Question

the area of a square is 49 sq.cm .A rectangular has the same perimeter as the square.if the length of the rectangle is 9.3.cm ,what is it's breadth?Also find which had greater area?​

Answer 1

Answer:

The breadth is 4.7 cm

And the square has the greater area

(Area of rectangle= 43.71 sq.cm

Answer 2

✯ Given:

• Area of a square = 49 cm².
• Rectangle has same perimeter as the square.
• Length of rectangle = 9.3.cm

✯ To Find:

• Breadth of rectangle.
• Which area is greater.

✯ Concept:

We are given area of a square. We will be finding the side of the square and it's perimeter. Since perimeter of rectangle is also same and length of the rectangle is given, we will use perimeter of rectangle formula to find the breadth of the rectangle. After finding breadth we will use area of rectangle formula and compare which area is greater.

✯ Solution:

To find the area of square, we use the formula:

${\tt \longmapsto \underline{\boxed{\tt a^2 = Area_{(Square)}}}}$

Here, a represents the side of the square.

${\tt \longmapsto a^2 = 49}$

${\tt \longmapsto a = \sqrt{49}}$

Square root of 49 is 7.

$\pink{\underline{\boxed{\tt a = 7cm}}}$

We have found the side of the square. We know that:

${\tt \longmapsto \underline{\boxed{\tt 4a = Perimeter_{(Square)}}}}$

Since we know the side, apply its value:

${\tt \longmapsto Perimeter_{(square)} = 4 \times 7}$

On multiplying in RHS, we get the perimeter as:

$\pink{\underline{\boxed{\tt Perimeter_{(square)} = 28cm}}}$

Use the perimeter of rectangle formula:

${\tt \longmapsto \underline{\boxed{\tt 2(l+b) = Perimeter_{(Rectangle)}}}}$

Here, l and b represents length and breadth of the rectangle respectively.

${\tt \longmapsto 2(l+b) = 28}$

Why perimeter of rectangle is considered to be 28cm? This is so because, according to the question; Perimeter of rectangle is equal to that of the square.

${\tt \longmapsto 2(9.3+b) = 28}$

Shift 2 from LHS to RHS by changing its sign.

${\tt \longmapsto (9.3+b) = \dfrac{28}{2}}$

On dividing on RHS, we get:

${\tt \longmapsto 9.3+b = 14}$

Shift 9.3 from LHS to RHS by changing its sign.

${\tt \longmapsto b = 14-9.3}$

On subtracting on RHS, we get the breadth as:

$\pink{\underline{\boxed{\tt b = 4.7cm}}}$

Let us find the area of rectangle as we know the breadth value. We know that:

${\tt \longmapsto \underline{\boxed{\tt Area_{(rectangle)} = lb}}}$

Apply the values of length and breadth.

${\tt \longmapsto Area_{(rectangle)} = 9.3 \times 4.7}$

On multiplying, 9.3 and 4.7, we get the area of rectangle as:

$\pink{\underline{\boxed{\tt Area_{(rectangle)} = 43.71cm^2}}}$

Hence, we have found the area of rectangle as 43.71cm². But we can see that area of square is 49, which is greater than that of its rectangle. Hence, we can say that:

Area of square has greater area.