Free Heron's Formula 01 Practice Test - 9th Grade 

Given, $a=35\text{cm};b=54\text{cm};c=61\text{cm}.$

Semi-perimeter
$s=\frac{a+b+c}{2}=\frac{35+54+61}{2}=75\text{cm}$

$A\left(△\right)=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{75(75-35)(75-54)(75-61)}=\sqrt{75\times 40\times 21\times 14}$
$A\left(△\right)=420\sqrt{5}{\text{cm}}^2.$

Now, $A\left(△\right)=\frac{1}{2}\times \text{base}\times \text{altitude}$

The altitude corresponding to shortest base is longest.
The shortest side = 35 cm.

$\therefore 420\sqrt{5}=\frac{1}{2}\times 35\times \text{altitude}\text{altitude}=\frac{2\times 420\sqrt{5}}{35}=24\sqrt{5}\text{cm}.$

The given statement is false. Heron's formula can be used in finding the area of quadrilateral as quadrilateral is formed by two triangles.

Let each of its equal sides be x.
Then, $2x+8=18$ $\Rightarrow x=5cm$
Hence, the sides are $\text{5 cm, 5 cm and 8 cm}$.
Using the Heron's formula,
$Area\left(A\right)=\sqrt{s(s-a)(s-b)(s-c)}$
where s = semi- perimeter and a,b,c are the three sides of the triangle.
$Semi-Perimeter\left(s\right)=\frac{\text{perimeter}}{2}$
$=\frac{18}{2}$ $=9cm$
$A=\sqrt{9(9-5)(9-5)(9-8)}$
$=\sqrt{9\times 4\times 4\times 1}$
$=\sqrt{144}c{m}^2$
$=12c{m}^2$

Area of square = $Sid{e}^2$
$\Rightarrow$ 36 = $Sid{e}^2$
$\Rightarrow$ Side = 6 cm

Side of triangle = 6 cm

Semi perimeter of the triangle s $=\frac{6+6+6}{2}cm=9cm$

Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{9(9-6{)}^3}$
$=\sqrt{243}$
$=9\sqrt{3}cm$

$S=\frac{Perimeter}{2}=\frac{(a+b+c)}{2}=\frac{(4+6+6)}{2}=8cm$

Area of the triangle =$\sqrt{s(s-a)(s-b)(s-c)}$ = $\sqrt{8(8-4)(8-6)(8-6)}$ = $\sqrt{128}=8\sqrt{2}{cm}^2$

Area of a triangle is also find using $\frac{1}{2}\times base\times height$

 $\Rightarrow 8\sqrt{2}=\frac{1}{2}\times 4\times height$

$\Rightarrow height=4\sqrt{2}cm$.

Let the side of the equilateral triangle = a cm.

Semi Perimeter, s = $\frac{3a}{2}$

$s-a=\frac{3a}{2}-a=\frac{3a-2a}{2}=\frac{a}{2}$

Now by heron’s formula

Area of the triangle= $\sqrt{s(s-a)(s-b)(s-c)}$ 

$\Rightarrow 9\sqrt{3}=\sqrt{\frac{3a}{2}\times \frac{a}{2}\times \frac{a}{2}\times \frac{a}{2}}$

$\Rightarrow \sqrt{243}=\sqrt{\frac{3a^4}{16}}$

Squaring both sides,
$\Rightarrow 243=\frac{3a^4}{16}$

$\Rightarrow a^4=81\times 16$

$\Rightarrow a=6cm$

Since ABCD is a parallelogram, its opposite sides will be equal.

AB = 3 cm, BC = 4 cm and AC = 5 cm

Now in triangle ABC,

Area of the triangle using heron’s formula,

$s=\frac{Perimeter}{2}=\frac{3+4+5}{2}=6cm$

Area = $\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{6(6-3)(6-4)(6-5)}$
= $\sqrt{36}$
= $6c{m}^2$

Area of the parallelogram = 2 (Area of triangle ABC)

$\Rightarrow \text{The area of the parallelogram}=2\times 6=12c{m}^2$

Using formula $\frac{1}{2}\times \text{base}\times \text{height}$ we can find the area of any triangle irrespective of its type if we know the base and the height of the given triangle.