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
$1,000 \text{ metres}$