AREA AND PERIMETER OF PLANE FIGURES - Questions & Answers
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) Length of a side and corresponding altitude of a triangle is doubled. The area of triangle will become :
(i) double (ii) half (iii) four times (iv) one-fourth
Step 1: The original area of the triangle is calculated as: 1/2 × base × height
Step 2: The new base becomes 2 × base and the new height becomes 2 × height.
Step 3: The new area = 1/2 × (2 × base) × (2 × height)
Step 4: New area = 4 × (1/2 × base × height) = 4 × Original Area
Step 5: Therefore, the area becomes four times.
Answer: (iii) four times
Step 2: The new base becomes 2 × base and the new height becomes 2 × height.
Step 3: The new area = 1/2 × (2 × base) × (2 × height)
Step 4: New area = 4 × (1/2 × base × height) = 4 × Original Area
Step 5: Therefore, the area becomes four times.
Answer: (iii) four times
(b) If each side of an equilateral triangle is halved, its area will be :
(i) halved (ii) four times (iii) unaltered (iv) one-fourth
Step 1: The original area of an equilateral triangle is: (√3 / 4) × side²
Step 2: The new side is: side / 2
Step 3: The new area = (√3 / 4) × (side / 2)²
Step 4: New area = (1/4) × (√3 / 4) × side² = 1/4 of the Original Area
Step 5: Therefore, the area becomes one-fourth.
Answer: (iv) one-fourth
Step 2: The new side is: side / 2
Step 3: The new area = (√3 / 4) × (side / 2)²
Step 4: New area = (1/4) × (√3 / 4) × side² = 1/4 of the Original Area
Step 5: Therefore, the area becomes one-fourth.
Answer: (iv) one-fourth
(c) The area of a triangle is 37.5 cm². If its base is 12.5 cm; the corresponding altitude is :
(i) 3 cm (ii) 25 cm (iii) 6 cm (iv) 12 cm
Step 1: The formula for area is: Area = 1/2 × base × altitude
Step 2: Substitute the given values: 37.5 = 1/2 × 12.5 × altitude
Step 3: Multiply both sides by 2: 75 = 12.5 × altitude
Step 4: Divide by 12.5: altitude = 75 / 12.5 = 6 cm
Answer: (iii) 6 cm
Step 2: Substitute the given values: 37.5 = 1/2 × 12.5 × altitude
Step 3: Multiply both sides by 2: 75 = 12.5 × altitude
Step 4: Divide by 12.5: altitude = 75 / 12.5 = 6 cm
Answer: (iii) 6 cm
(d) ABC is a triangle with AB = AC = 12 cm and ∠A = 90°, the area of the triangle ABC is :
(i) 144 cm² (ii) 36 cm² (iii) 72 cm² (iv) 108 cm²
Step 1: Since ∠A = 90°, it is a right-angled triangle.
Step 2: The sides containing the right angle act as the base and height (AB and AC).
Step 3: Area = 1/2 × product of perpendicular sides = 1/2 × AB × AC
Step 4: Area = 1/2 × 12 × 12 = 72 cm²
Answer: (iii) 72 cm²
Step 2: The sides containing the right angle act as the base and height (AB and AC).
Step 3: Area = 1/2 × product of perpendicular sides = 1/2 × AB × AC
Step 4: Area = 1/2 × 12 × 12 = 72 cm²
Answer: (iii) 72 cm²
(e) The sides of a triangle are 9 cm, 12 cm and 15 cm; the area of the triangle is :
(i) 54 cm² (ii) 96 cm² (iii) 108 cm² (iv) 135 cm²
Step 1: Check if the sides form a right-angled triangle using Pythagoras theorem: 9² + 12² = 81 + 144 = 225.
Step 2: Since 15² is also 225, it is a right-angled triangle with base 9 cm and height 12 cm.
Step 3: Area = 1/2 × base × height
Step 4: Area = 1/2 × 9 × 12 = 54 cm²
Answer: (i) 54 cm²
Step 2: Since 15² is also 225, it is a right-angled triangle with base 9 cm and height 12 cm.
Step 3: Area = 1/2 × base × height
Step 4: Area = 1/2 × 9 × 12 = 54 cm²
Answer: (i) 54 cm²
2. Find the area of a triangle whose sides are 18 cm, 24 cm and 30 cm. Also, find the length of altitude corresponding to the largest side of the triangle.
Step 1: First, check for a right-angled triangle: 18² + 24² = 324 + 576 = 900.
Step 2: Since 30² is also 900, the triangle is right-angled.
