CHAPTER 10 – HERON’S FORMULA (Class 9, NCERT/CBSE)
Part A — 1-Mark Questions (20) with Solutions
Q1. State Heron’s (Hero’s) formula for area of a triangle with sides \(a,b,c\).
If \(s=\dfrac{a+b+c}{2}\), then \( \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}\).
Q2. Define the semi-perimeter \(s\) of a triangle.
\( s=\dfrac{\text{perimeter}}{2}=\dfrac{a+b+c}{2}\).
Q3. For an equilateral triangle of side \(a\), write area using Heron’s formula.
\(s=\tfrac{3a}{2}\Rightarrow \text{Area}=\sqrt{s(s-a)^3}=\dfrac{\sqrt3}{4}a^2\).
Q4. If sides are \(3,4,5\), find area quickly.
Right triangle; area \(=\tfrac12\cdot 3\cdot 4=6\). Heron also gives \(s=6\Rightarrow \sqrt{6\cdot3\cdot2\cdot1}=6\).
Q5. A triangle has perimeter \(28\) cm and sides \(6\) cm and \(9\) cm. Find the third side.
\(28-(6+9)=13\) cm.
Q6. If \(a=7,b=8,c=9\), compute \(s\).
\(s=\dfrac{7+8+9}{2}=12\).
Q7. When is Heron’s formula especially useful?
When heights are not easily available but all three sides are known.
Q8. Area of a triangle with sides \(5,5,6\) is…
\(s=8\Rightarrow \sqrt{8\cdot3\cdot3\cdot2}=12\ \text{cm}^2\).
Q9. For sides \(13,14,15\), area equals…
\(s=21\Rightarrow \sqrt{21\cdot8\cdot7\cdot6}=84\ \text{units}^2\).
Q10. If area \(=A\) and base \(=b\), height \(h\) equals…
\(h=\dfrac{2A}{b}\).
Q11. Feasibility check: can \(2,3,6\) be triangle sides?
No. \(2+3\not>6\) violates triangle inequality.
Q12. If sides are in ratio \(3:4:5\) and perimeter \(=72\) cm, find sides.
Sum ratio \(=12\Rightarrow\) factor \(=6\Rightarrow\) sides \(=18,24,30\) cm.
Q13. Show Heron’s formula reduces to right-triangle area for \(8,15,17\).
\(s=20\Rightarrow \sqrt{20\cdot12\cdot5\cdot3}=60=\tfrac12\cdot8\cdot15\).
Q14. If \(a=b=c=10\) cm, find area.
\(\dfrac{\sqrt3}{4}a^2=25\sqrt3\ \text{cm}^2\).
Q15. Semi-perimeter of triangle with sides \(9,10,17\).
\(s=\dfrac{36}{2}=18\).
Q16. If \(s-a=3, s-b=4, s-c=5\) and \(s=12\), area is…
\(\sqrt{12\cdot3\cdot4\cdot5}= \sqrt{720}=12\sqrt5\).
Q17. Units: if sides in metres, area is in…
\(\text{m}^2\) (square metres).
Q18. If sides \(= (a,b,c)\) scaled by \(k\), area scales by…
Factor \(k^2\).
Q19. Area for \(a=6,b=7,c=11\).
\(s=12\Rightarrow \sqrt{12\cdot6\cdot5\cdot1}= \sqrt{360}=6\sqrt{10}\).
Q20. If \(s=30\) and \((s-a)(s-b)(s-c)=2000\), area \(=\) ?
\(\sqrt{30\times 2000}=\sqrt{60000}=100\sqrt{6}\).
Part B — 2-Mark Questions (20) with Solutions
Q1. Find area when sides are \(7,8,9\) cm.
\(s=12\Rightarrow A=\sqrt{12\cdot5\cdot4\cdot3}=\sqrt{720}=12\sqrt5\ \text{cm}^2\).
Q2. Triangle sides \(8,15,17\) cm. Verify Heron and right-triangle area match.
Heron: \(s=20\Rightarrow A=\sqrt{20\cdot12\cdot5\cdot3}=60\). Right triangle: \(\tfrac12\cdot8\cdot15=60\).
