Question

The ratio of base to height of a parallelogram is 2:3. Or the area of a parallelogram is 64 sq.em the find is base and height0​

Answer 1

Correct Question :

The ratio of base to height of a parallelogram is 2:2 and the area of a parallelogram is 64 cm².Find the base and the height.

Given :

The ratio of base to height of a parallelogram is 2:3 and the area of a parallelogram is 64 cm².

To Find :

The base and height.

Solution :

Analysis :

Here the formula of area of parallelogram is used. We first have to take a common ratio then form a equation and then equating that equation we can get the base and the height.

Required Formula :

Area of parallelogram = base × height

Explanation :

Let us assume that the common ratio is,

• Base = 2x
• Height = 3x

• Area = 64 cm²

We know that if we are given the area of parallelogram and is asked to find the base and the height then our required formula is,

Area of parallelogram = base × height

where,

• Area = 64 cm²
• Base = 2x
• Height = 2x

Using the required formula and substituting the required values,

⇒ Area of parallelogram = base × height

⇒ 64 = 2x × 2x

⇒ 64 = 4x²

⇒ 64/4 = x²

⇒ 16 = x²

⇒ √16 = x

⇒ 4 = x

∴ x = 4.

The base and height :

• Base = 2x = 2 × 4 = 8 cm.
• Height = 2x = 2 × 4 = 8 cm.

The base and height is 8 cm.

Verification :

⇒ Area of parallelogram = base × height

⇒ 64 = 8 × 8

⇒ 64 = 64

∴ LHS = RHS.

• Hence verified.

Answer 2

Step-by-step explanation:

$\frak {Given} \begin{cases} &\sf{Given\ ratio\ of\ base\ and\ height\ =\ \bf 2\ :\ 3.} \\ &\sf{Area\ =\ \bf 64cm^2.} \end{cases}$

To find:- Base and Height ?

☯️ By taking the common ratio let the base of the parallelogram be 2x and the height of the parallelogram be 2x.

__________________

$\frak{\underline{\underline{\dag As\ we\ know\ that:-}}}$

$\sf : \implies {Area\ of\ parallelogram\ =\ b\ \times\ h.}$

Here:-

• b, is for base = 2x.
• h, is for height = 2x.
• area is given as 64cm².

__________________

$\frak{\underline{\underline{\dag By\ substituting\ the\ values,\ we\ get:-}}}$

$\sf : \implies {Area\ of\ parallelogram\ =\ b\ \times\ h} \\ \\ \\ \sf : \implies {64\ =\ 2x\ \times\ 2x} \\ \\ \\ \sf : \implies {64\ =\ 4x^2} \\ \\ \\ \sf : \implies {\cancel \dfrac{64}{4}\ =\ x^2} \\ \\ \\ \sf : \implies {16\ =\ x^2} \\ \\ \\ \sf : \implies {\sqrt{16}\ =\ x} \\ \\ \\ \sf : \implies {4\ =\ x} \\ \\ \\ \sf : \implies {\pink{\underline{\boxed{\bf x\ =\ 4.}}}}\bigstar$

$\sf \therefore {\underline{The\ required\ value\ of\ x\ is\ \bf 4.}}$

So here:-

$\sf : \implies {Base\ =\ 2x\ =\ 2\ \times\ 4\ =\ \bf 8cm.}$

$\sf : \implies {Height\ =\ 2x\ =\ 2\ \times\ 4\ =\ \bf 8cm.}$

Hence:-

$\sf \therefore {\underline{The\ base\ and\ height\ is\ \bf 8cm.}}$