Exercise 11.1: Heron's Formula — NCERT Solutions

The sides are $a=8$ cm, $b=11$ cm, $c=13$ cm.

Step 1 — Semi-perimeter: $s=\frac{a+b+c}{2}=\frac{8+11+13}{2}=\frac{32}{2}=16\text{ cm}$

Step 2 — Heron's formula: $\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{16(16-8)(16-11)(16-13)}$ $=\sqrt{16\times 8\times 5\times 3}$ $=\sqrt{1920}=\sqrt{64\times 30}=8\sqrt{30}\text{ cm}^2$

So the area is $8\sqrt{30}\approx 43.8$ cm$^2$.

Step 1 — Find the sides. Let the sides be $3x$, $5x$, $7x$. $3x+5x+7x=300\Rightarrow 15x=300\Rightarrow x=20$ So the sides are $60$ m, $100$ m, $140$ m.

Step 2 — Semi-perimeter: $s=\frac{300}{2}=150\text{ m}$

Step 3 — Heron's formula: $\text{Area}=\sqrt{150(150-60)(150-100)(150-140)}$ $=\sqrt{150\times 90\times 50\times 10}$ $=\sqrt{6750000}$ $=1500\sqrt{3}\text{ m}^2\approx 2598\text{ m}^2$

Given: each side $a=b=c=12$ cm.

Step 1 — Semi-perimeter: $s=\frac{12+12+12}{2}=18\text{ cm}$

Step 2 — Heron's formula: $\text{Area}=\sqrt{18(18-12)(18-12)(18-12)}=\sqrt{18\times 6\times 6\times 6}$ $=\sqrt{3888}=36\sqrt{3}\text{ cm}^2$

Check with $\dfrac{\sqrt3}{4}a^2=\dfrac{\sqrt3}{4}\times 144=36\sqrt3$ cm$^2$. ✓

So the area is $36\sqrt{3}\approx 62.4$ cm$^2$.

Step 1 — Find the base. Equal sides $=10$ cm each, perimeter $=32$ cm. $\text{base}=32-(10+10)=12\text{ cm}$ Sides: $10$, $10$, $12$ cm.

Step 2 — Semi-perimeter: $s=\frac{32}{2}=16\text{ cm}$

Step 3 — Heron's formula: $\text{Area}=\sqrt{16(16-10)(16-10)(16-12)}$ $=\sqrt{16\times 6\times 6\times 4}=\sqrt{2304}=48\text{ cm}^2$

Step 1 — Semi-perimeter: $s=\frac{7+24+25}{2}=\frac{56}{2}=28\text{ cm}$

Step 2 — Heron's formula: $\text{Area}=\sqrt{28(28-7)(28-24)(28-25)}$ $=\sqrt{28\times 21\times 4\times 3}=\sqrt{7056}=84\text{ cm}^2$

Verification: Since $7^2+24^2=49+576=625=25^2$, the triangle is right-angled with legs $7$ and $24$. $\text{Area}=\tfrac12\times 7\times 24=84\text{ cm}^2\ \checkmark$

Step 1 — Find one side. Perimeter $=180$ cm, so each side $=\dfrac{180}{3}=60$ cm.

Step 2 — Semi-perimeter: $s=\frac{180}{2}=90\text{ cm}$

Step 3 — Heron's formula: $\text{Area}=\sqrt{90(90-60)(90-60)(90-60)}=\sqrt{90\times 30\times 30\times 30}$ $=\sqrt{2430000}=900\sqrt{3}\text{ cm}^2\approx 1558.8\text{ cm}^2$

The diagonal $AC=41$ m splits $ABCD$ into $\triangle ABC$ and $\triangle ACD$.

Triangle $ABC$: sides $9$, $40$, $41$. Since $9^2+40^2=81+1600=1681=41^2$, it is right-angled at $B$. $\text{Area}_{ABC}=\tfrac12\times 9\times 40=180\text{ m}^2$

Triangle $ACD$: sides $AC=41$, $CD=28$, $DA=15$. $s=\frac{41+28+15}{2}=42\text{ m}$ $\text{Area}_{ACD}=\sqrt{42(42-41)(42-28)(42-15)}$ $=\sqrt{42\times 1\times 14\times 27}=\sqrt{15876}=126\text{ m}^2$

Total area: $180+126=306\text{ m}^2$

Step 1 — Split by the diagonal. The diagonal $48$ m divides the rhombus into two congruent triangles, each with sides $30$, $30$, $48$.

Step 2 — Area of one triangle: $s=\frac{30+30+48}{2}=54\text{ m}$ $\text{Area}=\sqrt{54(54-30)(54-30)(54-48)}=\sqrt{54\times 24\times 24\times 6}$ $=\sqrt{186624}=432\text{ m}^2$

Step 3 — Total area: $2\times 432=864\text{ m}^2$

Step 4 — Other diagonal. Area of rhombus $=\tfrac12 d_1 d_2$: $864=\tfrac12\times 48\times d_2\Rightarrow d_2=\frac{864\times 2}{48}=36\text{ m}$

Step 1 — Semi-perimeter: $s=\frac{50+80+50}{2}=\frac{180}{2}=90\text{ m}$

Step 2 — Area (Heron's formula): $\text{Area}=\sqrt{90(90-50)(90-80)(90-50)}$ $=\sqrt{90\times 40\times 10\times 40}=\sqrt{1440000}=1200\text{ m}^2$

Step 3 — Cost: $1200\times 7=\textrm{₹}\,8400$

Step 1 — Third side. Perimeter $=144$ m. $\text{third side}=144-(48+52)=44\text{ m}$ Sides: $48$, $52$, $44$ m.

Step 2 — Semi-perimeter: $s=\frac{144}{2}=72\text{ m}$

Step 3 — Area (Heron's formula): $\text{Area}=\sqrt{72(72-48)(72-52)(72-44)}$ $=\sqrt{72\times 24\times 20\times 28}=\sqrt{967680}\approx 983.7\text{ m}^2$

Step 4 — Altitude on the longest side ($52$ m): $\text{Area}=\tfrac12\times 52\times h\Rightarrow h=\frac{2\times 983.7}{52}\approx 37.8\text{ m}$

Setup. Let $ABCD$ be the trapezium with $AB=25$ m, $CD=10$ m ($AB\parallel CD$), $AD=13$ m, $BC=14$ m. Draw $CE\parallel AD$ meeting $AB$ at $E$.

Then $AECD$ is a parallelogram, so $CE=AD=13$ m and $AE=CD=10$ m. $EB=AB-AE=25-10=15\text{ m}$

Triangle $EBC$ has sides $CE=13$, $BC=14$, $EB=15$. $s=\frac{13+14+15}{2}=21\text{ m}$ $\text{Area}_{EBC}=\sqrt{21(21-13)(21-14)(21-15)}=\sqrt{21\times 8\times 7\times 6}=\sqrt{7056}=84\text{ m}^2$

Height of trapezium. Using base $EB=15$: $84=\tfrac12\times 15\times h\Rightarrow h=\frac{168}{15}=11.2\text{ m}$

Area of trapezium: $=\tfrac12(AB+CD)\times h=\tfrac12(25+10)\times 11.2=\tfrac12\times 35\times 11.2=196\text{ m}^2$