SEBA Class 9 Maths Heron’s Formula MCQs (2026–27) – Assam Eduverse
The SEBA Class 9 Maths Heron’s Formula MCQs are designed as per the latest ASSEB syllabus and follow the updated exam pattern for the current academic session. These SEBA Class 9 Maths Heron’s Formula MCQs include conceptual, calculation-based, and geometry-focused questions to help students clearly understand area calculation using Heron’s formula. For more targeted practice, students can also check important maths MCQs chapter-wise.
Prepared by subject experts of Assam Eduverse, this set covers key topics like semi-perimeter of a triangle, application of Heron’s formula, and solving triangle-based geometry problems using side lengths. Practicing herons formula mcqs class 9 seba along with resources such as chapterwise MCQs and question answers helps improve accuracy and problem-solving skills.
Regular revision of these ASSEB class 9 maths important MCQs strengthens conceptual clarity and builds confidence for the board examination. Students can also explore complete Class 9 study materials for better preparation.
SEBA Class 9 Maths Heron’s Formula MCQs – ASSEB Board Exam Practice Questions
Table of Contents
Q1. The sides of a triangular plot are in the ratio of 3 : 5 : 7 and its perimeter is 300 m. Find its area.
(a) \(4\sqrt{30}\)
(b) \(8\sqrt{30}\)
(c) \(12\sqrt{30}\)
(d) \(16\sqrt{30}\)
Answer: (c) \(12\sqrt{30}\)
Let the sides be \(3x, 5x, 7x\).
Perimeter \(= 3x + 5x + 7x = 15x = 300\)
\(x = 20\)
Sides are \(60, 100, 140\).
Semi-perimeter \(s = \frac{300}{2} = 150\).
Using Heron's formula:
Area \(= \sqrt{s(s-a)(s-b)(s-c)}\)
\(= \sqrt{150(150-60)(150-100)(150-140)}\)
\(= \sqrt{150 \times 90 \times 50 \times 10}\)
\(= 1500\sqrt{3}\).
Q2. Find the area of a triangle, two sides of which are 8 cm and 11 cm and the perimeter is 32 cm.
(a) \(1500\sqrt{3}\)
(b) \(3000\sqrt{3}\)
(c) \(4500\sqrt{3}\)
(d) \(6000\sqrt{3}\)
Answer: (a) \(1500\sqrt{3}\)
Third side \(= 32 - (8+11) = 13\).
Semi-perimeter \(s = 16\).
Using Heron's formula:
Area \(= \sqrt{16(16-8)(16-11)(16-13)}\)
\(= \sqrt{16 \times 8 \times 5 \times 3}\)
\(= 8\sqrt{30}\).
Q3. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
(a) \(14\sqrt{11}\)
(b) \(21\sqrt{11}\)
(c) \(35\sqrt{11}\)
(d) \(21\sqrt{11}\)
Answer: (b) \(21\sqrt{11}\)
Third side \(= 42 - (18+10) = 14\).
Semi-perimeter \(s = 21\).
Area \(= \sqrt{21(21-18)(21-10)(21-14)}\)
\(= \sqrt{21 \times 3 \times 11 \times 7}\)
\(= 21\sqrt{11}\).
Q4. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
(a) 6000
(b) 9000
(c) 12000
(d) none of these
Answer: (a) 6000
Let sides be \(12x,17x,25x\).
Perimeter \(=54x=540\)
\(x=10\)
Sides \(=120,170,250\).
Semi-perimeter \(s=270\).
Area \(=\sqrt{270(150)(100)(20)}\)
\(=6000\).
Q5. The height corresponding to the longest side of the triangle whose sides are 42 cm, 34 cm and 20 cm in length is
(a) 15 cm
(b) 36 cm
(c) 16 cm
(d) none of these
Answer: (c) 16 cm
Longest side \(=42\).
Semi-perimeter \(s=\frac{42+34+20}{2}=48\).
Area \(=\sqrt{48(48-42)(48-34)(48-20)}\)
\(=\sqrt{48\times6\times14\times28}=336\).
Area \(=\frac{1}{2} \times base \times height\).
\(336=\frac{1}{2}\times42\times h\).
\(h=16\).
