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Class 9 Maths Chapter 6 HOTS Questions - Triangles

Question 1. ABCD is a square. P is any point inside it, such that ΔPQR is another square. Prove that AP = CR

Class 9 Maths Chapter 6 HOTS Questions - Triangles

Hint: Join AP and CR.
In ΔADP and ΔCDR, we have :
AD = CD                      [sides of a square]
∠ADP = ∠CDR                     [each = 90° - ∠PDC]
DP = DR                    [side of a square]
⇒ ΔADP ≌ ΔCDR                     [SAS congruence]
⇒ AP = CR                     [C.P.C.T]

 Question 2. E and F are the midpoints of sides AB, AC of ΔABC. CE and BF are produced to X and Y respectively, such that EX = CE and FY = BF. AX and AY are joined. Find in your figure, a triangle congruent to ΔAEX and demonstrate the congruency. Show that XAY is a straight line.

Class 9 Maths Chapter 6 HOTS Questions - Triangles

Hint: Prove ΔAEX ≌ ΔBEC                    [By SAS congruency]
⇒ ∠ XAE = ∠ CBE                    [c.p.c.t.]
⇒ ∠XAB = ∠CBA
But they form a pair of co-interior angles.
⇒ XA || BC                    ...(1)
Similarly, ΔAFY ≌ ΔCFB
⇒ AY || BC                    ...(2)
from (1) and (2) XAY is a st. line.

Question 3. In the adjacent figure, BA || DF and CA || EG. If BD = EC then prove that BG = DF and EG = CF.

Class 9 Maths Chapter 6 HOTS Questions - Triangles

Hint: In ΔGBE and ΔFDC ∠ABC = ∠FDE and ∠DED = ∠ACB
also BE = DC
∴ ΔGBE ≌ ΔFDC                    [ASA congruency]
⇒ BG = DF and EG = CF

 Question 4. ABCD is a square. M is the midpoint of AB and PQ ⊥ CM meets AD at P. CB produced meet at Q. Prove that (i) PA = BQ and (ii) CP = AB + PA

Class 9 Maths Chapter 6 HOTS Questions - Triangles

Hint: Prove, ΔAMP ≌ ΔBMQ                    [ASA cong.]
⇒ MP = MQ and PA = QB                    [c.p.c.t.]
⇒ PA = BQ
Again, prove, ∆CMP ≌ ΔCMQ                    [SAS cong.]
⇒ CP = CQ                     [c.p.c.t.]
⇒ CP = CB + BQ = AB + PA

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FAQs on Class 9 Maths Chapter 6 HOTS Questions - Triangles

1. What are the different types of triangles?
Ans. There are several types of triangles based on their side lengths and angles. The different types of triangles include equilateral triangles (all sides and angles are equal), isosceles triangles (two sides and two angles are equal), scalene triangles (no sides or angles are equal), acute triangles (all angles are less than 90 degrees), right triangles (one angle is exactly 90 degrees), and obtuse triangles (one angle is greater than 90 degrees).
2. How do you find the area of a triangle?
Ans. The area of a triangle can be found using the formula A = (base * height) / 2. The base of the triangle is the length of one of its sides, and the height is the perpendicular distance from the base to the opposite vertex. Simply multiply the base and height, and divide the result by 2 to obtain the area.
3. What is the Pythagorean theorem and how is it used in triangles?
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 other two sides. This theorem is often used to find the length of one side of a right triangle when the lengths of the other two sides are known.
4. How do you determine if a set of side lengths can form a triangle?
Ans. To determine if a set of side lengths can form a triangle, you can use the triangle inequality theorem. According to this theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. So, if the sum of the lengths of the two shorter sides is greater than the length of the longest side, then a triangle can be formed.
5. How do you find the missing angles in a triangle?
Ans. The sum of the angles in a triangle is always 180 degrees. To find a missing angle, subtract the sum of the two known angles from 180 degrees. For example, if two angles in a triangle are 60 degrees and 40 degrees, the missing angle can be found by subtracting 60 degrees and 40 degrees from 180 degrees, resulting in a missing angle of 80 degrees.
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