Question

*Length of a rectangle is 5cm and breadth is 3 cm then find the ratio of its perimeter to it's area.*

1️⃣ 8:15
2️⃣ 16:15
3️⃣ 15:16
4️⃣ 15:8​

Answer 1

Solution:

Length of rectangle, L = 5 cm

Breadth of rectangle, B = 3 cm

Perimeter of rectangle, P = 2 * (L + B) = 2 * (5 + 3) = 2 * 8 = 16 cm

Area of rectangle, A = L * B = 5 * 3 = 15 cm²

Ratio of rectangle's perimeter to rectangle's area = P : A = 16 : 15

2️⃣ 16:15 is the correct answer.

Extra Information =>

Rectangle =>

A quadrilateral which has four right angles is a rectangle. It has 4 edges and 4 vertices. Alternate side of a rectangle are equal. The length of diagonals are also equal.

Perimeter =>

The path which surrounds a two-dimensional figure is called perimeter.

Area =>

Area is the space occupied by a surface of an object or by a flat shape.

Answer 2

${\large{\bold{\rm{\underline{Let's \; understand \; the \; question \; 1^{st}}}}}}$

● This question says that length of a rectangle is 5 cm and it's breadth is 3 cm then we have to find the ratio of it's perimeter to it's area. So it's cleared that which formulas we have to use here !. Let's solve this question !.

${\large{\bold{\rm{\underline{Given \; that}}}}}$

● Length of rectangle = 5 cm

● Breadth of rectangle = 3 cm

${\large{\bold{\rm{\underline{To \; find}}}}}$

● The ratio of it's (rectangle) perimeter to it's area.

${\large{\bold{\rm{\underline{Solution}}}}}$

● The ratio of it's (rectangle) perimeter to it's area = 16:15

${\large{\bold{\rm{\underline{Using \; concepts}}}}}$

● Formula to find area of rectangle.

● Formula to find perimeter of rectangle.

${\large{\bold{\rm{\underline{Using \; formula}}}}}$

● Area of rectangle = l × b

● Perimeter of rectangle = 2(l+b)

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

~ Let us find area of this rectangle first,

★ Length of rectangle = 5 cm

★ Breadth of rectangle = 3 cm

★ Area of rectangle = l × b

• Area of rectangle = 5 × 3

• Area of rectangle = 15 cm²

${\frak{Henceforth, \: 15 \: cm^{2} \: is \: area \: of \: rectangle}}$

~ Let us find perimeter of this rectangle now,

★ Length of rectangle = 5 cm

★ Breadth of rectangle = 3 cm

★ Perimeter of rectangle = 2(l+b)

• Perimeter of rectangle = 2(5+3)

• Perimeter of rectangle = 2(8)

• Perimeter of rectangle = 2 × 8

• Perimeter of rectangle = 16 cm

${\frak{Henceforth, \: 16 \: cm \: is \: perimeter \: of \: rectangle}}$

~ Not at last let us find the ratio of it's (rectangle) perimeter to it's area

★ Perimeter of rectangle = 16 cm

★ Area of rectangle = 15 cm²

So, perimeter : area

Henceforth, 16:15

${\frak{Henceforth, \: 16:15 \: is \: ratio \: of \: perimeter \: to \: area}}$

${\large{\bold{\rm{\underline{Additional \; knowledge}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Circumference \: of \: circle \: = \: 2 \pi r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: triangle \: = \: \dfrac{1}{2} \times breadth \times height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: paralloelogram \: = \: Breadth \times Height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle \: = \: \pi b^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: triangle \: = \: (1st \: + \: 2nd \: + 3rd) \: side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: paralloelogram \: = \: 2(a+b)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cylinder \: = \: \pi r^{2}h}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Surface \: area \: of \: cylinder \: = \: 2 \pi rh + 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Lateral \: area \: of \: cylinder \: = \: 2 \pi rh}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Base \: area \: of \: cylinder \: = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Height \: of \: cylinder \: = \: \dfrac{v}{\pi r^{2}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: cylinder \: = \:\sqrt frac{v}{\pi h}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: cuboid \: = \: 2(l \times b + b \times h + l \times h}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto LSA \: of \: cuboid \: = \: 2h(l+b)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cuboid \: = \: L \times B \times H}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diagonal \: of \: cuboid \: = \: \sqrt 3l}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: cuboid \: = \: 12 \times Sides}}}$