Step 3: The base and height are the two smaller sides (18 cm and 24 cm).
Step 4: Area = 1/2 × 18 × 24 = 216 cm².
Step 5: The largest side is the hypotenuse, which is 30 cm.
Step 6: Let the altitude to the largest side be 'h'. Area = 1/2 × 30 × h.
Step 7: 216 = 15 × h.
Step 8: h = 216 / 15 = 14.4 cm.
Answer: Area is 216 cm² and the altitude is 14.4 cm.
Step 2: Since 30² is also 900, the triangle is right-angled.
Step 3: The base and height are the two smaller sides (18 cm and 24 cm).
Step 4: Area = 1/2 × 18 × 24 = 216 cm².
Step 5: The largest side is the hypotenuse, which is 30 cm.
Step 6: Let the altitude to the largest side be 'h'. Area = 1/2 × 30 × h.
Step 7: 216 = 15 × h.
Step 8: h = 216 / 15 = 14.4 cm.
Answer: Area is 216 cm² and the altitude is 14.4 cm.
3. The lengths of the sides of a triangle are in the ratio 3 : 4 : 5. Find the area of the triangle if its perimeter is 144 cm.
Step 1: Let the lengths of the sides be 3x, 4x, and 5x.
Step 2: The perimeter is the sum of the sides: 3x + 4x + 5x = 144.
Step 3: 12x = 144, which means x = 12.
Step 4: The sides are: 3×12 = 36 cm, 4×12 = 48 cm, and 5×12 = 60 cm.
Step 5: Since 36² + 48² = 1296 + 2304 = 3600 = 60², it is a right-angled triangle.
Step 6: Area = 1/2 × base × height = 1/2 × 36 × 48.
Step 7: Area = 18 × 48 = 864 cm².
Answer: The area of the triangle is 864 cm².
Step 2: The perimeter is the sum of the sides: 3x + 4x + 5x = 144.
Step 3: 12x = 144, which means x = 12.
Step 4: The sides are: 3×12 = 36 cm, 4×12 = 48 cm, and 5×12 = 60 cm.
Step 5: Since 36² + 48² = 1296 + 2304 = 3600 = 60², it is a right-angled triangle.
Step 6: Area = 1/2 × base × height = 1/2 × 36 × 48.
Step 7: Area = 18 × 48 = 864 cm².
Answer: The area of the triangle is 864 cm².
4. ABC is a triangle in which AB = AC = 4 cm and ∠A = 90°. Calculate :
(i) the area of ∆ ABC.
(ii) the length of perpendicular from A to BC.
Step 1: (i) Because ∠A is 90°, AB and AC are the base and the height.
Step 2: Area of ∆ABC = 1/2 × AB × AC = 1/2 × 4 × 4 = 8 cm².
Step 3: (ii) To find the perpendicular from A to BC, we first find the length of BC using Pythagoras theorem.
Step 4: BC = √(AB² + AC²) = √(4² + 4²) = √(16 + 16) = √32 = 4√2 cm.
Step 5: Let the perpendicular from A to BC be 'h'. Area can also be written as 1/2 × BC × h.
Step 6: 8 = 1/2 × 4√2 × h = 2√2 × h.
Step 7: h = 8 / 2√2 = 4 / √2 = 2√2 cm.
Answer: (i) Area is 8 cm², (ii) Perpendicular is 2√2 cm.
Step 2: Area of ∆ABC = 1/2 × AB × AC = 1/2 × 4 × 4 = 8 cm².
Step 3: (ii) To find the perpendicular from A to BC, we first find the length of BC using Pythagoras theorem.
Step 4: BC = √(AB² + AC²) = √(4² + 4²) = √(16 + 16) = √32 = 4√2 cm.
Step 5: Let the perpendicular from A to BC be 'h'. Area can also be written as 1/2 × BC × h.
Step 6: 8 = 1/2 × 4√2 × h = 2√2 × h.
Step 7: h = 8 / 2√2 = 4 / √2 = 2√2 cm.
Answer: (i) Area is 8 cm², (ii) Perpendicular is 2√2 cm.
5. The area of an equilateral triangle is 36√3 sq. cm. Find its perimeter.
Step 1: The area of an equilateral triangle is given by the formula: (√3 / 4) × side².
Step 2: Substitute the given area: 36√3 = (√3 / 4) × side².
Step 3: Divide both sides by √3: 36 = (1 / 4) × side².
Step 4: Multiply by 4: side² = 36 × 4 = 144.
