Question

ABCD is a rectangular ground of length 40 m and breadth 30 m. Ramesh reached at C walking from A to B and B to C.But shyam reached to C walking from A to C directly.who walked the shorter distance and by how much?​

Answer 1

Answer:

• Shyam walked the shorter distance by 20 m.

Step-by-step explanation:

Given that:

• ABCD is a rectangular ground of length 40 m and breadth 30 m.

To Find:

• Who walked the shorter distance and by how much?

Let us assume:

In ABCD.

• Length = AB = CD = 40 m
• Breadth = BC = AD = 30 m

Finding the distance covered by Ramesh:

Ramesh reached at C walking from A to B and B to C.

⟿ Distance = AB + BC

⟿ Distance = 40 + 30

⟿ Distance = 70

∴ Distance covered by Ramesh = 70 m

Finding the distance covered by Shyam:

But Shyam reached to C walking from A to C directly.

Distance = AC

Applying pythagoras theorm.

⟿ Distance = √(AB² + BC²)

⟿ Distance = √(40² + 30²)

⟿ Distance = √(1600 + 900)

⟿ Distance = √(2500)

⟿ Distance = 50

∴ Distance covered by Shyam = 50 m

Finding the difference between the distance covered by Ramesh and Shyam:

⟿ Difference = 70 - 50

⟿ Difference = 20

∴ Difference between the distance covered by Ramesh and Shyam = 20 m

Hence,

• Shyam walked the shorter distance by 20 m.

Answer 2

Given :-

ABCD  is a rectangular ground of length 40 m and breadth 30 m. Ramesh reached at C walking from A to B and B to C. But Shyam reached to C walking from A to C directly

To Find :-

Who walked the shorter distance and by how much

Solution :-

Distance covered by Ramesh

40 + 30

70 m

Distance covered by Shyam

$\sf H^2 = P^2 + B^2$

$\sf H^2 = 30^2+40^2$

$\sf H^2 = 900 + 1600$

$\sf H^2 = 2500$

$\sf H = \sqrt{2500}$

$\sf H =50$

HENCE,

Ramesh cover more distance by (70 - 50) = 20 m