Mensuration
- Two equal circles intersect so that their centres, and the points at which they intersect form a square of side 1 cm. The area (in sq.cm) of the portion that is common to the circles is
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Distance between centres = diagonal of square = √2cm.∴ Ex radii = 1 cm. √2
Required area = Area of ex-circle – area of square= π - 1 2
Correct Option: B
Distance between centres = diagonal of square = √2cm.∴ Ex radii = 1 cm. √2
Required area = Area of ex-circle – area of square= π - 1 2
- Diagonals of a Trapezium ABCD with AB || CD intersect each other at the point O. If AB = 2CD, then the ratio of the areas of ∆AOB and ∆COD is
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In ∆COD and ∆AOB,
∠OAB = ∠OCD
∠OBA = ∠ODC
By AA–similarity, ∆AOB ~ ∆COD∴ Area of ∆AOB = 4 Area of ∆CD 1
= 4 : 1Correct Option: A
In ∆COD and ∆AOB,
∠OAB = ∠OCD
∠OBA = ∠ODC
By AA–similarity, ∆AOB ~ ∆COD∴ Area of ∆AOB = 4 Area of ∆CD 1
= 4 : 1
- The perimeter of two squares are 24 cm and 32 cm. The perimeter (in cm) of a third square equal in area to the sum of the areas of these squares is :
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Side of square, whose perimeter is 24 cm = 24/4 = 6 cm
So, Area of the square = 62 = 36 cm²
Again, side of square, whose perimeter is 32cm
= 32/4 = 8 cm
So, Area of this square = 82 = 64 cm²
According to the question Area of new square = 64 + 36 = 100cm²
∴ Side of the new square = 100 = 10 cm
Hence, Perimeter of new square = 10 × 4 = 40 cmCorrect Option: B
Side of square, whose perimeter is 24 cm = 24/4 = 6 cm
So, Area of the square = 62 = 36 cm²
Again, side of square, whose perimeter is 32cm
= 32/4 = 8 cm
So, Area of this square = 82 = 64 cm²
According to the question Area of new square = 64 + 36 = 100cm²
∴ Side of the new square = 100 = 10 cm
Hence, Perimeter of new square = 10 × 4 = 40 cm
- If the ratio of areas of two squares is 225 : 256, then the ratio of their perimeter is :
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Ratio of area = 225 225 ⇒ Ratio of side = √ 225 = 15 225 16 ∴ Ratio of perimeter = 4 × 15 = 15 ⇒ 15 : 16 4 × 16 16 Correct Option: C
Ratio of area = 225 225 ⇒ Ratio of side = √ 225 = 15 225 16 ∴ Ratio of perimeter = 4 × 15 = 15 ⇒ 15 : 16 4 × 16 16
- The perimeter of two squares are 40 cm and 24 cm. The perimeter of a third square , whose area is equal to the difference of the area of these squares, is
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Side of the first square = 40 = 10 cm 4 Side of the second square = 24 = 6 cm 4
Difference of the area of these squares = (10 × 10 – 6 × 6) cm²
= (100 – 36) cm² = 64 cm²
∴ Area of the third square = 64 cm²
⇒ Side of third square = √64 = 8 cm
∴ Perimeter of this square = (4 × 8) cm = 32 cmCorrect Option: B
Side of the first square = 40 = 10 cm 4 Side of the second square = 24 = 6 cm 4
Difference of the area of these squares = (10 × 10 – 6 × 6) cm²
= (100 – 36) cm² = 64 cm²
∴ Area of the third square = 64 cm²
⇒ Side of third square = √64 = 8 cm
∴ Perimeter of this square = (4 × 8) cm = 32 cm