Q3. Two sides \(18\) cm, \(10\) cm; perimeter \(42\) cm. Find area.
Third side \(14\) cm; \(s=21\). \(A=\sqrt{21\cdot3\cdot11\cdot7}=21\sqrt{11}\ \text{cm}^2\).
Q4. Sides in ratio \(12:17:25\) and perimeter \(540\) cm. Find area.
Factor \(=10\Rightarrow\) sides \(120,170,250\) cm; \(s=270\). \(A=\sqrt{270\cdot150\cdot100\cdot20}=9000\ \text{cm}^2\).
Q5. Isosceles triangle with sides \(12,12,6\) cm. Find area.
\(s=15\Rightarrow A=\sqrt{15\cdot3\cdot3\cdot9}=9\sqrt{15}\ \text{cm}^2\).
Q6. Find area of triangle with sides \(9,10,17\).
\(s=18\Rightarrow A=\sqrt{18\cdot9\cdot8\cdot1}=36\ \text{units}^2\).
Q7. Equilateral triangle perimeter \(=180\) cm. Area?
Side \(=60\) cm; \(A=\dfrac{\sqrt3}{4}\cdot 60^2=900\sqrt3\ \text{cm}^2\).
Q8. If sides \(a,b,c\) are integers and \(s=20\) with \(s-a=12,s-b=5,s-c=3\), find area.
\(A=\sqrt{20\cdot12\cdot5\cdot3}=\sqrt{3600}=60\ \text{units}^2\).
Q9. Triangle sides \(5,6,7\). Find area and height to side \(7\).
\(s=9\Rightarrow A=\sqrt{9\cdot4\cdot3\cdot2}= \sqrt{216}=6\sqrt6\). Height \(h=\dfrac{2A}{7}=\dfrac{12\sqrt6}{7}\).
Q10. Sides \(a,b,c\) scaled by \(k\). Show area scales by \(k^2\).
\(s\to ks'\). \(A'=\sqrt{ks' (ks'-ka)(ks'-kb)(ks'-kc)}=k^2\sqrt{s'(s'-a)(s'-b)(s'-c)}=k^2A\).
Q11. If \(A=\sqrt{30\cdot 7\cdot 5\cdot 2}\), find the sides sum.
Here \(s=30\Rightarrow a+b+c=60\).
Q12. Find area of triangle with sides \(x, x, 2x\) if possible.
Not possible: violates triangle inequality (\(x+x>2x\) gives equality only; area \(=0\)).
Q13. Triangle sides \(10,13,13\). Find area.
\(s=18\Rightarrow A=\sqrt{18\cdot8\cdot5\cdot5}=\sqrt{3600}=60\ \text{units}^2\).
Q14. Sides \(a=4,b=13,c=15\). Find area.
\(s=16\Rightarrow A=\sqrt{16\cdot12\cdot3\cdot1}=\sqrt{576}=24\).
Q15. For \(6,10,14\) does a triangle exist? If yes, area?
No: \(6+10=16\not>14\) (equal okay), actually \(16>14\) so triangle exists. \(s=15\Rightarrow A=\sqrt{15\cdot9\cdot5\cdot1}= \sqrt{675}=15\sqrt3\).
Q16. Triangle sides \(a,b,c\) with \(s-a=2, s-b=3, s-c=4\) and \(s=12\). Find area.
\(A=\sqrt{12\cdot2\cdot3\cdot4}=\sqrt{288}=12\sqrt2\).
Q17. Find area when sides \(= (11,13,20)\).
\(s=22\Rightarrow A=\sqrt{22\cdot11\cdot9\cdot2}=\sqrt{4356}=66\).
Q18. If \(A=84\) and base \(=14\), height \(=\ ?\)
\(h=\dfrac{2A}{b}=\dfrac{168}{14}=12\).
Q19. Isosceles triangle \(b=c=10\), base \(=12\). Find area.
\(s=16\Rightarrow A=\sqrt{16\cdot6\cdot6\cdot4}=\sqrt{2304}=48\).
Q20. If \(AB=24, BC=32, CA=40\) m, verify Heron equals right-triangle area.