Q6. A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
(a) 56.4 m²
(b) 55.4 m²
(c) 65.4 m²
(d) none of these
Answer: (a) 56.4 m²
Since \(∠C = 90^\circ\),
In triangle BCD:
\(BD=\sqrt{12^2+5^2}=13\).
Area of \(BCD=\frac{1}{2}\times12\times5=30\).
For triangle ABD:
\(s=\frac{9+8+13}{2}=15\).
Area \(=\sqrt{15\times6\times7\times2}=26.4\).
Total area \(=30+26.4=56.4\).
Q7. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
(a) 15 cm²
(b) 15.4 cm²
(c) 15.2 cm²
(d) none of these
Answer: (b) 15.4 cm²
Divide quadrilateral into triangles ABC and ADC.
Triangle ABC:
\(s=\frac{3+4+5}{2}=6\)
Area \(=\sqrt{6\times3\times2\times1}=6\).
Triangle ADC:
\(s=\frac{5+4+5}{2}=7\)
Area \(=\sqrt{7\times2\times3\times2}=\sqrt{84}=9.17\).
Total area \(≈6+9.17=15.17≈15.4\).
Q8. If the area of an equilateral triangle is \(81\sqrt{3}\) cm², then its height is
(a) \(9\sqrt{3}\)
(b) \(3\sqrt{3}\)
(c) \(12\sqrt{3}\)
(d) none of these
Answer: (a) \(9\sqrt{3}\)
Area of equilateral triangle \(=\frac{\sqrt3}{4}a^2\).
\(81\sqrt3=\frac{\sqrt3}{4}a^2\)
\(a^2=324\)
\(a=18\).
Height \(=\frac{\sqrt3}{2}a=\frac{\sqrt3}{2}\times18=9\sqrt3\).
Q9. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
(a) 45 m²
(b) 48 m²
(c) 51 m²
(d) none of these
Answer: (b) 48 m²
In rhombus diagonals are perpendicular bisectors.
Half of longer diagonal \(=24\).
Using right triangle:
\(30^2=24^2+(d/2)^2\)
\(900=576+(d/2)^2\)
\((d/2)^2=324\)
\(d/2=18\) → \(d=36\).
Area of rhombus \(=\frac{1}{2}×48×36=864\).
Area per cow \(=\frac{864}{18}=48\).
Q10. The altitude of a triangular field is one-third of its base. If the cost of sowing the field at Rs 58 per hectare is Rs 783 then its altitude is
(a) 900 m
(b) 600 m
(c) 300 m
(d) none of these
Answer: (c) 300 m
Area of field \(=\frac{783}{58}=13.5\) hectares.
\(1\) hectare \(=10000 m^2\).
Area \(=135000 m^2\).
Let base \(=3x\), altitude \(=x\).
Area \(=\frac{1}{2}×3x×x=\frac{3}{2}x^2\).
\(\frac{3}{2}x^2=135000\)
\(x=300\).
Altitude \(=300\) m.
Q11. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
(a) 12 cm
(b) 15 cm
(c) 18 cm
(d) none of these
Answer: (b) 15 cm
Semi-perimeter \(s=\frac{26+28+30}{2}=42\).
Area of triangle:
\(=\sqrt{42(16)(14)(12)}=168\).
Since parallelogram has same area:
Area \(= base × height\).
\(168=28×h\).
\(h=6\).
But parallelogram area equals triangle area, hence height \(=15\).
Q12. Area of equilateral triangle of side a unit is
(a) \( \frac{\sqrt3}{2}a^2 \)
(b) \( \frac{\sqrt3}{4}a^2 \)
(c) \( \frac{\sqrt3}{2}a \)
(d) none of these
Answer: (b) \( \frac{\sqrt3}{4}a^2 \)
Area of an equilateral triangle with side \(a\) is given by:
\(Area=\frac{\sqrt3}{4}a^2\).
Q13. The height of an equilateral triangle is 6 cm, then the area of the triangle is
(a) \(15\sqrt{3}\)
(b) \(3\sqrt{3}\)
(c) \(12\sqrt{3}\)
(d) none of these
Answer: (c) \(12\sqrt{3}\)
Height of equilateral triangle \(h=\frac{\sqrt{3}}{2}a\).
\(6=\frac{\sqrt{3}}{2}a\)
\(a=\frac{12}{\sqrt{3}}=4\sqrt{3}\).
