Question

find the area of a rhombus having each side equal to 13 cm and one whose diagonal is 24 cm

Answer 1

Hello tannu!!

Thanks for asking the question:—

But getting through the question let’s discuss some important points regarding rhombus!!

1) Rhombus is a quadrilateral whose all 4 sides are equal.

2) it’s area is half of product of its diagonals.

3) The diagonals of rhombus bisect each other at 90 degree..and equally!!

So let’s go through the question:—

Given,

Side=13

Diagonal=24

Therefore, from above property.

Let’s draW a diagonal therefore, it will divide the line segment in equal proportion.

Thus, we know it bisect in 90degree..

We apply Pythagoras theorem,

13sq=12sq+x2

X=5.

Therefore ,length of diagonal=2x=10cm

Using area formul 1/2 x10x24

Area=)) 120cm sq.

Hope it helps you!!

Thanks!!

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Answer 2

Answer: 120 cm²

Step-by-step explanation:

Let us name the rhombus ABCD. (Refer to the image.)

All sides of a rhombus are equal.

Side of the rhombus = AD = 13 cm

Diagonals of a rhombus perpendicularly bisect each other. Let the point of intersection of the diagonals be = O. (Refer to the image.)

Let us assume that the diagonal AC = 24 cm. (Given in the question)

AO = OC (∵ Diagonals bisect each other.)

AO = OC = 1/2 of AC

∴ AO = 12 cm

ΔAOD is a right-angled triangle. (∵ Diagonals are perpendicular to each other.)

According to pythagoras theorem,

AO² + OD² = AD²

(12cm)² + OD² = (13cm)²

144 cm² + OD² = 169 cm²

OD² = 169 cm² - 144 cm²

OD² = 25 cm²

OD = √25 cm²

∴ OD = 5 cm

Diagonal BD = 2 × OD (∵ Diagonals bisect each other.)

∴ BD = 2 × 5 cm = 10 cm

Area of rhobmus = 1/2 × Product of the diagonals

Area of ABCD = 1/2 × 24 cm × 10 cm

∴ Area of ABCD = 120 cm²