Question

Find the area of a rhombus one of whose side is 25 m and and one of whose diagonal is 48 m.

Answer 1

this is example
What is the area of a rhombus whose side is 30 cm and one diagonal is 40 cm?

Tehis rhombus can be cut along its diagonals to creäte (4) congruënt right triangles, each with a hypoteneuse of 30cm and a leg of 20cm.
The remaining leg is √(30^2cm-20^2cm) = √(900cm-400cm) = √500cm = ±22.36068cm.
Two of these triangles together make a rectangle measuring ±22.36068cm × 20cm, so all four give us ±22.36068cm × 20cm × 2 = ±894.4272cm^2.

AnRedr alternative way to answer:
Because 500 = 5×100, √500 = √5×√100 = √5×10a
√5 ×10cm×20cm×2 = √5×400cm^2

I hope this will help you

Answer 2

Answer:

336 m²

Step-by-step explanation:

Observe the attached figure

Let Diagonal AC = 48 cm

We know that diagonals of Rhombus bisect each other at 90°

∴ In ΔAEF

∠AED = 90°,

AE = 1/2 AC = 24 m

AD = 25 m (side given)

Now, ∵ AED is right triangle, applying Pythagoras Theorem

AD² = AE² + ED²

25² = 24² + ED²

ED² = 25² - 24²

ED² = (25 + 24)(25 - 24)     [∵ a² - b² = (a + b) (a - b)]

ED² = 49 × 1

ED² = 7²

ED = 7 m

∴ BD = 2ED = 14 m

Hence length of another diagonal is 14 m

We know that area of Rhombus = 1/2 × diagonal 1 × diagonal 2

                                                     = 1/2 × 48 × 14

                                                     = 336 m²

∴ Area of given Rhombus = 336 m²