Step 5: side = √144 = 12 cm.
Step 6: Perimeter = 3 × side = 3 × 12 = 36 cm.
Answer: The perimeter is 36 cm.
Step 2: Substitute the given area: 36√3 = (√3 / 4) × side².
Step 3: Divide both sides by √3: 36 = (1 / 4) × side².
Step 4: Multiply by 4: side² = 36 × 4 = 144.
Step 5: side = √144 = 12 cm.
Step 6: Perimeter = 3 × side = 3 × 12 = 36 cm.
Answer: The perimeter is 36 cm.
6. Find the area of an isosceles triangle with perimeter 36 cm and base 16 cm.
Step 1: In an isosceles triangle, two sides are equal. Let them be 'a'.
Step 2: Perimeter = a + a + base = 2a + 16 = 36.
Step 3: 2a = 36 - 16 = 20, so a = 10 cm.
Step 4: Draw a perpendicular from the top vertex to the base. It divides the base equally into 8 cm and 8 cm.
Step 5: This forms a right triangle with hypotenuse 10 cm and base 8 cm. Let the height be 'h'.
Step 6: h = √(10² - 8²) = √(100 - 64) = √36 = 6 cm.
Step 7: Area = 1/2 × total base × height = 1/2 × 16 × 6 = 48 cm².
Answer: The area is 48 cm².
Step 2: Perimeter = a + a + base = 2a + 16 = 36.
Step 3: 2a = 36 - 16 = 20, so a = 10 cm.
Step 4: Draw a perpendicular from the top vertex to the base. It divides the base equally into 8 cm and 8 cm.
Step 5: This forms a right triangle with hypotenuse 10 cm and base 8 cm. Let the height be 'h'.
Step 6: h = √(10² - 8²) = √(100 - 64) = √36 = 6 cm.
Step 7: Area = 1/2 × total base × height = 1/2 × 16 × 6 = 48 cm².
Answer: The area is 48 cm².
7. The base of an isosceles triangle is 24 cm and its area is 192 sq. cm. Find its perimeter.
Step 1: Formula for the area is 1/2 × base × height.
Step 2: 192 = 1/2 × 24 × height.
Step 3: 192 = 12 × height, which means height = 192 / 12 = 16 cm.
Step 4: The perpendicular height divides the isosceles triangle into two right triangles with a base of 24 / 2 = 12 cm.
Step 5: Using Pythagoras theorem, the equal side (hypotenuse) = √(height² + half-base²).
Step 6: Equal side = √(16² + 12²) = √(256 + 144) = √400 = 20 cm.
Step 7: Perimeter = sum of all sides = 20 + 20 + 24 = 64 cm.
Answer: The perimeter is 64 cm.
Step 2: 192 = 1/2 × 24 × height.
Step 3: 192 = 12 × height, which means height = 192 / 12 = 16 cm.
Step 4: The perpendicular height divides the isosceles triangle into two right triangles with a base of 24 / 2 = 12 cm.
Step 5: Using Pythagoras theorem, the equal side (hypotenuse) = √(height² + half-base²).
Step 6: Equal side = √(16² + 12²) = √(256 + 144) = √400 = 20 cm.
Step 7: Perimeter = sum of all sides = 20 + 20 + 24 = 64 cm.
Answer: The perimeter is 64 cm.
8. The given figure shows a right-angled triangle ABC and an equilateral triangle BCD. Find the area of the shaded portion.
Step 1: In right-angled ∆ABC (right angled at B), AB = 8 cm and hypotenuse AC = 16 cm.
Step 2: Find base BC using Pythagoras theorem: BC = √(AC² - AB²) = √(16² - 8²).
Step 3: BC = √(256 - 64) = √192 = 8√3 cm.
Step 4: Area of right-angled ∆ABC = 1/2 × AB × BC = 1/2 × 8 × 8√3 = 32√3 cm².
Step 5: ∆BCD is an equilateral triangle on base BC. Side of ∆BCD = 8√3 cm.
Step 6: Area of equilateral ∆BCD = (√3 / 4) × side² = (√3 / 4) × (8√3)² = (√3 / 4) × 192 = 48√3 cm².
Step 7: Area of the shaded portion = Area of ∆BCD - Area of ∆ABC.
Step 8: Shaded Area = 48√3 - 32√3 = 16√3 cm².
Answer: The area of the shaded portion is 16√3 cm² (or approx 27.71 cm²).
Step 2: Find base BC using Pythagoras theorem: BC = √(AC² - AB²) = √(16² - 8²).