Right triangle (\(24^2+32^2=40^2\)). Heron: \(s=48\Rightarrow A=\sqrt{48\cdot24\cdot16\cdot8}=384\). Also \(\tfrac12\cdot24\cdot32=384\ \text{m}^2\).
Part C — 3-Mark Questions (20) with Solutions
Q1. A triangular garden has sides \(40\text{ m},32\text{ m},24\text{ m}\). Find its area by Heron and verify with base–height.
\(s=48\Rightarrow A=\sqrt{48\cdot8\cdot24\cdot16}=384\ \text{m}^2\). Since it’s right-angled, \(A=\tfrac12\cdot32\cdot24=384\ \text{m}^2\).
Q2. Two sides of a triangular plot are \(8\) cm and \(11\) cm; perimeter \(32\) cm. Find area.
Third side \(=13\) cm; \(s=16\). \(A=\sqrt{16\cdot8\cdot5\cdot3}=8\sqrt{30}\ \text{cm}^2\).
Q3. A triangular park has sides \(120,80,50\) m. Find area and cost of fencing at ₹20/m, leaving a \(3\) m gate.
\(s=125\Rightarrow A=\sqrt{125\cdot5\cdot45\cdot75}= \dfrac{375\sqrt{15}}{2}\ \text{m}^2\). Perimeter \(=250\) m, wire \(=247\) m; cost \(=247\times 20=\) ₹4,940.
Q4. In a triangle, sides are in ratio \(3:5:7\); perimeter \(300\) m. Find the area.
Factor \(20\Rightarrow\) sides \(60,100,140\) m; \(s=150\). \(A=\sqrt{150\cdot90\cdot50\cdot10}=\dfrac{1500\sqrt3}{2}=750\sqrt3\ \text{m}^2\).
Q5. Find area and height to side \(a=14\) for triangle \( (a,b,c)=(14,13,15)\).
\(s=21\Rightarrow A=\sqrt{21\cdot7\cdot8\cdot6}=\sqrt{7056}=84\). Height \(h=\dfrac{2A}{a}=\dfrac{168}{14}=12\).
Q6. Sides \( (9,10,17)\). Find area and check obtuse/acute by cosine law cue.
\(A=36\). Since \(17^2=289\) and \(9^2+10^2=181\), \(289>181\Rightarrow\) obtuse (largest angle opposite side \(17\)).
Q7. A triangular hoarding has sides \(22\text{ m},120\text{ m},122\text{ m}\). Find painted area and 3-month rent at ₹5,000 per m\(^2\) per year.
\(s=132\Rightarrow A=\sqrt{132\cdot10\cdot12\cdot110}=\sqrt{1320\cdot1320}=1320\ \text{m}^2\). Rent for 3 months \(=1320\times 5000\times \tfrac{3}{12}=\) ₹16,50,000.
Q8. For triangle with sides \(15,11,6\) m, find the colored wall area.
\(s=16\Rightarrow A=\sqrt{16\cdot1\cdot5\cdot10}=\sqrt{800}=20\sqrt{2}\ \text{m}^2\).
Q9. If sides are \(x+1, x+2, x+3\) and area \(= \sqrt{s(s-a)(s-b)(s-c)}=30\). Given \(x=5\) fits. Verify.
Sides \(6,7,8\Rightarrow s=10.5\). \(A=\sqrt{10.5\cdot4.5\cdot3.5\cdot2.5}= \sqrt{413.4375}\approx 20.35\neq 30\). So \(x=5\) doesn’t fit; area depends on \(x\). (Concept check.)
Q10. A triangular grass patch has sides \(9\) m, \(12\) m, \(15\) m. Find area and number of \(1\ \text{m}^2\) turf rolls needed.
\(s=18\Rightarrow A=\sqrt{18\cdot9\cdot6\cdot3}=\sqrt{2916}=54\ \text{m}^2\). Rolls needed \(=54\).
Q11. Sides \( (25,26,27)\). Find area and altitude to side \(26\).
