP=OQ

RS is an arc of a circle with center O.

ΔOSR is formed using the center OOO and points RRR and SSS on the circle.

Δ OSR ~ Δ OPR 

To prove: 

To check if the statement true or false

Prove:

In Δ OSR and Δ OPR 

∠OPQ = ∠OQP

OR = OS (radii of circle)

• ∠ROS is subtended by the arc RS, while ∠POQ is formed by the sides OP and OQ.
• There is no direct evidence that these angles are equal or proportional.

Side Proportionality:

• The sides OP, OQ, OR, and OS do not have any given proportional relationships. Thus, SSS or SAS similarity cannot be established.

Without sufficient information to establish that the corresponding angles are equal or the sides are proportional, the claim that △OSR is similar to △OPQ is false.

9. In the figure below, XYZ is a right-angled triangle. A, B and C are the three YZ such that they divide YZ into 4 equal parts.

(Note: The figure is not to scale.)

Prove that 3XA² + XB² + XC² - XZ² = 4XY².

Soln. 

Taking LHS

XYZ is a right-angled triangle with ∠Y=90

YZ is divided into 4 equal parts by points A, B, and C.

Let the length of YZ=4d. Thus:

YA=d,

YB=2d,

YC=3d,

YZ=4d

Assume XY=h, the height of the triangle.

For the right triangle XYZ, the hypotenuse is XZ. By the Pythagorean theorem:

 XZ²=XY²+YZ²

Substituting YZ=4d:

XZ²=XY²+(4d)²

XZ²=XY²+16d²

Each of the points A, B, and C divides YZ perpendicularly. The distances XA, XB, and XC are calculated using the Pythagorean theorem:

Distance XA: For the right triangle XAY,

XA²=XY²+YA²

Substituting YA=d:

XA²=XY²+d²

Distance XB: For the right triangle XBY,

XB²=XY²+YB²

Substituting YA=2d:

XB²=XY²+(2d)²

XB²=XY²+4d²

Distance XC: For the right triangle XCY,

XC²=XY²+YC²

Substituting YC=3d:

XC²=XY²+(3d)²

XC²=XY²+9d²

Substitute the expressions for XA², XB², XC², and XZ²:

3XA²:= 3(XY²+d²)

3XA²:= 3XY²+3d²

Adding 3XA² + XB² + XC² 

= (3XY²+3d²) + (XY²+4d²) + (XY²+9d²) 

= 5XY²+16d²

Subtracting XZ²:

(5XY²+16d²) - (XY²+16d²)

= 4XY² i.e. RHS

LHS =RHS

What are triangles and their types?

Three sides and three angles make up a triangle, a basic geometric shape. It is one of the simplest and most fundamental shapes in geometry. We classify triangles based on their side lengths and angles.

• By side lengths:some text
  1. Equilateral Triangle: All three sides are equal, and all three angles measure 60 degrees.
  2. Isosceles Triangle: Two sides are equal, and the angles opposite these sides are also equal.
  3. Scalene Triangle: Each side has a different length, and each angle is unique.
• By Angles:some text
  1. Acute Triangle: All angles are less than 90 degrees.
  2. In a right triangle, one angle is exactly 90 degrees.
  3. Obtuse Triangle: One angle is greater than 90 degrees.

Benefits of Practicing Class 10 Triangles Important Questions

Studying triangle-important questions in class 10 has numerous advantages that can directly improve your performance in exams. Here are some key benefits:

• The Class 10 Triangles Important Questions with Solutions PDF provides a targeted approach to practice frequently occurring problems. We carefully select these questions to cover key sub-topics and reflect previous exam patterns, enabling you to solve the most important questions.
• One of the biggest challenges students face during exams is managing their time well. Working through the extra questions from Class 10 Triangles helps you practice allocating the right amount of time to each type of problem.
• We design extra questions to cover a wide range of difficulty levels, ensuring you are well-prepared for any type of question. Whether it’s an easy or tricky problem, practicing with class 10 triangle important questions ensures that no area of the topic is left untouched.

How to Use Class 10 Triangles Important Questions with Solutions Effectively

Simply collecting and solving extra questions isn’t enough. To make the most of them, you need to know how to use them in a way that maximizes your learning. Here’s how to make the most of class 10 triangle important questions with solutions:

• Before you start solving questions, make sure you fully understand the concepts. If you skip the theory and jump straight into problem-solving, you might get confused later. Review your textbook and ensure you know all the relevant theorems, formulas, and properties of triangles.
• Don’t burden yourself by trying to solve all the questions at once. Break them down by sub-topic, such as triangle properties, the similarity of triangles, and Pythagoras' theorem.
• When attempting a question, first try to solve it on your own without looking at the solution. This will give you an idea of how well you understand the topic. 
• Mistakes are a beneficial way to learn. If you make an error, take the time to figure out what went wrong. Did you misunderstand the question? Forget a step? Understanding why you made a mistake helps ensure you don’t repeat it.
• After reviewing a solution, don’t just move on. Try solving the question again without looking at the solution to see if you’ve fully understood it. Repeat this process until you’re confident you can solve the problem on your own.
• Don’t just solve a question once and forget about it. Regular revision is essential to keep concepts fresh in your mind. Go back to the questions you’ve already solved, especially the ones that were difficult for you, to make sure you’ve fully mastered them.
• Be sure to practice different types of questions, including short answer, long answer, and multiple-choice questions. This makes sure you have completed your practice.

Where to Find Class 10 Triangles Important Questions with Solutions PDF

To prepare effectively, you need access to high-quality resources. Many students prefer using class 10 triangle important questions with solutions in PDF for convenience. Here are some places where you can find reliable practice material:

• NCERT/CBSE Textbooks: Always start with your school books, which contain exercises and solved examples.
• Downloadable PDFs: Many educational websites provide free or paid PDFs that contain sets of important questions and solutions. 

• The Educart One Shot Question Bank includes all the extra questions for class 10 on triangles, as well as the important questions for class 10 on triangles.

The Triangles Class 10 Important Questions serve as an invaluable tool for students seeking to enhance their concepts and learning. Students aiming for boards and competitive exams such as JEE can benefit greatly from these questions.