Step 3: BC = √(256 - 64) = √192 = 8√3 cm.
Step 4: Area of right-angled ∆ABC = 1/2 × AB × BC = 1/2 × 8 × 8√3 = 32√3 cm².
Step 5: ∆BCD is an equilateral triangle on base BC. Side of ∆BCD = 8√3 cm.
Step 6: Area of equilateral ∆BCD = (√3 / 4) × side² = (√3 / 4) × (8√3)² = (√3 / 4) × 192 = 48√3 cm².
Step 7: Area of the shaded portion = Area of ∆BCD - Area of ∆ABC.
Step 8: Shaded Area = 48√3 - 32√3 = 16√3 cm².
Answer: The area of the shaded portion is 16√3 cm² (or approx 27.71 cm²).
9. Find the area and the perimeter of quadrilateral ABCD, given below; if, AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and ∠DBC = 90°.
Step 1: In right-angled triangle DBC, hypotenuse is DC = 13 cm, and one side BD = 12 cm.
Step 2: Find BC using Pythagoras theorem: BC = √(DC² - BD²) = √(13² - 12²) = √(169 - 144) = √25 = 5 cm.
Step 3: Perimeter of quadrilateral ABCD = AB + BC + CD + DA = 8 + 5 + 13 + 10 = 36 cm.
Step 4: Area of ∆DBC = 1/2 × base × height = 1/2 × 12 × 5 = 30 cm².
Step 5: For ∆ABD, sides are 8 cm, 10 cm, and 12 cm. Find semi-perimeter (s) = (8 + 10 + 12) / 2 = 15 cm.
Step 6: Use Heron's formula for ∆ABD: Area = √(s(s-a)(s-b)(s-c)) = √(15 × (15-8) × (15-10) × (15-12)).
Step 7: Area = √(15 × 7 × 5 × 3) = √1575 = 15√7 cm².
Step 8: Total Area of ABCD = Area of ∆DBC + Area of ∆ABD = 30 + 15√7 cm².
Answer: Perimeter is 36 cm, Area is (30 + 15√7) cm².
Step 2: Find BC using Pythagoras theorem: BC = √(DC² - BD²) = √(13² - 12²) = √(169 - 144) = √25 = 5 cm.
Step 3: Perimeter of quadrilateral ABCD = AB + BC + CD + DA = 8 + 5 + 13 + 10 = 36 cm.
Step 4: Area of ∆DBC = 1/2 × base × height = 1/2 × 12 × 5 = 30 cm².
Step 5: For ∆ABD, sides are 8 cm, 10 cm, and 12 cm. Find semi-perimeter (s) = (8 + 10 + 12) / 2 = 15 cm.
Step 6: Use Heron's formula for ∆ABD: Area = √(s(s-a)(s-b)(s-c)) = √(15 × (15-8) × (15-10) × (15-12)).
Step 7: Area = √(15 × 7 × 5 × 3) = √1575 = 15√7 cm².
Step 8: Total Area of ABCD = Area of ∆DBC + Area of ∆ABD = 30 + 15√7 cm².
Answer: Perimeter is 36 cm, Area is (30 + 15√7) cm².
10. The base of a triangular field is three times its height. If the cost of cultivating the field at ₹ 36.72 per 100 m² is ₹ 49,572; find its base and height.
Step 1: Total cost of cultivation is ₹ 49,572.
Step 2: The rate of cultivation is ₹ 36.72 per 100 m², which means ₹ 0.3672 per 1 m².
Step 3: Total Area = Total Cost / Rate per m² = 49572 / 0.3672 = 135,000 m².
Step 4: Let the height of the field be 'h' meters. The base is given as 3 times the height, so base = 3h.
Step 5: Area = 1/2 × base × height = 1/2 × 3h × h = 1.5h².
Step 6: Equate the areas: 1.5h² = 135,000.
Step 7: h² = 135,000 / 1.5 = 90,000.
Step 8: h = √90,000 = 300 m.
Step 9: Base = 3 × h = 3 × 300 = 900 m.
Answer: Base is 900 m and height is 300 m.
Step 2: The rate of cultivation is ₹ 36.72 per 100 m², which means ₹ 0.3672 per 1 m².
Step 3: Total Area = Total Cost / Rate per m² = 49572 / 0.3672 = 135,000 m².
Step 4: Let the height of the field be 'h' meters. The base is given as 3 times the height, so base = 3h.
Step 5: Area = 1/2 × base × height = 1/2 × 3h × h = 1.5h².