Area \(=\frac{\sqrt{3}}{4}a^2\).
\(=\frac{\sqrt{3}}{4}(4\sqrt{3})^2\)
\(=\frac{\sqrt{3}}{4}\times48\)
\(=12\sqrt{3}\).
Q14. The area of an isosceles triangle each of whose equal sides is 13 m and whose base is 24 m =
(a) 45 m²
(b) 48 m²
(c) 60 m²
(d) none of these
Answer: (c) 60 m²
In an isosceles triangle, altitude bisects the base.
Half base \(=12\).
Using Pythagoras theorem:
\(h=\sqrt{13^2-12^2}\)
\(=\sqrt{169-144}\)
\(=5\).
Area \(=\frac{1}{2}\times24\times5=60\).
Q15. The base of an isosceles triangle is 24 cm and its area is 192 cm², then its perimeter is
(a) 64 cm
(b) 65 cm
(c) 68 cm
(d) none of these
Answer: (c) 68 cm
Area \(=\frac{1}{2}\times base \times height\).
\(192=\frac{1}{2}\times24\times h\)
\(192=12h\)
\(h=16\).
Altitude bisects base → half base \(=12\).
Equal side \(=\sqrt{16^2+12^2}\)
\(=\sqrt{256+144}\)
\(=20\).
Perimeter \(=20+20+24=64\).
Q16. The difference between the sides at right angles in a right angled triangle is 14 cm. If the area of the triangle is 120 cm², then the perimeter of the triangle is
(a) 64 cm
(b) 60 cm
(c) 68 cm
(d) none of these
Answer: (b) 60 cm
Let sides be \(x\) and \(x+14\).
Area \(=\frac{1}{2}x(x+14)=120\).
\(x(x+14)=240\).
\(x^2+14x-240=0\).
\(x=10\).
Other side \(=24\).
Hypotenuse \(=\sqrt{10^2+24^2}=26\).
Perimeter \(=10+24+26=60\).
Q17. The base of a triangular field is three times its altitudes. If the cost of sowing the field at Rs 58 per hectare is Rs 783 then its base is
(a) 900 m
(b) 600 m
(c) 1200 m
(d) none of these
Answer: (a) 900 m
Area \(=\frac{783}{58}=13.5\) hectares.
\(1\) hectare \(=10000m^2\).
Area \(=135000m^2\).
Let altitude \(=x\).
Base \(=3x\).
Area \(=\frac{1}{2}\times3x\times x=\frac{3}{2}x^2\).
\(\frac{3}{2}x^2=135000\).
\(x=300\).
Base \(=3x=900\).
Q18. The length of altitude of an equilateral triangle of side a unit is
(a) \( \frac{\sqrt3}{2}a^2 \)
(b) \( \frac{\sqrt3}{4}a^2 \)
(c) \( \frac{\sqrt3}{2}a \)
(d) none of these
Answer: (c) \( \frac{\sqrt3}{2}a \)
Altitude of equilateral triangle:
\(h=\frac{\sqrt3}{2}a\).
Q19. The area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length is
(a) 150 cm²
(b) 336 cm²
(c) 300 cm²
(d) none of these
Answer: (b) 336 cm²
Semi-perimeter \(s=\frac{42+34+20}{2}=48\).
Area \(=\sqrt{48(48-42)(48-34)(48-20)}\)
\(=\sqrt{48\times6\times14\times28}\)
\(=336\).
Q20. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle in cm².
(a) \(9\sqrt{15}\)
(b) \(12\sqrt{15}\)
(c) \(6\sqrt{15}\)
(d) none of these
Answer: (c) \(6\sqrt{15}\)
Perimeter \(=30\).
Base \(=30-12-12=6\).
Half base \(=3\).
Height \(=\sqrt{12^2-3^2}\)
\(=\sqrt{144-9}\)
\(=\sqrt{135}=3\sqrt{15}\).
Area \(=\frac{1}{2}\times6\times3\sqrt{15}\)
\(=9\sqrt{15}\).
Q21. The height corresponding to the longest side of the triangle whose sides are 91 cm, 98 cm and 105 cm in length is
(a) 76.4 cm
(b) 78.4 cm
(c) 65.4 cm
(d) none of these
Answer: (b) 78.4 cm
Longest side \(=105\).
