A circle of area 154m² is circumscribing the equilateral triangle of side 7m. Find the difference of the two geometrical figures.
* If possible then please attach figure in answer too.*
Answers
Answer:
Given Area of circle=154cm
2
⇒πr
2
=154cm
2
⇒r
2
=
22
154×7
=49cm
⇒r=
49
=7cm
ABC is an equilateral △, h is the altitude of △ABC
0 is the incenter of △ABC, and this is the point of intersection of the angular bisectors. Hence, these bisectors are also the altitude and medians whose point of intersection divides the medians in the ratio 2:1
∴∠ADB=90° & OD=
3
1
AD & OD is radius of circle. Then,
r=
3
h
⇒h=3r=3×7=21cm
Let each side of an equilateral triangle be 'a', then altitude of an equilateral triangle is (
2
3
) times its side. So that
h=
2
3
a⇒a=
3
2h
=
3
2×21
=
3
2×21
3
=14
3
cm
Perimeter of triangle ABC=3a=3×14
3
cm=42
3
cm=42×1.73
=72.66cm
2
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Answer:
Given Area of circle=154cm
2
⇒πr
2
=154cm
2
⇒r
2
=
22
154×7
=49cm
⇒r=
49
=7cm
ABC is an equilateral △, h is the altitude of △ABC
0 is the incenter of △ABC, and this is the point of intersection of the angular bisectors. Hence, these bisectors are also the altitude and medians whose point of intersection divides the medians in the ratio 2:1
∴∠ADB=90° & OD=
3
1
AD & OD is radius of circle. Then,
r=
3
h
⇒h=3r=3×7=21cm
Let each side of an equilateral triangle be 'a', then altitude of an equilateral triangle is (
2
3
) times its side. So that
h=
2
3
a⇒a=
3
2h
=
3
2×21
=
3
2×21
3
=14
3
cm
Perimeter of triangle ABC=3a=3×14
3
cm=42
3
cm=42×1.73
=72.66cm
2