Question

Q.1. The perimeter of a rhombus is 52 cm. If one diagonal is 24 cm find. [5].
I) length of its other diagonal.
II) its area

Answer 1

Answer:

Perimeter (P) = 52 cm

One diagonal (d1) = 24 cm

I) P = 2 √(d1^2 + d2^2)

=> 52 = 2 √( 576 + d2^2)

=> 26 = √ (576 + d2^2)

=> 676 = 576 + d2^2

=>d2^2 = 100

=> d2 = 10

Therefore, the other diagonal is 10 cm.

II) Area = (1/4) d1 √( P^2 - 4 P^2)

= 120 cm^2

Answer 2

Given:-

Perimeter of the rhombus = 52 cm

One of the diagonals = 24 cm

To find:-

I) length of it's other diagonal

II) it's area.

Note:-

Refer to the attachment provided.

Solution:-

We know,

Perimeter of a rhombus = 4 × side

Hence,

$\sf{52 = 4\times side}$

= $\sf{\dfrac{52}{4} = side}$

= $\sf{side = 13\:cm}$

Now,

In the figure we can see that,

AC = 24 cm

OA = $\sf{\dfrac{1}{2}\times 24\:cm}$

OA = 12 cm

We know, that the diagonals of a rhombus intersect each other at an angle of 90°

Hence,

In ∆ AOB,

Perpendicular = OA = 12 cm

Hypotenuse = AB = 13 cm $\sf{[\because side\:of\:rhombus = 13\:cm]}$

Base = OB

According to Pythagoras theorem,

$\sf{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}$

$\sf{\implies (Base)^2 = (Hypotenuse)^2 - (Perpendicular)^2}$

= $\sf{OB = \sqrt{(13)^2 - (12)^2}}$

= $\sf{OB = \sqrt{169 - 144}}$

= $\sf{OB = \sqrt{25}}$

= $\sf{OB = 5\:cm}$

Since,

OB = 5 cm

Therefore,

BD = OB × 2 = $\sf{5\times 2}$ = 10 cm

I)Therefore, the measure of other diagonal of the rhombus is 10 cm.

II) We know,

Area of the rhombus = $\sf{\dfrac{1}{2}\times (Product\:of\:the\:diagonals)}$

Area = $\sf{\dfrac{1}{2}\times 24\times 10}$

Area = $\sf{120\:cm^2}$

Therefore the area of the Rhombus is 120 cm².

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Remember!!

The two diagonals intersect each other in such a way that they are half pf their original length.

For example,

In this figure, AD intersects AC in such a way that AO is 1/2 the length of the total length of the diagonal i.e., AC.

Same goes with the diagonal BD. AC intersects BD in such a way that OB is 1/2 the length of BD.

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