A circle is circumscribed by the rhombus
which in turn is made up by joining the
midpoints of a rectangle whose sides
12cm and 16cm respectively. What is the
area of the circle?
एक वृत समचतुर्भुज से घिरा है। जिसे एक
आयत जिसकी भुजा 12cm और16cm है के
मध्य बिन्दुओ को मिलाकर बनाया गया है।तब
वृत का क्षेत्रफल ज्ञात करे।
Answer= (576π/25)cm²
Answers
Answer:
72.41 cm²
Step-by-step explanation:
Referring to the figure attached below, we have
ABCD is a rectangle with dimensions 16 cm x 12 cm.
Joining the midpoints E, F, G & H of the rectangle, a rhombus EFGH is formed.
Let HF and EG, the diagonals of the rhombus, be denoted as “d1” and “d2” respectively.
From the figure we can say, AB = HF = 16 cm and BC = EG = 12 cm.
We know that the formula of the radius of a circle which is inscribed in a rhombus is given by,
Radius, r = \frac{d1 * d2}{2 * \sqrt{d1^2 +d2^2} }
2∗
d1
2
+d2
2
d1∗d2
Therefore, substituting the given values in the above formula, we get
r = \frac{16 * 12}{2 * \sqrt{16^2 + 12^2}} = \frac{192}{2 * \sqrt{400} } = \frac{192}{2 * 20} = \frac{192}{40} = \frac{24}{5}
2∗
16
2
+12
2
16∗12
=
2∗
400
192
=
2∗20
192
=
40
192
=
5
24
= 4.8 cm
Thus, substituting the value of r, we get
The area of the circle inscribed inside the rhombus as,
= πr²
= (22/7) * (4.8)²
= (22/7) * 23.04
= 72.41 cm²