Question

15. The length and diagonal of
a rectangle are 40 cm and 41
cm. Find its breadth
O 7 cm
O 9 cm
O 11 cm
O 13 cm​

Answer 1

Step-by-step explanation:

Answer is this first find 3rd side then find perimeter

Thanks a lot

Answer 2

Given :

• Length and Diagonal of Rectangle is 40 cm and 41 cm respectively .

To Find :

• Breadth of Rectangle .

Solution :

In Δ ADC :

$\longmapsto\tt{AD=40\:cm}$

$\longmapsto\tt{AC=41\:cm}$

Using Pythagoras Theorem :

$\longmapsto\tt{{(AC)}^{2}={(AD)}^{2}+{(DC)}^{2}}$

$\longmapsto\tt{{(41)}^{2}={(40)}^{2}+{(DC)}^{2}}$

$\longmapsto\tt{1681=1600+{(BC)}^{2}}$

$\longmapsto\tt{1681-1600={(BC)}^{2}}$

$\longmapsto\tt{81={(BC)}^{2}}$

$\longmapsto\tt{\sqrt{81}=BC}$

$\longmapsto\tt\bf{9\:cm=BC}$

So , The Breadth of Rectangle is 9 cm ...

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• Area of Rectangle = length × breadth
• Perimeter of Rectangle = 2 (l+b)
• Diagonal of Rectangle = √(l)² + (b)²
• Area of Sqaure = Side × Side
• Perimeter of Square = 4 × Side
• Diagonal of Square = √2a

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