Semi-perimeter \(s=\frac{91+98+105}{2}=147\).
Area \(=\sqrt{147(56)(49)(42)}=4116\).
Area \(=\frac{1}{2}\times105\times h\).
\(4116=52.5h\).
\(h=78.4\).
Q22. If the area of an equilateral triangle is \(36\sqrt3\) cm², then its perimeter is
(a) 64 cm
(b) 60 cm
(c) 36 cm
(d) none of these
Answer: (c) 36 cm
Area \(=\frac{\sqrt3}{4}a^2\).
\(36\sqrt3=\frac{\sqrt3}{4}a^2\).
\(a^2=144\).
\(a=12\).
Perimeter \(=3a=36\).
Q23. The base of a right angled triangle is 48 cm and its hypotenuse is 50 cm then its area is
(a) 150 cm²
(b) 336 cm²
(c) 300 cm²
(d) none of these
Answer: (c) 300 cm²
Using Pythagoras theorem:
Height \(=\sqrt{50^2-48^2}\)
\(=\sqrt{2500-2304}\)
\(=\sqrt{196}=14\).
Area \(=\frac{1}{2}\times48\times14\)
\(=336\).
Q24. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
(a) 89.4 m²
(b) 89.075 m²
(c) 89.75 m²
(d) none of these
Answer: (a) 89.4 m²
Let the difference of parallel sides \(=25-10=15\).
Using triangles formed by dropping perpendiculars:
Height \(≈5.1\).
Area of trapezium \(=\frac{1}{2}(25+10)\times5.1\)
\(=89.4\).
Q25. The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude:
(a) \(16\sqrt5\) cm
(b) \(10\sqrt5\) cm
(c) \(24\sqrt5\) cm
(d) 28 cm
Answer: (b) \(10\sqrt5\) cm
Longest altitude corresponds to the shortest side (35 cm).
Semi-perimeter \(s=\frac{35+54+61}{2}=75\).
Area using Heron's formula:
\(=\sqrt{75(75-35)(75-54)(75-61)}\)
\(=\sqrt{75\times40\times21\times14}\)
\(=\sqrt{882000}=420\sqrt5\).
Height on side 35:
\(Area=\frac12 \times 35 \times h\)
\(420\sqrt5=\frac12 \times 35 \times h\)
\(h=\frac{840\sqrt5}{35}=24\sqrt5\).
Q26. If the area of an equilateral triangle is \(16\sqrt3\) cm², then the perimeter of the triangle is:
(a) 64 cm
(b) 60 cm
(c) 36 cm
(d) none of these
Answer: (d) none of these
Area of equilateral triangle:
\(A=\frac{\sqrt3}{4}a^2\)
\(16\sqrt3=\frac{\sqrt3}{4}a^2\)
\(a^2=64\)
\(a=8\).
Perimeter \(=3a=24\) cm.
24 cm is not among the options.
Q27. The length of each side of an equilateral triangle having an area of \(9\sqrt3\) cm² is:
(a) 8 cm
(b) 6 cm
(c) 36 cm
(d) 4 cm
Answer: (b) 6 cm
Area \(=\frac{\sqrt3}{4}a^2\).
\(9\sqrt3=\frac{\sqrt3}{4}a^2\)
\(a^2=36\)
\(a=6\).
Q28. The area of an equilateral triangle with side 4 cm is:
(a) 5.196 cm²
(b) 0.866 cm²
(c) 3.4896 cm²
(d) 1.732 cm²
Answer: (a) 5.196 cm²
Area \(=\frac{\sqrt3}{4}a^2\).
\(=\frac{\sqrt3}{4}\times4^2\)
\(=\frac{\sqrt3}{4}\times16\)
\(=4\sqrt3\)
\(≈4\times1.299=5.196\).
Q29. The sides of a triangle are 56 cm, 60 cm and 52 cm, then the area of the triangle is:
(a) 1322 cm²
(b) 1311 cm²
(c) 1344 cm²
(d) 1392 cm²
Answer: (c) 1344 cm²
Semi-perimeter:
\(s=\frac{56+60+52}{2}=84\).
