Question

Each side of a rhombus is 15 cm and the length of one of its diagonals is 24cm.The area
of the rhombus is
(a) 432 cm
(d) 144 cm2
(b) 216 cm 2
(c) 180 cm2

From class 8. Plz explain with step by step....​

Answer 1

» Question :

Each side of a rhombus is 15 cm and the length of one of its diagonals is 24cm. Then find the area

of the rhombus .

» To Find :

The Area of the Rhombus.

» Given :

• One of the Diagonal of the Rhombus $\rightarrow D_{1} = 24$

• Equal side of the Rhombus = $15 cm$

» We Know :

Pythagoras theorem :

$\sf{\underline{\boxed{h^{2} = p^{2} + b^{2}}}}$

Where ,

• h = Hypotenuse
• b = Base
• p = Height

Area of a Rhombus :

$\sf{\underline{\boxed{A_{R} = \dfrac{1}{2} \times Product\:of\:its\:parallel\:sides}}}$

» Concept :

To Find the area of the Rhombus ,first we have to find the other parallel side as they directly proportional to each other.

$\sf{\boxed{A_{R} \propto D_{R}}}$

By looking at the figure ABE ,we can conclude that the figure is right-angled triangle .

So, by using the Pythagoras theorem ,we can find the length BE.

• Hypotenuse (h) ➝ AB = 15 cm
• Height (p) ➝ AE = 12 cm

Let the base be x cm

Formula :

$\sf{\underline{\boxed{h^{2} = p^{2} + b^{2}}}}$

Substituting ,the values in it ,we get :

$\sf{\Rightarrow 15^{2} = 12^{2} + x^{2}}$

$\sf{\Rightarrow 15^{2} - 12^{2} = x^{2}}$

$\sf{\Rightarrow \sqrt{15^{2} - 12^{2}}= x}$

$\sf{\Rightarrow \sqrt{225 - 144}= x}$

$\sf{\Rightarrow \sqrt{81}= x}$

$\sf{\Rightarrow 9 cm = x}$

Hence ,the base BE is 9 cm.

Now ,we know that the two diagonalas of a Rhombus is equally divided by each other at right-angles , so the other diagonal of the Rhombus is :

$D_{R} = 2 \times BE$

$\Rightarrow D_{R} = 2 \times 9$

$\Rightarrow D_{R} = 18 cm$

Hence, the other diagonal is 18 cm.

Now ,by the two Diagonals we find the area if the Rhombus.

» Solution :

Area of the Rhombus :

• Diagonal of the Rhombus $\rightarrow D_{1} = 24$

• Diagonal of the Rhombus $\rightarrow D_{2} = 18 cm$

Formula :

$\sf{\underline{\boxed{A_{R} = \dfrac{1}{2} \times Product\:of\:its\:parallel\:sides}}}$

Substituting ,the values in it ,we get :

$\sf{\Rightarrow A_{R} = \dfrac{1}{2} \times 24 \times 18}$

$\sf{\Rightarrow A_{R} = \dfrac{1}{\cancel{2}} \times 24 \times \cancel{18}}$

$\sf{\Rightarrow A_{R} = 24 \times 9}$

$\sf{\Rightarrow A_{R} = 216 cm^{2}}$

Hence, the area of the Rhombus is 216 cm².

» Diagram :

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0) \thicklines\qbezier( - 2.828,0)( - 2.828,0)(0,2.828) \qbezier(2.828,0)(2.828,0)(0,2.828)\qbezier(2.828,0)(2.828,0)(0, - 2.828)\qbezier(0, - 2.828)( +0, - 2.828)( - 2.828,0) \put( - 2.9,0){ \line(1,0){5.6}}\put( - 0.1, - 2.828){ \line(0,1){5.6}} \put( - 2,0.2){ \bf12 \ cm }\put( 0.5,0.2){ \bf12 \ cm } \put( - 0.1, 0.4){ \line(1,0){0.4}}\put( 0.3, 0.4){ \line(0, - 1){0.4}}\put( - 3,0.3){ \bf A }\put( - 1,2.4){ \bf B }\put(2.7,0.2){ \bf C }\put(0.2, - 2.9){ \bf D }\put( - 2.6,1.7){ \bf15\ cm }\qbezier( - 1, -1.3)( - 1, - 1.3)( - 1.5, -1.8)\qbezier( 2, 1.5)( 2, 1.5)( 1.4, 0.9)\qbezier( - 1.8, 1.5)( - 1.8, 1.5)( - 1.3, 1.2)\qbezier( 1.8, - 1.5)( 1.8, - 1.5)( 1.3, - 0.9)\end{picture}$

» Additional information :

• Surface area of a Cube = 6(a)².

• Curved surface area = 4(a)²

• Surface area of a Cuboid = 2(lb + lh + bh)

• Curved surface area of a Cuboid = 2(l + b)h

Answer 2

Step-by-step explanation:

$\bf{\underline{\underline\red{Given:-}}}$

• Each side of a rhombus is 15 cm

• The length of one of its diagonals is 24cm

$\bf{\underline{\underline\blue{To\:Find:-}}}$

• The area of the rhombus

$\bf{\underline{\underline\green{Solution:-}}}$

Let ABCD be the rhombus

Given:-

Diagonal BD = 24cm

Length of sides of rhombus

→ AB = BC = CD = AD = 15cm

As we know that

The diagonals of a rhombus bisect each other at right angles ( 90° )

It divided the diagonal into two equal halves.

So:-

$\rm \implies BO = \dfrac{1}{2} \times BD$

$\rm \implies BO = \dfrac{1}{2} \times 24$

$\rm \implies BO = 12cm$

In ∆AOB

By applying Pythagoras theorem:-

$\boxed{ \rm { {Base}^{2} + {Height}^{2} = Hypotenuse}^{2} }$

Here:-

• Base = BO

• Height = AO

• Hypotenuse = AB

→ AO² + BO² = AB²

→ AO² = AB² - BO²

→ AO² = 15² - 12²

→ AO² = 225 - 144

→ AO² = 81

→$\rm AO = \sqrt{81}$

→ AO = 9cm

Since:-

→ AC = 2 × AO

→ AC = 2 × 9

→ AC = 18cm

We have:-

One diagonal = 24cm

Other diagonal = 18cm

$\boxed{ \rm \leadsto Area \: of \: rhombus \: = \frac{1}{2} \times d_{1} \times d_{2} }$

Here:-

$\rm d_{1} \: and \: d_{2} \: are \: the \: diagonals \: of \: the \: rhombus$

Area of rhombus

$= \dfrac{1}{2} \times 24 \times 18$

= 12 × 18

= 216cm²

$\boxed{ \rm \therefore Area \: of \: rhombus = {216cm}^{2} }$