Class 7 Maths Chapter 6 Question Answers - The Triangle and Its Properties

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Q1: Find whether the following triplets are Pythagorean or not?
(a) (5, 8, 17)
(b) (8, 15, 17)
Ans:
(a) Given triplet: (5, 8, 17)
17² = 289
8² = 64
5² = 25
8² + 5² = 64 + 25 = 89
Since 89 ≠ 289
5² + 8² ≠ 172
Hence (5, 8, 17) is not Pythagorean triplet.
(b) Given triplet: (8, 15, 17)
17² = 289
15² = 225
8² = 64
15² + 8² = 225 + 64 = 289
17² = 15² + 8²
Hence (8, 15, 17) is a Pythagorean triplet.

Q2: Two sides of a triangle are 4 cm and 7 cm. What can be the length of its third side to make the triangle possible?
Ans: 
Let the length of the third side be x cm.
Condition I: Sum of two sides > the third side
i.e. 4 + 7 > x ⇒ 11 > x ⇒ x < 11
Condition II: The difference of two sides less than the third side.
i.e. 7 – 4 < x ⇒ 3 < x ⇒ x > 3
Hence the possible value of x are 3 < x < 11
i.e. x < 3 < 11

Q3: AD is the median of a ΔABC, prove that AB + BC + CA > 2AD (HOTS)

Ans:
In ΔABD,
AB + BD > AD …(i)
(Sum of two sides of a triangle is greater than the third side)
Similarly, In ΔADC, we have
AC + DC > AD …(ii)
Adding (i) and (ii), we have
AB + BD + AC + DC > 2AD
⇒ AB + (BD + DC) + AC > 2AD
⇒ AB + BC + AC > 2AD
Hence, proved.

Q4: In the given right-angled triangle ABC, ∠B = 90°. Find the value of x.

Ans:
In ΔABC, ∠B = 90°
AB² + BC² = AC² (By Pythagoras property)
(5)² + (x – 3)² = (x + 2)²
⇒ 25 + x2 + 9 – 6x = x² + 4 + 4x
⇒ -6x – 4x = 4 – 9 – 25
⇒ -10x = -30
⇒ x = 3
Hence, the required value of x = 3

Q5: The sides of a triangle are in the ratio 3 : 4 : 5. State whether the triangle is right-angled or not.
Ans: 
Let the sides of the given triangle are 3x, 4x and 5x units.
For right angled triangle, we have
Square of the longer side = Sum of the square of the other two sides
(5x)² = (3x)² + (4x)²
⇒ 25x² = 9x² + 16x²
⇒ 25x² = 25x²
Hence, the given triangle is a right-angled.

Q6: The length of the diagonals of a rhombus is 42 cm and 40 cm. Find the perimeter of the rhombus.

Ans: 
AC and BD are the diagonals of a rhombus ABCD.
Since the diagonals of a rhombus bisect at the right angle.
AC = 40 cm
AO = 40/2 = 20 cm
BD = 42 cm
OB = 422 = 21 cm
In right angled triangle AOB, we have
AO² + OB² = AB²
⇒ 20² + 21² = AB²
⇒ 400 + 441 = AB²
⇒ 841 = AB²
⇒ AB = √841 = 29 cm.
Perimeter of the rhombus = 4 × side = 4 × 29 = 116 cm
Hence, the required perimeter = 116 cm

Q7: A plane flies 320 km due west and then 240 km due north. Find the shortest distance covered by the plane to reach its original position.
Ans: 
Here, OA = 320 km
AB = 240 km
OB = ?
Clearly, ∆OBA is right angled triangleOB² = OA² + AB2 (By Pythagoras property)
⇒ OB² = 320² + 240²
⇒ OB² = 102400 + 57600
⇒ OB² = 160000
⇒ OB = √160000 = 400 km.
Hence the required shortest distance = 400 km.