\(s=39\Rightarrow A=\sqrt{39\cdot14\cdot13\cdot12}=\sqrt{85272}= \) \( \boxed{ \, \text{approx } 292\ } \) (exact factorization: \(39\cdot14\cdot13\cdot12= (3\cdot13)\cdot(2\cdot7)\cdot13\cdot(3\cdot4)= (3^2\cdot 13^2\cdot 2^2\cdot 7\cdot 4/4)= \) compute cleanly: \(39\cdot14=546\), \(13\cdot12=156\); product \(=852,?\,\) \(546\cdot156=852,? \) → \(546\times156=852?6=852?6\) which equals \(85,176\). So \(A=\sqrt{85,176}= \mathbf{292}\). Altitude to \(26\): \(h=\dfrac{2A}{26}=\dfrac{584}{26}=22.4615...\).
Q12. For a steel plate triangular piece \( (30,34,40)\) cm, mass per cm\(^2\) is \(0.2\) g. Find mass.
\(s=52\Rightarrow A=\sqrt{52\cdot22\cdot18\cdot12}=\sqrt{247,?}\). Compute: \(52\cdot22=1144\), \(18\cdot12=216\), product \(=247,?\,\) \(1144\cdot216=247,? =247,?\,\) Exactly \(247,?=247,? \) \(1144*216=1144*(200+16)=228,800+18,304=247,104\). \(A=\sqrt{247,104}= \mathbf{ 497.10...}\) but factor: \(247,104= (144)\cdot(1,716)=12^2\cdot 1716=144\cdot 1716\). \(\sqrt{247,104}=12\sqrt{1716}=12\cdot 6\sqrt{ 47.666}\approx 497.1\ \text{cm}^2\). Mass \(=0.2\times 497.1\approx 99.42\text{ g}\).
Q13. Prove: For sides \(a,b,c\), \(16A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\).
Square Heron’s formula: \(A^2=s(s-a)(s-b)(s-c)\). Multiply and divide RHS by 16: \(16A^2=(2s)(2s-2a)(2s-2b)(2s-2c)\Rightarrow\) the identity.
Q14. In a triangle, sides are \( (x, x+1, x+2)\) with area \(60\). Find \(x\) (integer solution).
Trial: \(x=13\Rightarrow s= \tfrac{39}{2}=19.5\). \(A=\sqrt{19.5\cdot6.5\cdot5.5\cdot4.5}\approx 59.9\approx 60\). So \(x=13\) (exact gives \(A=\sqrt{ (39/2)(13/2)(11/2)(9/2)}=\sqrt{ (39\cdot13\cdot11\cdot9)/16 }= \sqrt{ 50193/16 }= \dfrac{\sqrt{50193}}{4}=60\)).
Q15. Altitude to side \(a\) is \(h_a\). Show \(A=\dfrac12 a h_a = \sqrt{s(s-a)(s-b)(s-c)}\).
Both expressions equal the triangle’s area; equality holds universally; it’s a formula equivalence.
Q16. A plot triangular \( (50,60,70)\) m. Find area and seed packets if each covers \(25\ \text{m}^2\).
\(s=90\Rightarrow A=\sqrt{90\cdot40\cdot30\cdot20}=\sqrt{2,160,000}= \mathbf{1469.69...}\) Exactly: factor \(= 90\cdot40=3600\), \(30\cdot20=600\), product \(=2,160,000\Rightarrow A= \,1470\ \text{m}^2\) (approx). Packets \(= \lceil 1470/25\rceil= \lceil 58.8\rceil=59\).
Q17. For triangle \( (6,8,10)\), find radius \(r\) of its incircle using \(A=rs\).
\(A=\sqrt{12\cdot6\cdot4\cdot2}=24\). \(s=12\Rightarrow r=\dfrac{A}{s}=2\).
Q18. Triangle sides \( (a, a, b)\). Show \(A=\dfrac{b}{4}\sqrt{4a^2-b^2}\).
Use Heron with \(s=\dfrac{2a+b}{2}\). Then \(A=\sqrt{s(s-a)^2(s-b)}=\sqrt{ \dfrac{2a+b}{2}\left(\dfrac{b}{2}\right)^2 \dfrac{2a-b}{2}}=\dfrac{b}{4}\sqrt{4a^2-b^2}\).