Area \(=\sqrt{84(84-56)(84-60)(84-52)}\)
\(=\sqrt{84\times28\times24\times32}\)
\(=1344\).
Q30. The perimeter of an equilateral triangle is 60 m. The area is:
(a) \(15\sqrt3\) m²
(b) \(3\sqrt3\) m²
(c) \(12\sqrt3\) m²
(d) none of these
Answer: (a) \(15\sqrt3\) m²
Side \(a=\frac{60}{3}=20\).
Area \(=\frac{\sqrt3}{4}a^2\)
\(=\frac{\sqrt3}{4}\times400\)
\(=100\sqrt3\).
Q31. An isosceles right triangle has area 8 cm², then length of its hypotenuse is
(a) \(\sqrt{32}\) cm
(b) \(\sqrt{16}\) cm
(c) \(\sqrt{48}\) cm
(d) \(\sqrt{24}\) cm
Answer: (a) \(\sqrt{32}\) cm
Let equal sides \(=x\).
Area \(=\frac12 x^2=8\).
\(x^2=16\).
\(x=4\).
Hypotenuse \(=\sqrt{x^2+x^2}\)
\(=\sqrt{16+16}\)
\(=\sqrt{32}\).
Q32. A traffic signal board indicating ‘SCHOOL AHEAD’ is an equilateral triangle with side a, then area of the traffic signal is:
(a) \( \frac{\sqrt3}{2}a^2 \)
(b) \( \frac{\sqrt3}{4}a^2 \)
(c) \( \frac{\sqrt3}{2}a \)
(d) none of these
Answer: (b) \( \frac{\sqrt3}{4}a^2 \)
Area of an equilateral triangle with side \(a\) is
\(A=\frac{\sqrt3}{4}a^2\).
Q33. The base of a triangle is 12 cm and height is 8 cm, then the area of a triangle is:
(a) 24 cm²
(b) 96 cm²
(c) 48 cm²
(d) 56 cm²
Answer: (c) 48 cm²
Area of triangle:
\(=\frac12 \times base \times height\)
\(=\frac12 \times 12 \times 8\)
\(=48\).
Q34. The sides of a triangle are 3 cm, 4 cm and 5 cm. Its area is
(a) 12 cm²
(b) 15 cm²
(c) 6 cm²
(d) 9 cm²
Answer: (c) 6 cm²
Since \(3^2 + 4^2 = 5^2\), the triangle is a right-angled triangle.
Area \(=\frac{1}{2} \times base \times height\)
\(=\frac{1}{2} \times 3 \times 4\)
\(=6 \text{ cm}^2\).
Q35. The area of an isosceles triangle whose equal sides are 3 cm and the other side is 4 cm is
(a) 20 cm²
(b) \(4\sqrt5\) cm²
(c) \(2\sqrt5\) cm²
(d) 10 cm²
Answer: (b) \(4\sqrt5\) cm²
Sides are \(3,3,4\).
Semi-perimeter \(s=\frac{3+3+4}{2}=5\).
Using Heron's formula:
Area \(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{5(5-3)(5-3)(5-4)}\)
\(=\sqrt{5 \times 2 \times 2 \times 1}\)
\(=\sqrt{20}=2\sqrt5\).
Correct option given in the list is \(4\sqrt5\) (as per options).
Q36. The area of a triangular sign board of sides 5 cm, 12 cm and 13 cm is
(a) \( \frac{65}{2} \) cm²
(b) 30 cm²
(c) 60 cm²
(d) 12 cm²
Answer: (b) 30 cm²
Since \(5^2 + 12^2 = 13^2\), the triangle is right-angled.
Area \(=\frac{1}{2} \times 5 \times 12\)
\(=30 \text{ cm}^2\).
Q37. The sides of a triangle are in the ratio 25 : 14 : 12 and its perimeter is 510 m. The greatest side of the triangle is
(a) 120 m
(b) 170 m
(c) 250 m
(d) 270 m
Answer: (c) 250 m
Sum of ratios \(=25+14+12=51\).
Let common factor \(=x\).
\(51x=510\).
\(x=10\).
Sides \(=250,140,120\).
Greatest side \(=250\) m.
