The Baudhayana Pythagoras Theorem Class 8 Extra Questions Maths Part 2 Chapter 2

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Class 8 The Baudhayana Pythagoras Theorem Extra Questions

Class 8 Maths Chapter 2 The Baudhayana Pythagoras Theorem Extra Questions

The Baudhayana Pythagoras Theorem Extra Questions Class 8

Question 1.
The length of the diagonal of a square is 6 cm. Find the sides of the square.
Solution:
Let ABCD be the square, and let AC be the diagonal of length 6 cm.
Then triangle ABC is the right-angled triangle such that AB = BC (∵ all sides of a square are equal)
By Pythagoras theorem,
AC² = AB² + BC²
⇒ AC² = 2AB² (∵ AB = BC)
⇒ AC = √2AB
⇒ AB = \(\frac{1}{\sqrt{2}}\) × AC
= \(\frac{1}{\sqrt{2}}\) × 6
= 3√2 cm

Question 2.
The hypotenuse of a right-angled triangle is 13 cm.
(a) If the sidelengths of the triangle are positive integers, find the length of the perpendicular legs.
(b) If both perpendicular legs are doubled, find the length of the new hypotenuse.
Solution:
(a) Let the lengths of the perpendicular legs be a cm and b cm, where a and b are positive integers.
According to the Baudhayana theorem, in a right-angled triangle:
a² + b² = c²
Here, c = 13 cm
a² + b² = 13² = 169
Since the sidelengths of the triangle are positive integers, (a, b, c) is a Baudhayana triple.
One of the primitive Baudhayana triplets is (5, 12, 13).
Since the given hypotenuse is 13 cm, the corresponding perpendicular legs must be 5 cm and 12 cm.

(b) New lengths of the perpendicular legs:
2 × 5 = 10 cm
2 × 12 = 24 cm
Applying Baudhayana’s theorem again:
(New hypotenuse)² = 10² + 24²
= 100 + 576
= 676
New hypotenuse = √676 = 26 cm

Question 3.
An architect designs a square courtyard whose area is S square units. Using the diagonal of this courtyard as the side, a second square is constructed. Again, a third square is constructed using the diagonal of the second square. Find the area of the third square in terms of S.
Solution:
Let the side of the original square be x units.

Area of first square = x² = S
⇒ x = √S
Diagonal of the first square = x√2 = Side of the second square
Area of the second square = (x√2)²
= 2x²
= 2S
Diagonal of the second square = √2S × √2 = √4S = Side of third square
Area of the third square = (√4S)² = 4S

Question 4.
Two rescue helicopters take off simultaneously from the same base station. One helicopter flies due north at a speed of 300 km/h, while the other flies due west at a speed of 400 km/h. After 1\(\frac {1}{2}\) hours, what is the distance between the two helicopters?
Solution:
Distance travelled by the first helicopter in 1\(\frac {1}{2}\) hours = 300 × \(\frac {3}{2}\) km = 450 km
Distance travelled by the second helicopter = 400 × \(\frac {3}{2}\) km = 600 km
Position of the two helicopters after 1\(\frac {1}{2}\) hours would be A and B as shown in the figure.

That is, OA = 450 km and OB = 600 km
From ∆AOB, we have
AB² = OA² + OB²
= (450)² + (600)²
= (150)² × 3² + (150)² × 4²
= 150² × (3² + 4²)
= 150² × 5²
⇒ AB = 150 × 5 = 750 km
Thus, the two helicopters will be 750 km apart after 1\(\frac {1}{2}\) hours.

Question 5.
A painter uses a ladder of length 10 m to reach a vertical wall. Initially, the ladder reaches a height of 6 m on the wall. If the foot of the ladder is moved 2 m towards the wall, find the distance by which the top of the ladder moves up the wall.
Solution:
The ladder, the wall, and the ground form a right-angled triangle ∆DLW.

By Baudhayana’s theorem:
DW² + WL² = DL²
⇒ DW² + 6² = 10²
⇒ DW² = 100 – 36 = 64
⇒ DW = 8 m
When the foot of the ladder is moved 2 m towards the wall:
The new base (RW) = 8 – 2 = 6 m
Applying Baudhayana’s theorem in the right-angled triangle ∆REW, we get
6² + WE² = 10²
⇒ WE² = 100 – 36 = 64
⇒ WE = 8 m
⇒ LE = WE – WL = 8 – 6 = 2 m
Therefore, the distance by which the top of the ladder would slide upwards on the wall if the foot of the ladder is moved 2 m towards the wall is 2 m.

Question 6.
Find the hounds for the hypotenuse of an isosceles right triangle whose equal sides have a length of 6 units.
Solution:
a = 6
c² = 2 × 6²
= 2 × 36
= 72
We know 8² = 64 and 9² = 81
64 < 72 < 81
∴ 72 lies between 82 and 92.
The length of the hypotenuse of the given triangle lies between 8 and 9 units.

Question 7.
If the hypotenuse of an isosceles right triangle is √50 units, find its other two sides.
Solution:
Given c = √50
Now, c² = 2a²
⇒ (√50)² = 2a²
⇒ 50 = 2a²
⇒ a² = 25
⇒ a = 5
Each of the other two sides has a length of 5 units.

Question 8.
If a right-angled triangle has shorter sides of lengths 3 cm and 4 cm, then what is the length of its hypotenuse?
Solution:
Here, a = 3 cm, b = 4 cm
c² = 3² + 4²
= 9 + 16
= 25
= 5²
c = 5

∴ Length of the hypotenuse is 5 cm.

Question 9.
If a right-angled triangle has a short side of length 5 cm and a hypotenuse of length 13 cm, what is the length of the third side?
Solution:
Here, a = 5 cm, c = 13 cm
13² = 5² + b²
⇒ b² = 169 – 25
⇒ b² = 144
⇒ b² = 12²
⇒ b = 12 cm

∴ Length of the third side is 12 cm.

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