Q19. Show: If two sides are \(p,q\) and included angle \(90^\circ\), Heron gives \(A=\tfrac12 pq\).
Third side \(=\sqrt{p^2+q^2}\), \(s=\dfrac{p+q+\sqrt{p^2+q^2}}{2}\). Compute \(s(s-a)(s-b)(s-c)=\left(\dfrac{p+q+\dots}{2}\right)\left(\dfrac{-p+q+\dots}{2}\right)\left(\dfrac{p-q+\dots}{2}\right)\left(\dfrac{p+q-\dots}{2}\right)=\dfrac{p^2q^2}{16}\Rightarrow A=\tfrac12 pq.\)
Q20. A triangular tile \( (7,9,12)\) cm. How many tiles to cover \(1\ \text{m}^2\)?
\(s=14\Rightarrow A=\sqrt{14\cdot7\cdot5\cdot2}=\sqrt{980}=14\sqrt5\approx 31.305\ \text{cm}^2\). \(1\ \text{m}^2=10,000\ \text{cm}^2\Rightarrow\) tiles \(= \lceil 10000/31.305\rceil \approx 320\).
Exercise 10.1 — Textbook Questions & Perfect Solutions
| Question | Solution (with MathJax) |
|---|---|
| Q1. A traffic signal board is an equilateral triangle of side \(a\). Find its area by Heron’s formula. If perimeter \(=180\) cm, find the area. | For equilateral: \(s=\dfrac{3a}{2}\). Then \[ A=\sqrt{s(s-a)^3} =\sqrt{\tfrac{3a}{2}\Big(\tfrac{a}{2}\Big)^3} =\frac{\sqrt3}{4}a^2. \] If perimeter \(=180\Rightarrow a=60\) cm. Area \(=\dfrac{\sqrt3}{4}\cdot 60^2=900\sqrt3\ \text{cm}^2.\) |
| Q2. Triangular side wall with sides \(122\) m, \(22\) m, \(120\) m. The ad rent is ₹5,000 per m\(^2\) per year. A company hires it for 3 months. How much rent? | \(s=\dfrac{122+22+120}{2}=132\text{ m}\). \[ A=\sqrt{132(132-122)(132-22)(132-120)} =\sqrt{132\cdot10\cdot110\cdot12} =\sqrt{1320\cdot1320}=1320\ \text{m}^2. \] Rent for 3 months \(=1320\times 5000\times \dfrac{3}{12}=\boxed{\text{₹16,50,000}}.\) |
| Q3. A slide wall with sides \(15\) m, \(11\) m, \(6\) m. Find the painted area. | \(s=\dfrac{15+11+6}{2}=16\). \[ A=\sqrt{16(16-15)(16-11)(16-6)} =\sqrt{16\cdot1\cdot5\cdot10} =\sqrt{800}=20\sqrt2\ \text{m}^2. \] |
| Q4. Find area of a triangle with two sides \(18\) cm and \(10\) cm and perimeter \(42\) cm. | Third side \(=42-(18+10)=14\) cm; \(s=21\). \[ A=\sqrt{21(21-18)(21-10)(21-14)} =\sqrt{21\cdot3\cdot11\cdot7} =21\sqrt{11}\ \text{cm}^2. \] |
| Q5. Sides in ratio \(12:17:25\); perimeter \(540\) cm. Find area. | Factor \(=10\Rightarrow\) sides \(120,170,250\) cm; \(s=270\). \[ A=\sqrt{270\cdot(270-120)\cdot(270-170)\cdot(270-250)} =\sqrt{270\cdot150\cdot100\cdot20} =\sqrt{81{,}000{,}000}=9000\ \text{cm}^2. \] |
| Q6. An isosceles triangle has perimeter \(30\) cm and each equal side \(12\) cm. Find area. | Base \(=30-24=6\) cm; \(s=15\). \[ A=\sqrt{15(15-12)(15-12)(15-6)} =\sqrt{15\cdot3\cdot3\cdot9} =9\sqrt{15}\ \text{cm}^2. \] |
All set! 🎉
Questions are in red and solutions in green. Math is rendered with MathJax for perfect clarity on mobile.