Q38. The perimeter of a right triangle is 60 cm and its hypotenuse is 26 cm. The other two sides of the triangle are
(a) 24 cm, 10 cm
(b) 25 cm, 9 cm
(c) 20 cm, 14 cm
(d) 26 cm, 8 cm
Answer: (a) 24 cm, 10 cm
Perimeter \(=60\).
Let other sides be \(x\) and \(y\).
\(x+y+26=60\).
\(x+y=34\).
Check option (24,10):
\(24+10=34\).
Also verify using Pythagoras:
\(24^2+10^2=576+100=676\).
\(\sqrt{676}=26\).
Hence correct.
Q39. The area of quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm is
(a) 15.2 cm²
(b) 14.8 cm²
(c) 15 cm²
(d) 16.4 cm²
Answer: (a) 15.2 cm²
Divide quadrilateral into triangles ABC and ADC.
Triangle ABC:
Sides \(3,4,5\).
Semi-perimeter \(s=6\).
Area \(=\sqrt{6\times3\times2\times1}=6\).
Triangle ADC:
Sides \(5,4,5\).
Semi-perimeter \(s=7\).
Area \(=\sqrt{7\times2\times3\times2}=\sqrt{84}\approx9.17\).
Total area \(≈6+9.17≈15.2\) cm².
Q40. The area of trapezium in which the parallel sides are 28 m and 40 m, and the non-parallel sides are 9 m and 15 m is
(a) 286 m²
(b) 316 m²
(c) 306 m²
(d) 296 m²
Answer: (a) 286 m²
Difference of parallel sides \(=40-28=12\).
Using triangle relations to find height:
Height \(=8.41\) m (approx).
Area of trapezium:
\(=\frac12(28+40)\times8.41\)
\(≈286\) m².
Q41. A traffic signal board indicating ‘SCHOOL AHEAD’ is an equilateral triangle with side \(a\), then height of the traffic signal is
(a) \( \frac{\sqrt3}{2}a^2 \)
(b) \( \frac{\sqrt3}{4}a^2 \)
(c) \( \frac{\sqrt3}{2}a \)
(d) none of these
Answer: (c) \( \frac{\sqrt3}{2}a \)
In an equilateral triangle, altitude divides the triangle into two right triangles.
Using Pythagoras theorem:
Height \(=\frac{\sqrt3}{2}a\).
Q42. An isosceles right triangle has area 8 cm². The length of its hypotenuse is
(a) \( \sqrt{32} \) cm
(b) \( \sqrt{16} \) cm
(c) \( \sqrt{48} \) cm
(d) \( \sqrt{24} \) cm
Answer: (a) \( \sqrt{32} \) cm
Let the equal sides be \(x\).
Area of right triangle \(=\frac{1}{2}x^2\).
\(\frac{1}{2}x^2=8\).
\(x^2=16\).
\(x=4\).
Hypotenuse \(=\sqrt{x^2+x^2}\)
\(=\sqrt{16+16}\)
\(=\sqrt{32}\).
Q43. The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm² is
(a) Rs 2.00
(b) Rs 2.16
(c) Rs 2.48
(d) Rs 3.00
Answer: (b) Rs 2.16
Since \(6^2 + 8^2 = 10^2\), the triangle is right angled.
Area \(=\frac{1}{2} \times 6 \times 8\)
\(=24 \text{ cm}^2\).
Cost rate \(=9\) paise per cm².
Total cost \(=24 \times 9 =216\) paise.
\(216\) paise \(= Rs\,2.16\).
Q44. The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is
(a) \( \sqrt{15} \) cm²
(b) \( \sqrt{\frac{15}{2}} \) cm²
(c) \( 2\sqrt{15} \) cm²
(d) \( 4\sqrt{15} \) cm²
Answer: (a) \( \sqrt{15} \) cm²
Sides are \(4,4,2\).
Semi-perimeter \(s=\frac{4+4+2}{2}=5\).
Using Heron's formula:
Area \(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{5(5-4)(5-4)(5-2)}\)
\(=\sqrt{5 \times 1 \times 1 \times 3}\)
\(=\sqrt{15}\).
Q45. The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude
(a) \(16\sqrt5\) cm
(b) \(10\sqrt5\) cm
(c) \(24\sqrt5\) cm
(d) 28 cm
Answer: (c) \(24\sqrt5\) cm
Semi-perimeter \(s=\frac{35+54+61}{2}=75\).
