please solve 11th question...
Please solve it as fast as possible as because tomorrow is my pre-board
please friends
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Aishwarya0739:
please solve it fast friends
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given ABCD is a square, which is inscribed in a circle of radius 8 cm.
since the diagonals of the square subtends right angle at circumference,
therefore the diagonals of square are the diameters of the circle.
BD=2*8=16 cm
let the sides of the square be x cm.
in the right angled triangle BCD,
BC²+CD²=BD²
x²+x²=(16)²
2x²=256
x²=128cm²
THUS, area of square is 128cm²
since the diagonals of the square subtends right angle at circumference,
therefore the diagonals of square are the diameters of the circle.
BD=2*8=16 cm
let the sides of the square be x cm.
in the right angled triangle BCD,
BC²+CD²=BD²
x²+x²=(16)²
2x²=256
x²=128cm²
THUS, area of square is 128cm²
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if it is totally inscribed...then diameter of sphere will be the base of square...
for sphere,
radius=8cm,
diameter=8*2cm=16cm,
so we get edge(side) of square...so area,
side*side,
16*16=256cm^;
hope answer helps you...
please mark as brainliest
for sphere,
radius=8cm,
diameter=8*2cm=16cm,
so we get edge(side) of square...so area,
side*side,
16*16=256cm^;
hope answer helps you...
please mark as brainliest
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