Step 6: Equate the areas: 1.5h² = 135,000.
Step 7: h² = 135,000 / 1.5 = 90,000.
Step 8: h = √90,000 = 300 m.
Step 9: Base = 3 × h = 3 × 300 = 900 m.
Answer: Base is 900 m and height is 300 m.
EXERCISE 19(B)
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) The area of the given figure is :
(i) AC × BD (ii) 1/2 × AC × BD (iii) 1/2 × AB × BD (iv) 1/2 × (AD + BC) × BD
Step 1: The given figure is a quadrilateral whose diagonals AC and BD intersect at right angles.
Step 2: When diagonals intersect at 90°, the area is half the product of its diagonals.
Step 3: Area = 1/2 × AC × BD.
Answer: (ii) 1/2 × AC × BD
Step 2: When diagonals intersect at 90°, the area is half the product of its diagonals.
Step 3: Area = 1/2 × AC × BD.
Answer: (ii) 1/2 × AC × BD
(b) If two adjacent sides and a diagonal of a rectangle are x, y and d respectively. The area of the rectangle is :
(i) x × y (ii) 1/2 × d² (iii) 1/2 × x × d (iv) 1/2 × y × d
Step 1: In a rectangle, the adjacent sides represent its length and breadth.
Step 2: Let length be x and breadth be y.
Step 3: Area of a rectangle = length × breadth = x × y.
Answer: (i) x × y
Step 2: Let length be x and breadth be y.
Step 3: Area of a rectangle = length × breadth = x × y.
Answer: (i) x × y
(d) The perimeter of a square is 72 cm, its area is :
(i) 900√2 cm² (ii) 30√3 cm² (iii) 324 cm² (iv) 356 cm²
Step 1: Perimeter of a square = 4 × side.
Step 2: 72 = 4 × side, which means side = 72 / 4 = 18 cm.
Step 3: Area of a square = side × side = 18 × 18 = 324 cm².
Answer: (iii) 324 cm²
Step 2: 72 = 4 × side, which means side = 72 / 4 = 18 cm.
Step 3: Area of a square = side × side = 18 × 18 = 324 cm².
Answer: (iii) 324 cm²
(e) Area of a rhombus is 360 cm². If one diagonal of it is 20 cm; the other diagonal is :
(i) 24 cm (ii) 18 cm (iii) 40 cm (iv) 36 cm
Step 1: Area of a rhombus = 1/2 × (product of diagonals).
Step 2: 360 = 1/2 × 20 × (other diagonal).
Step 3: 360 = 10 × (other diagonal).
Step 4: other diagonal = 360 / 10 = 36 cm.
Answer: (iv) 36 cm
Step 2: 360 = 1/2 × 20 × (other diagonal).
Step 3: 360 = 10 × (other diagonal).
Step 4: other diagonal = 360 / 10 = 36 cm.
Answer: (iv) 36 cm
2. Find the area of a quadrilateral one of whose diagonals is 30 cm long and the perpendiculars from the other two vertices are 19 cm and 11 cm respectively.
Step 1: Area of a quadrilateral can be found by splitting it into two triangles along the diagonal.
Step 2: The formula is Area = 1/2 × diagonal × (sum of perpendiculars from opposite vertices).
Step 3: Substitute the values: Area = 1/2 × 30 × (19 + 11).
Step 4: Area = 15 × 30 = 450 cm².
Answer: The area of the quadrilateral is 450 cm².
Step 2: The formula is Area = 1/2 × diagonal × (sum of perpendiculars from opposite vertices).
Step 3: Substitute the values: Area = 1/2 × 30 × (19 + 11).
Step 4: Area = 15 × 30 = 450 cm².
Answer: The area of the quadrilateral is 450 cm².
3. The diagonals of a quadrilateral are 16 cm and 13 cm. If they intersect each other at right angles; find the area of the quadrilateral.
Step 1: For any quadrilateral where the diagonals intersect at 90°, the area is half their product.
Step 2: Area = 1/2 × (diagonal 1) × (diagonal 2).
Step 3: Area = 1/2 × 16 × 13.
Step 4: Area = 8 × 13 = 104 cm².
Answer: The area of the quadrilateral is 104 cm².
Step 2: Area = 1/2 × (diagonal 1) × (diagonal 2).
Step 3: Area = 1/2 × 16 × 13.
Step 4: Area = 8 × 13 = 104 cm².
Answer: The area of the quadrilateral is 104 cm².
