Mensuration
- If a and b are the lengths of the sides of a right triangle whose hypotenuse is 10 and whose area is 20, then the value of (a + b)² is
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Using Rule 1,
BC = a units, AB = b units
AC = √a² + b² = 10
⇒ a² + b² = 100 .......(i)
Area of ∆ABC= 1 × base × height 2 = 1 ab 2 ∴ 1 ab = 20 2
⇒ ab = 40 square units........(ii)
7there4; (a + b)² = a² + b² + 2ab
= 100 + 2 × 40 = 180 square unitsCorrect Option: B
Using Rule 1,
BC = a units, AB = b units
AC = √a² + b² = 10
⇒ a² + b² = 100 .......(i)
Area of ∆ABC= 1 × base × height 2 = 1 ab 2 ∴ 1 ab = 20 2
⇒ ab = 40 square units........(ii)
7there4; (a + b)² = a² + b² + 2ab
= 100 + 2 × 40 = 180 square units
- A wire is bent into the form of a circle, whose area is 154 cm². If the same wire is bent into the form of an equilateral triangle, the approximate area of the equilateral triangle is
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Using Rule 14,
If the radius of circle be r cm, then
πr² = 154⇒ 22 r² = 154 7 ⇒ r² = 154 × 7 7 × 7 22
∴r = 7 cm
∴ Length of wire = 2πr= 2 × 22 7 × = 44 cm 7
Perimeter of equilateral triangle∴ Side of equilateral triangle = 44 cm 3 ∴ Area of equilateral triangle = √3 × side² 4 ∴ Area of equilateral triangle = √3 × 44 × 44 4 3 3 = 1.732 × 44 × 11 = 838.288 9 9
≈ 93.14 sq.cm.Correct Option: A
Using Rule 14,
If the radius of circle be r cm, then
πr² = 154⇒ 22 r² = 154 7 ⇒ r² = 154 × 7 7 × 7 22
∴r = 7 cm
∴ Length of wire = 2πr= 2 × 22 7 × = 44 cm 7
Perimeter of equilateral triangle∴ Side of equilateral triangle = 44 cm 3 ∴ Area of equilateral triangle = √3 × side² 4 ∴ Area of equilateral triangle = √3 × 44 × 44 4 3 3 = 1.732 × 44 × 11 = 838.288 9 9
≈ 93.14 sq.cm.
- If the ratio of the altitudes of two triangles be 3 : 4 and the ratio of their corresponding areas be 4 : 3, then the ratio of their corresponding lengths of bases is
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Using Rule 1,
Area of triangle = 1 × base × height = 1 bh 2 2 ∴ 1 b1h1 = 2 4 1 b2h2 3 2 ⇒ b1 × 3 = 4 b2 × 4 3 ⇒ b1 = 4 × 4 = 16 b2 3 × 3 9 Correct Option: B
Using Rule 1,
Area of triangle = 1 × base × height = 1 bh 2 2 ∴ 1 b1h1 = 2 4 1 b2h2 3 2 ⇒ b1 × 3 = 4 b2 × 4 3 ⇒ b1 = 4 × 4 = 16 b2 3 × 3 9
- Let A be the area of a square whose each side is 10 cm. Let B be the area of a square whose diagonals are 14 cm each. Then (A – B) is equal to
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A = 10² = 100 sq. cm.
B = 1 × 14² = 98 sq. cm. 2
∴ A – B = 100 – 98 = 2 sq. cm.Correct Option: C
A = 10² = 100 sq. cm.
B = 1 × 14² = 98 sq. cm. 2
∴ A – B = 100 – 98 = 2 sq. cm.
- Two sides of a parallelogram are 20 cm and 25 cm. If the altitude corresponding to the side of length 25 cm is 10 cm, then the altitude corresponding to the other pair of sides is
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Using Rule 1,
Area of parallelogram = base × height
= 25 × 10 = 250 sq. cm.
If the required altitude be x cm, then
x × 20 = 250⇒ x = 250 = 12.5 cm. 20 Correct Option: C
Using Rule 1,
Area of parallelogram = base × height
= 25 × 10 = 250 sq. cm.
If the required altitude be x cm, then
x × 20 = 250⇒ x = 250 = 12.5 cm. 20
