Extra Practice Questions
Heron’s Formula (R.D. Sharma Level – Class 9)
1. The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3:2. Find the area of the triangle.
Let the equal sides be 3x and the base be 2x.
Perimeter = 3x + 3x + 2x = 8x = 32 cm => x = 4 cm.
Sides are a=12 cm, b=12 cm, c=8 cm.
Semi-perimeter (s) = 32/2 = 16 cm.
Area = √(16(16-12)(16-12)(16-8)) = √(16 × 4 × 4 × 8) = 32√2 cm².
Answer: The area is 32√2 cm².
2. Find the area of a quadrilateral ABCD whose sides are AB=9m, BC=40m, CD=28m, DA=15m and the angle between the first two sides is a right angle.
Given ∠B = 90°. The quadrilateral consists of two triangles: ΔABC and ΔADC.
Step 1: Find the area of right ΔABC.
Area(ΔABC) = 1/2 × base × height = 1/2 × 9 × 40 = 180 m².
Step 2: Find the diagonal AC.
AC² = AB² + BC² = 9² + 40² = 81 + 1600 = 1681 => AC = 41 m.
Step 3: Find the area of ΔADC using Heron’s Formula.
Sides are a=28, b=15, c=41. s = (28+15+41)/2 = 84/2 = 42 m.
Area(ΔADC) = √(42(42-28)(42-15)(42-41)) = √(42 × 14 × 27 × 1) = 126 m².
Step 4: Find the total area.
Area(ABCD) = Area(ΔABC) + Area(ΔADC) = 180 + 126 = 306 m².
Answer: The area of the quadrilateral is 306 m².
… (18 more questions follow)
3. The sides of a triangular field are 51m, 37m and 20m. Find the number of rose beds that can be prepared in the field, if each rose bed occupies a space of 6 sq. m.
Sides a=51, b=37, c=20. s = (51+37+20)/2 = 108/2 = 54 m.
Area = √(54(54-51)(54-37)(54-20)) = √(54 × 3 × 17 × 34) = 306 m².
Number of rose beds = Total Area / Area per bed = 306 / 6 = 51.
Answer: 51 rose beds can be prepared.
4. Find the cost of leveling a triangular ground at the rate of ₹20 per square meter if its sides are 20m, 21m and 13m.
Sides a=20, b=21, c=13. s = (20+21+13)/2 = 54/2 = 27 m.
Area = √(27(27-20)(27-21)(27-13)) = √(27 × 7 × 6 × 14) = 126 m².
Cost of leveling = Area × Rate = 126 × 20 = ₹2520.
Answer: The cost is ₹2520.
5. The perimeter of a right-angled triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle.
Let the other two sides be a and b. Perimeter a+b+25=60 => a+b=35.
By Pythagoras theorem, a²+b²=25²=625.
We know (a+b)² = a²+b²+2ab => 35² = 625 + 2ab => 1225 = 625 + 2ab => 2ab=600 => ab=300.
Area of a right triangle = 1/2 × base × height = 1/2 × ab = 1/2 × 300 = 150 cm².
Answer: The area of the triangle is 150 cm².
6. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30m and its longer diagonal is 48m, how much area of grass field will each cow be getting?
A rhombus is made of two congruent triangles. Consider one triangle with sides 30m, 30m and 48m.
s = (30+30+48)/2 = 108/2 = 54 m.
Area of one triangle = √(54(54-30)(54-30)(54-48)) = √(54 × 24 × 24 × 6) = 432 m².
Total area of rhombus = 2 × 432 = 864 m².
Area per cow = Total Area / Number of cows = 864 / 18 = 48 m².
Answer: Each cow will get 48 m² of grass field.
7. The length of the sides of a triangle are 5cm, 12cm and 13cm. Find the length of the perpendicular from the opposite vertex to the side whose length is 13cm.
The sides 5, 12, 13 form a Pythagorean triplet (5²+12² = 25+144 = 169 = 13²). It is a right-angled triangle.
Area = 1/2 × base × height = 1/2 × 5 × 12 = 30 cm².
Now, let the base be the hypotenuse (13 cm) and the perpendicular be ‘p’.
Area = 1/2 × 13 × p = 30 => p = 60/13 cm.
Answer: The length of the perpendicular is 60/13 cm.
8. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours, each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
There are 5 pieces of each color.
First, find the area of one triangular piece. Sides are 20, 50, 50.
s = (20+50+50)/2 = 120/2 = 60 cm.
Area = √(60(60-20)(60-50)(60-50)) = √(60 × 40 × 10 × 10) = 200√6 cm².
Cloth required for each colour = 5 × Area of one piece = 5 × 200√6 = 1000√6 cm².
Answer: 1000√6 cm² of cloth of each colour is required.
9. The perimeter of a triangular field is 450m and its sides are in the ratio 13:12:5. Find the area of the field.
Let sides be 13x, 12x, 5x. Perimeter = 13x+12x+5x = 30x = 450m => x=15m.
Sides are: a=13*15=195m, b=12*15=180m, c=5*15=75m.
s = 450/2 = 225m.
Area = √(225(225-195)(225-180)(225-75)) = √(225 × 30 × 45 × 150) = 6750 m².
Answer: The area of the field is 6750 m².
10. Find the percentage increase in the area of a triangle if its each side is doubled.
Let original sides be a, b, c. Original semi-perimeter s = (a+b+c)/2.
Original Area A₁ = √(s(s-a)(s-b)(s-c)).
New sides are 2a, 2b, 2c. New semi-perimeter s’ = (2a+2b+2c)/2 = 2s.
New Area A₂ = √(2s(2s-2a)(2s-2b)(2s-2c)) = √(16 * s(s-a)(s-b)(s-c)) = 4A₁.