TEST YOURSELF
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) The base of a right triangle is 8 cm and its hypotenuse is 10 cm; the area of the triangle is :
(i) 24 cm² (ii) 40 cm² (iii) 48 cm² (iv) 80 cm²
Step 1: Use Pythagoras theorem to find the height (perpendicular).
Step 2: height = √(hypotenuse² - base²) = √(10² - 8²) = √(100 - 64).
Step 3: height = √36 = 6 cm.
Step 4: Area = 1/2 × base × height = 1/2 × 8 × 6 = 24 cm².
Answer: (i) 24 cm²
Step 2: height = √(hypotenuse² - base²) = √(10² - 8²) = √(100 - 64).
Step 3: height = √36 = 6 cm.
Step 4: Area = 1/2 × base × height = 1/2 × 8 × 6 = 24 cm².
Answer: (i) 24 cm²
(b) The area of an equilateral triangle is 4√3 cm, its perimeter is :
(i) 16 cm (ii) 4 cm (iii) 12 cm (iv) 8 cm
Step 1: Set the area formula equal to the given value: (√3 / 4) × side² = 4√3.
Step 2: Cancel √3 from both sides: (1/4) × side² = 4.
Step 3: Multiply by 4: side² = 16, which means side = 4 cm.
Step 4: Perimeter of an equilateral triangle = 3 × side = 3 × 4 = 12 cm.
Answer: (iii) 12 cm
Step 2: Cancel √3 from both sides: (1/4) × side² = 4.
Step 3: Multiply by 4: side² = 16, which means side = 4 cm.
Step 4: Perimeter of an equilateral triangle = 3 × side = 3 × 4 = 12 cm.
Answer: (iii) 12 cm
(c) If the perimeter of a square is 80 cm, its area is 80 cm².
(i) true (ii) false (iii) none of these two
Step 1: First, calculate the side from the perimeter. Perimeter = 4 × side = 80.
Step 2: side = 80 / 4 = 20 cm.
Step 3: Calculate the area using the side: Area = side × side = 20 × 20 = 400 cm².
Step 4: The calculated area is 400 cm², not 80 cm². Therefore, the statement is false.
Answer: (ii) false
Step 2: side = 80 / 4 = 20 cm.
Step 3: Calculate the area using the side: Area = side × side = 20 × 20 = 400 cm².
Step 4: The calculated area is 400 cm², not 80 cm². Therefore, the statement is false.
Answer: (ii) false
(d) If the area of a trapezium is 32 cm² and distance between its parallel sides is 8 cm; the sum of the length of its parallel sides is :
(i) 4 cm (ii) 16 cm (iii) 8 cm (iv) 12 cm
Step 1: The area of a trapezium = 1/2 × (sum of parallel sides) × (distance between them).
Step 2: Substitute the known values: 32 = 1/2 × (sum of parallel sides) × 8.
Step 3: Simplify: 32 = 4 × (sum of parallel sides).
Step 4: Divide by 4: sum of parallel sides = 32 / 4 = 8 cm.
Answer: (iii) 8 cm
Step 2: Substitute the known values: 32 = 1/2 × (sum of parallel sides) × 8.
Step 3: Simplify: 32 = 4 × (sum of parallel sides).
Step 4: Divide by 4: sum of parallel sides = 32 / 4 = 8 cm.
Answer: (iii) 8 cm
2. AD is altitude of an isosceles triangle ABC in which AB = AC = 30 cm and BC = 36 cm. A point O is marked on AD in such a way that ∠BOC = 90°. Find the area of quadrilateral ABOC.
Step 1: In an isosceles triangle, the altitude AD bisects the base BC. So, BD = DC = 18 cm.
Step 2: Apply Pythagoras theorem in right ∆ABD to find altitude AD.
Step 3: AD = √(AB² - BD²) = √(30² - 18²) = √(900 - 324) = √576 = 24 cm.
Step 4: For ∆BOC, since OB = OC by symmetry and ∠BOC = 90°, it is an isosceles right-angled triangle.
Step 5: The altitude OD from O to BC also bisects the right angle, making ∠BOD = 45°.
Step 6: In right ∆BOD, tan(45°) = OD / BD => 1 = OD / 18, so OD = 18 cm.
Step 7: Area of the entire ∆ABC = 1/2 × base BC × height AD = 1/2 × 36 × 24 = 432 cm².
Step 8: Area of ∆BOC = 1/2 × base BC × height OD = 1/2 × 36 × 18 = 324 cm².
Step 9: Area of quadrilateral ABOC = Area of ∆ABC - Area of ∆BOC = 432 - 324 = 108 cm².