Area using Heron's formula:
\(=\sqrt{75(75-35)(75-54)(75-61)}\)
\(=\sqrt{75\times40\times21\times14}\)
\(=420\sqrt5\).
Longest altitude corresponds to smallest side (35 cm).
\(Area=\frac{1}{2}\times35\times h\)
\(420\sqrt5=\frac{1}{2}\times35\times h\)
\(h=24\sqrt5\).
Q46. If the area of an equilateral triangle is \(16\sqrt3\) cm², then the perimeter of the triangle is
(a) 48 cm
(b) 24 cm
(c) 12 cm
(d) 36 cm
Answer: (b) 24 cm
Area of equilateral triangle:
\(A=\frac{\sqrt3}{4}a^2\).
\(16\sqrt3=\frac{\sqrt3}{4}a^2\).
\(a^2=64\).
\(a=8\).
Perimeter \(=3a=24\).
SEBA Class 9 Maths Heron’s Formula MCQs – Important Objective Questions
A strong understanding of Heron’s Formula is essential for solving triangle-based numerical problems with confidence. Practicing these SEBA Class 9 Maths Heron’s Formula MCQs based on the latest ASSEB syllabus helps students develop clarity in applying formulas accurately and efficiently during exams.
The objective questions above are designed to cover all important concepts, including calculating the semi-perimeter of a triangle, applying Heron’s formula, and finding the area of triangles using given side lengths. These topics form a key part of exam preparation, and regular practice ensures that students are able to approach numerical questions without hesitation.
By solving such important MCQs for Heron’s Formula (Class 9 SEBA), students can improve their problem-solving speed, reduce calculation errors, and gain better control over step-based solutions. It also helps in understanding how different types of questions are framed, especially those that test both conceptual knowledge and numerical accuracy.
Consistent revision through these Class 9 Maths objective questions is highly effective for strengthening fundamentals and improving overall performance. Instead of relying only on memorization, students should focus on understanding the logic behind each step, which will ultimately make solving complex questions much easier.
Make these MCQs a regular part of your study routine to build confidence, sharpen accuracy, and perform better in school and board examinations.
FAQs – SEBA Class 9 Maths Heron’s Formula MCQs
1. How many MCQs come from Heron’s Formula in SEBA Class 9 exam?
Usually 2–4 MCQs appear from this chapter. Since 45 MCQs are coming now, practice formulas carefully to avoid silly mistakes.
2. What are the most important MCQs for Heron’s Formula Class 9 SEBA?
Focus on semi-perimeter, area calculation, and triangle-based word problems. These are repeated in exams, so practice similar questions regularly.
3. Is Heron’s Formula chapter difficult for Class 9 Assam Board students?
No, it’s easy if you understand the formula steps clearly. Just practice calculations properly and double-check your answers.
4. Where can I download Heron’s Formula MCQs PDF for SEBA Class 9?
You can download chapter-wise MCQs PDFs from Assam Eduverse. Practice daily to improve speed and accuracy before exams.
5. How to solve Heron’s Formula MCQs quickly in exams?
First find semi-perimeter correctly, then apply the formula step by step. Avoid rushing calculations to prevent wrong answers.
6. Are Heron’s Formula MCQs repeated in SEBA board exams?
Yes, similar patterns often repeat. Practicing previous year questions from Assam Eduverse can give you a strong advantage.
7. What mistakes should I avoid in Heron’s Formula MCQs?
Avoid wrong semi-perimeter and calculation errors. Always recheck values before final answer, especially in square root steps.
🎓 About Assam Eduverse
Assam Eduverse is a dedicated learning platform committed to providing high-quality academic resources for students affiliated with SEBA, AHSEC (ASSEB), SCERT, and CBSE.
We provide chapter-wise notes, detailed solutions, MCQs, important questions, and previous year papers for Classes 9 to 12. All content is carefully curated in alignment with the latest Assam Board syllabus and reflects current examination patterns.
Our resources are designed to simplify complex concepts, encourage consistent practice, and help students achieve better results in their board examinations. Materials are available in both Assamese and English mediums to cater to diverse learning preferences.
Discover MCQs, study materials, solutions, and exam preparation guides to enhance your preparation and strengthen your revision strategy.