SEBA Class 9 Maths Heron’s Formula MCQs (2026–27) – Assam Eduverse
The SEBA Class 9 Maths Heron’s Formula MCQs (2026–27) are prepared according to the latest ASSEB syllabus and the updated board exam pattern. These SEBA Class 9 Maths Heron’s Formula MCQs include conceptual objective questions, calculation-based MCQs, and geometry practice questions designed to strengthen students’ understanding of area calculation using Heron’s formula.
Prepared by subject experts of Assam Eduverse, these questions focus on important topics such as semi-perimeter of a triangle, application of Heron’s formula, finding the area of triangles using side lengths, and solving geometry problems involving triangles. Practicing herons formula mcqs class 9 seba and assam board class 9 maths objective questions helps students improve their problem-solving skills and numerical accuracy.
Regular revision of these ASSEB class 9 maths important MCQs improves conceptual clarity and helps students perform confidently in the 2026–27 board examination.
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 (2026–27) Important Objective Questions
The SEBA Class 9 Maths Heron’s Formula MCQs provided here are prepared according to the latest ASSEB syllabus 2026–27. These SEBA Class 9 Maths Heron’s Formula MCQs include conceptual objective questions and exam-oriented practice sets designed to strengthen triangle area calculations.
Students preparing for the board examination should regularly practice SEBA Class 9 Maths Heron’s Formula MCQs. These questions cover important topics such as semi-perimeter of a triangle, Heron’s formula, and finding the area of triangles using the lengths of three sides.
The herons formula mcqs class 9 seba provided here are prepared by subject experts to ensure alignment with the latest examination pattern. These ASSEB class 9 maths important MCQs help students understand formula-based geometry concepts clearly.
Regular revision using SEBA Class 9 Maths Heron’s Formula MCQs along with assam board class 9 maths objective questions improves numerical accuracy, conceptual clarity, and board exam performance.
Frequently Asked Questions (FAQ)
1. Are these SEBA Class 9 Maths Heron’s Formula MCQs based on the latest syllabus?
Yes, these MCQs follow the latest ASSEB syllabus for the 2026–27 academic session.
2. Are herons formula mcqs class 9 seba helpful for exam preparation?
Yes, these MCQs help students practice triangle area calculations using Heron’s formula.
3. Which topics are included in Heron’s Formula objective questions?
Topics include semi-perimeter calculation, Heron’s formula application, and finding the area of triangles.
4. Who prepared these ASSEB class 9 maths important MCQs?
These MCQs are prepared by subject experts of Assam Eduverse according to the updated syllabus.
5. Are assam board class 9 maths objective questions from Heron’s Formula important?
Yes, questions based on Heron’s Formula are frequently asked in Class 9 Mathematics exams.
6. Do these MCQs include numerical problems?
Yes, these MCQs include numerical problems related to area calculation using Heron’s formula.
7. Can practicing MCQs improve mathematics exam performance?
Yes, regular practice improves numerical accuracy and conceptual understanding.
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