Answer: The area of quadrilateral ABOC is 108 cm².
Step 2: Apply Pythagoras theorem in right ∆ABD to find altitude AD.
Step 3: AD = √(AB² - BD²) = √(30² - 18²) = √(900 - 324) = √576 = 24 cm.
Step 4: For ∆BOC, since OB = OC by symmetry and ∠BOC = 90°, it is an isosceles right-angled triangle.
Step 5: The altitude OD from O to BC also bisects the right angle, making ∠BOD = 45°.
Step 6: In right ∆BOD, tan(45°) = OD / BD => 1 = OD / 18, so OD = 18 cm.
Step 7: Area of the entire ∆ABC = 1/2 × base BC × height AD = 1/2 × 36 × 24 = 432 cm².
Step 8: Area of ∆BOC = 1/2 × base BC × height OD = 1/2 × 36 × 18 = 324 cm².
Step 9: Area of quadrilateral ABOC = Area of ∆ABC - Area of ∆BOC = 432 - 324 = 108 cm².
Answer: The area of quadrilateral ABOC is 108 cm².
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Quick Review Flashcards - Click to flip and test your knowledge!
Question
Term: Perimeter
Answer
Definition: The length of the boundary of a plane figure.
Question
Term: Area
Answer
Definition: The measure of the surface enclosed by the boundary of a plane figure.
Question
What is the difference between 'x square metre' and 'x metre square'?
Answer
'x square metre' refers to an area, while 'x metre square' refers to a square with sides of length x metres.
Question
What is the general formula for the area of a triangle given its base and corresponding height?
Answer
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
Question
In the context of a triangle, what does 'corresponding height' (or altitude) mean?
Answer
The length of the perpendicular drawn from the opposite vertex to the chosen base.
Question
How is the semi-perimeter s of a triangle with sides a, b, and c calculated for Heron's formula?
Answer
s = \frac{a + b + c}{2}
Question
State Heron's formula for the area of a triangle with sides a, b, c and semi-perimeter s.
Answer
\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}
Question
What is the mathematical relation between 1 \text{ m}^2 and \text{cm}^2?
Answer
1 \text{ m}^2 = 100 \times 100 \text{ cm}^2 (or 10,000 \text{ cm}^2)
Question
What is the formula for the perimeter of an equilateral triangle with side length a?
Answer
\text{Perimeter} = 3a
Question
What is the formula for the area of an equilateral triangle with side length a?
Answer
\text{Area} = \frac{\sqrt{3}}{4} a^2
Question
In an isosceles triangle, the perpendicular from the vertex to the base _____ the base.
Answer
bisects
Question
What is the formula for the area of an isosceles triangle with base b and equal sides a?
Answer
\text{Area} = \frac{1}{4} \times b \times \sqrt{4a^2 - b^2}
Question
The area of a right-angled triangle is equal to half the product of the sides containing the _____.
Answer
right angle
Question
Formula: Area of a general quadrilateral with diagonal d and perpendiculars h_1, h_2 from remaining vertices to d
Answer
\text{Area} = \frac{1}{2} \times d \times (h_1 + h_2)
Question
If the two diagonals of a quadrilateral intersect at right angles, what is the formula for its area?
Answer
\text{Area} = \frac{1}{2} \times (\text{product of the diagonals})
Question
What is the formula for the length of the diagonal d of a rectangle with length l and breadth b?
Answer
d = \sqrt{l^2 + b^2}
Question
What is the formula for the perimeter of a rectangle with length l and breadth b?
Answer
P = 2(l + b)
Question
What is the formula for the length of the diagonal d of a square with side length a?
Answer
d = a\sqrt{2}
Question
How can the diagonal d of a square be expressed in terms of its area A?
Answer
d = \sqrt{2 \times A}
Question
What is the formula for the area of a parallelogram?
Answer
\text{Area} = \text{base} \times \text{height}
Question
In a parallelogram, what distance defines its 'height'?
Answer
The perpendicular distance between its base and the side opposite to the base.
Question
What is the formula for the area of a rhombus using its diagonals d_1 and d_2?
Answer
\text{Area} = \frac{1}{2} \times d_1 \times d_2
Question
How is the side length of a rhombus related to its diagonals d_1 and d_2?
Answer
(\text{side})^2 = (\frac{d_1}{2})^2 + (\frac{d_2}{2})^2
Question
Why can the area of a rhombus be calculated using the formula \text{base} \times \text{height}?
Answer
Because a rhombus is also a type of parallelogram.