Increase in area = A₂ – A₁ = 4A₁ – A₁ = 3A₁.
Percentage increase = (Increase / Original Area) × 100 = (3A₁ / A₁) × 100 = 300%.
Answer: The area increases by 300%.
11. Find the area of a trapezium whose parallel sides are 25cm and 10cm, and the non-parallel sides are 14cm and 13cm.
Draw a line from the end of the smaller parallel side, parallel to one of the non-parallel sides, to form a triangle and a parallelogram.
Let the trapezium be ABCD with AB||DC. AB=25, DC=10. AD=13, BC=14. Draw CE||DA.
AECD is a parallelogram, so AE=10, CE=13. EB = 25-10 = 15.
Now consider ΔCEB with sides 13, 14, 15. s=(13+14+15)/2 = 21.
Area(ΔCEB) = √(21*8*7*6) = 84 cm².
Also, Area(ΔCEB) = 1/2 * base * height = 1/2 * 15 * h = 84 => h = 168/15 = 11.2 cm.
Area of Trapezium = 1/2 * (sum of parallel sides) * height = 1/2 * (25+10) * 11.2 = 196 cm².
Answer: The area is 196 cm².
12. Find the area of an equilateral triangle with perimeter 60m.
Perimeter = 3a = 60m => side a = 20m.
Area = (√3/4)a² = (√3/4)(20)² = 100√3 m².
Answer: The area is 100√3 m².
13. The sides of a quadrilateral, taken in order, are 5, 12, 14, 15 metres respectively, and the angle contained by the first two sides is a right angle. Find its area.
This is a quadrilateral with sides AB=5, BC=12, CD=14, DA=15 and ∠B=90°.
Area(ΔABC) = 1/2 * 5 * 12 = 30 m².
Diagonal AC = √(5²+12²) = 13 m.
For ΔADC, sides are 13, 14, 15. s=(13+14+15)/2 = 21.
Area(ΔADC) = √(21*8*7*6) = 84 m².
Total Area = 30 + 84 = 114 m².
Answer: The area is 114 m².
14. The difference between the sides of a right-angled triangle containing the right angle is 7 cm and its area is 60 cm². Find its perimeter.
Let sides be a and b. a-b=7 and Area=1/2*ab=60 => ab=120.
We know (a+b)² = (a-b)²+4ab = 7²+4(120) = 49+480 = 529. So, a+b=√529=23.
Solving a+b=23 and a-b=7 gives 2a=30 => a=15, and b=8.
Hypotenuse c = √(15²+8²) = √289 = 17 cm.
Perimeter = a+b+c = 15+8+17 = 40 cm.
Answer: The perimeter is 40 cm.
15. The perimeter of a rhombus is 100m and one of its diagonals is 40m. Find the area of the rhombus.
Perimeter = 4a = 100m => side a = 25m.
The diagonals of a rhombus bisect each other at right angles. Consider half of the rhombus, which is a triangle with sides 25, 25 and the diagonal 40.
Or better, consider a right triangle formed by half-diagonals and a side. Let half of the other diagonal be x.
x² + (40/2)² = 25² => x² + 20² = 25² => x² = 625 – 400 = 225 => x=15m.
The diagonals are d₁=40m and d₂=2x=30m.
Area = 1/2 * d₁ * d₂ = 1/2 * 40 * 30 = 600 m².
Answer: The area of the rhombus is 600 m².
16. From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 14cm, 10cm and 6cm. Find the area of the triangle.
Let the side of the equilateral triangle be ‘a’. The sum of the lengths of the perpendiculars from any interior point to the sides is equal to the altitude of the triangle.
Altitude (h) = 14 + 10 + 6 = 30 cm.
For an equilateral triangle, h = (√3/2)a.
30 = (√3/2)a => a = 60/√3 = 20√3 cm.
Area = 1/2 * base * height = 1/2 * a * h = 1/2 * (20√3) * 30 = 300√3 cm².
Answer: The area is 300√3 cm².
17. A field is in the shape of a trapezium. Its parallel sides are 25m and 10m. The non-parallel sides are 14m and 13m. Find the area of the field.
This is the same as Question 11.
Draw CE parallel to DA. AECD is a parallelogram, so AE=10m, CE=13m. The remaining side of the triangle is EB = 25-10 = 15m.
For ΔCEB with sides 13, 14, 15, the area (from Heron’s formula) is 84 m².
Also, Area = 1/2 * base * height => 84 = 1/2 * 15 * h => h = 11.2 m.
Area of Trapezium = 1/2 * (10+25) * 11.2 = 196 m².
Answer: The area is 196 m².
18. The perimeter of an equilateral triangle is 36 cm. Find its area.
Perimeter = 3a = 36 cm => side a = 12 cm.
Area = (√3/4)a² = (√3/4)(12)² = 36√3 cm².
Answer: The area is 36√3 cm².
19. Find the area of a regular hexagon of side ‘a’.
A regular hexagon is composed of 6 equilateral triangles of side ‘a’.
Area of one equilateral triangle = (√3/4)a².
Area of hexagon = 6 × Area of one triangle = 6 × (√3/4)a² = (3√3/2)a².
Answer: The area is (3√3/2)a².
20. The base of an isosceles triangle is 10 cm and one of its equal sides is 13 cm. Find its area.
Sides are a=13, b=13, c=10.
s = (13+13+10)/2 = 36/2 = 18 cm.
Area = √(18(18-13)(18-13)(18-10)) = √(18 × 5 × 5 × 8) = 60 cm².
Alternatively, drop a perpendicular from the vertex to the base. It bisects the base into 5 cm.
Height h = √(13² - 5²) = √144 = 12 cm.
Area = 1/2 * base * height = 1/2 * 10 * 12 = 60 cm².
Answer: The area is 60 cm².
