Answer :

Consider a triangle **ABCD**,

According to the problem, it is given that the sides are in the ratio **2:1**.

The perimeter of the rectangle **30**cm.

We know that the sides of the rectangles are represented by Length(**L**) and Breadth(**B**).

We also know that,

Perimeter(**P**) of the rectangle = **2 × (L + B)**.

From the problem, we came to know that the ratio **L: B = 2:1**.

Let us assume the values of Length and Breadth be **2x** and.

From the problem, we have **P = 30cm.**

Substituting the values we get,

⇒ 30 = 2 × (2x + x)

⇒ 30 = 2 × (3x)

⇒ 30 = 6x

⇒

⇒ **x = 5cm**.

Using the value of ‘x’ we find the value of length and breadth of Rectangle.

Now,

⇒ L = 2x

⇒ L = 2 × 5

⇒ **L = 10cm**.

⇒ **B = 5cm**.

From this, we can say that **AB** and **CD** are lengths, while **BC** and **DA** are breadths.

The lengths of **AB, BC, CD, DA** are **10cm, 5cm, 10cm, 5cm**.

Rate this question :

The sides of a reKarnataka Board - Mathematics Part II

| Construct squarKarnataka Board - Mathematics Part II

Construct rectangKarnataka Board - Mathematics Part II

All rectangles arKarnataka Board - Mathematics Part II

The side ofKarnataka Board - Mathematics Part II

In a rhombus ABCDKarnataka Board - Mathematics Part II

In a given squareKarnataka Board - Mathematics Part II

A rhombus is a paKarnataka Board - Mathematics Part II

| Construct squarKarnataka Board - Mathematics Part II

A rectangular yarKarnataka Board - Mathematics Part II