Question
State the formula for the area of a trapezium with parallel sides a and b and height h.
Answer
\text{Area} = \frac{1}{2}(a + b) \times h
Question
What is the length of the line segment joining the mid-points of the non-parallel sides of a trapezium with parallel sides a and b?
Answer
\text{Length} = \frac{a + b}{2}
Question
If the length of the segment joining the mid-points of non-parallel sides is L and height is h, what is the area of the trapezium?
Answer
\text{Area} = L \times h
Question
Concept: Circumference
Answer
Definition: The length of the boundary of a circle.
Question
The constant ratio between a circle's circumference and its diameter is represented by the Greek letter _____.
Answer
\pi (pi)
Question
What is the formula for the circumference C of a circle with radius r?
Answer
C = 2\pi r
Question
What is the common fractional approximation used for the value of \pi?
Answer
\frac{22}{7}
Question
What is the common decimal approximation used for the value of \pi?
Answer
3.14
Question
What is the formula for the area of a circle with radius r?
Answer
\text{Area} = \pi r^2
Question
If a circle is cut from a rectangular sheet of paper, the diameter of the largest possible circle is equal to the _____ of the sheet.
Answer
width (or shorter side)
Question
What is the formula for the area of a circular ring bounded by concentric circles with outer radius R and inner radius r?
Answer
\text{Area} = \pi(R^2 - r^2)
Question
How is the width of a uniform circular track calculated given outer radius R and inner radius r?
Answer
\text{Width} = R - r
Question
The distance covered by a wheel in one complete rotation is equal to its _____.
Answer
circumference
Question
How do you calculate the total distance covered by a wheel of radius r in n rotations?
Answer
\text{Distance} = 2\pi r \times n
Question
How do you determine the speed of a wheel in \text{km/h} if you know the distance covered in one minute?
Answer
Multiply the distance per minute by 60 to get distance per hour, then convert units to kilometres.
Question
To find the altitude corresponding to the largest side of a triangle using Heron's formula, you first find the total area and then use the formula _____.
Answer
\frac{1}{2} \times \text{largest side} \times \text{altitude} = \text{Area}
Question
If the area of a circle is numerically equal to its circumference, what is the radius of the circle?
Answer
2 units
Question
In a rectangular field of dimensions L \times B with an internal path of uniform width w, what are the dimensions of the inner rectangle?
Answer
(L - 2w) and (B - 2w)
Question
In a rectangular field of dimensions L \times B with an external path of uniform width w, what are the dimensions of the outer rectangle?
Answer
(L + 2w) and (B + 2w)
Question
If a wire is bent from one shape into another, which geometric property remains constant?
Answer
The perimeter (or length of the wire).
Question
Formula: Perimeter of a rhombus with side a
Answer
P = 4a
Question
Formula: Perimeter of a square with side a
Answer
P = 4a
Question
What is the area of a square in terms of its side a?
Answer
A = a^2
Question
How is the area of a shaded region between two shapes typically calculated?
Answer
By subtracting the area of the smaller (unshaded/inner) shape from the area of the larger (outer) shape.
Question
What property of diagonals is used to solve side length problems in a rhombus using Pythagoras' Theorem?
Answer
The diagonals of a rhombus bisect each other at right angles.
Question
If the side of a square is halved, its area becomes _____ of the original area.
Answer
one-fourth
Question
If the radius of a circle is doubled, its area becomes _____ times the original area.
Answer
four
Question
What is the formula for the area of a rectangle?
Answer
\text{Area} = \text{length} \times \text{breadth}
Question
The length of the line segment joining the mid-points of the non-parallel sides of a trapezium is often called the _____.
Answer
median
Question
If a rectangle has length 12 \text{ cm} and the same perimeter as a square of side 10.5 \text{ cm}, what is its width?
Answer
9 \text{ cm}
Question
A square has an area of 484 \text{ m}^2. What is the length of its side?
Answer
22 \text{ m}
Question
In a right-angled triangle, the side opposite the right angle is called the _____.
Answer
hypotenuse
Question
What is the area of a kite composed of two isosceles triangles on the same base with heights h_1 and h_2?
Answer
\text{Area} = \frac{1}{2} \times \text{base} \times (h_1 + h_2)
Question
If a circle has a diameter of 126 \text{ cm}, what is its radius?
Answer
63 \text{ cm}
Question
The area of a sector of a circle is proportional to the _____ at the centre.
Answer
angle
Question
How many metres are in 1 \text{ kilometre}?
Answer
1,000 \text{ metres}