Find the area of a rhombus if the measure of each side is 26 cm and one of its diagonals is 48 cm.
The answer which should come is 480 cm^2. I have checked from my book but I am unable to solve it.Please help me. No irrelevant answer allowd. who will give correct answer he will be marked as brainliest.
Answers
first diagonal d1 = 48
so the,Second diagonal d2 is =2*√a²-(d/2)²
=2*√26²-(48/2)²
d2=20
Area=1/2(d1*d2)=
1/2(48*20)
=480
Hope it helps uu !!!
The area of the rhombus is 480 cm².
Step-by-step explanation:
Referring to the figure attached below, we have
ABCD is a rhombus
Each side of a rhombus, AB = BC = CD = DA = 26cm
One of the diagonal, AC = 48 cm
We know that the diagonals of a rhombus bisect each other at right angles.
∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
and
∴ OA = OC = ½ * AC = ½ * 48 = 24 cm
Now, considering ∆ AOB and applying Pythagoras theorem, we get
AB² = OB² + OA²
⇒ OB = √[AB² – OA²]
⇒ OB = √[26² – 24²]
⇒ OB = √[100]
⇒ OB = 10 cm
∴ OB = OD = 10 cm
∴ Other diagonal of the rhombus, BD = 2 * OB = 10 * 2 = 20 cm
Thus,
The area of the rhombus ABCD,
= ½ * AC * BD
= ½ * 48 * 20
= 10 * 48
= 480 cm²
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