Tag: Exercise 10.1 Chapter 10 Herone’s law Class 9th Maths Solutions

Heron’s formula throughout:

Heron’s formula: for sides $a,b,c$ and $s={\displaystyle \frac{a+b+c}{2}}$

$$\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}\mathrm{.}$$

---

1. Equilateral triangle with side $a$.

For an equilateral triangle $a=b=c$
$s={\displaystyle \frac{3a}{2}}$. So

$$\text{Area}=\sqrt{\frac{3a}{2}(\frac{3a}{2}-a{)}^3}=\sqrt{\frac{3a}{2}(\frac{a}{2}{)}^3}=\frac{\sqrt{3}}{4}{a}^2\mathrm{}$$

If perimeter $=180$, each side $=180\mathrm{/}3=60$cm.
Area $={\displaystyle \frac{\sqrt{3}}{4}}\times {60}^2=900\sqrt{3}{\text{cm}}^2\approx 1558.849{\text{cm}}^2\mathrm{}$

---

2. Triangle with sides $122\text{ m},22\text{ m},120\text{ m}$. Rent at ₹5000 per m${}^2$ per year; hired for 3 months.

Compute $s={\displaystyle \frac{122+22+120}{2}}={\displaystyle \frac{264}{2}}=132$

$s-122=10,s-22=110,s-120=12$

Area $=\sqrt{132\cdot 10\cdot 110\cdot 12}$

Note $132\cdot 10=1320$ and $110\cdot 12=1320$, product $={1320}^2$.

So area $=1320{\text{m}}^2$

Annual earnings per months = $\frac{3}{12}=\frac{1}{\mathrm{4 }}$year.

Rent $=1320\times 5000\times \frac{1}{4}=1320\times 1250=\text{₹}1,650,000$

---

3. Triangle with sides $15\text{ m},11\text{ m},6\text{ m}$

$s={\displaystyle \frac{15+11+6}{2}}={\displaystyle \frac{32}{2}}=16$
$s-15=1,s-11=5,s-6=10$

Area $=\sqrt{16\cdot 1\cdot 5\cdot 10}=\sqrt{800}=20\sqrt{2}{\text{m}}^2\approx 28.284{\text{m}}^2\mathrm{}$

---

4. Triangle with two sides $18$ cm and $10$ cm, perimeter $42$ cm.

Third side $=42-(18+10)=14$ cm. So sides $18,10,14$

$s={\displaystyle \frac{18+10+14}{2}}={\displaystyle \frac{42}{2}}=21$
$s-18=3,s-10=11,s-14=7$

Area $=\sqrt{21\cdot 3\cdot 11\cdot 7}=\sqrt{4851}=21\sqrt{11}{\text{cm}}^2\approx 69.649{\text{cm}}^2\mathrm{}$

---

5. Sides in ratio $12:17:25$, perimeter $540$ cm.

Sum of ratios $=12+17+25=54$. So scale $x=540\mathrm{/}54=10$.

Sides $=120,170,250$

$s={\displaystyle \frac{120+170+250}{2}}={\displaystyle \frac{540}{2}}=270$
$s-120=150,s-170=100,s-250=20$

Area $=\sqrt{270\cdot 150\cdot 100\cdot 20}\mathrm{}$
Compute $270\cdot 150=40500,100\cdot 20=2000,$ product $=40500\times 2000=81,000,000$
$\sqrt{\mathrm{81,000,000}}=9000$

So area $=9000{\text{cm}}^2\mathrm{}$

---

6. Isosceles triangle: perimeter $30$ cm, equal sides $=12$ cm each.

Base $=30-2\times 12=30-24=6$

Alternatively $a=b=12,c=6$

$s={\displaystyle \frac{12+12+6}{2}}=15$
$s-12=3,s-12=3,s-6=9$

Area $=\sqrt{15\cdot 3\cdot 3\cdot 9}=\sqrt{1215}=9\sqrt{15}{\text{cm}}^2\approx 34.857{\text{cm}}^2\mathrm{}$

(Checks with height method: height $=\sqrt{{12}^2-3^2}=\sqrt{135}=3\sqrt{15}$, area $=\frac{1}{2}\times 6\times 3\sqrt{15}=9\sqrt{15}$)