Free Heron's Formula 03 Practice Test - 9th Grade 

Semi perimeter of a triangle = $\frac{(a+b+c)}{2}$ $\Rightarrow$ $\frac{(3+5+6)}{2}=7m$

Area of a triangle = $\sqrt{s(s-a)(s-b)(s-c)}$ $\Rightarrow$  $\sqrt{7(7-3)(7-5)(7-6)}$

$\Rightarrow$  $\sqrt{7\times 4\times 2\times 1}$
$\Rightarrow$  $\sqrt{14\times 4}$

$\Rightarrow \sqrt{56}=2\sqrt{14}{m}^2$

Since,cost of painting $1m^2$ of wall is ₹5,

The cost of painting the whole wall $=5\times 2\sqrt{14}=₹10\sqrt{14}.$

Let the side of the rhombus = a

Area of the triangle containing the diagonal = $\frac{1}{2}$ times area of the rhombus

One of the diagonal = 6 cm

Semi perimeter = $\frac{6+a+a}{2}=a+3$

Now the area of the triangle containing the diagonal =$\sqrt{s(s-a)(s-b)(s-c)}$ 

$3\sqrt{7}$ = $\sqrt{(a+3)(a+3-a)(a+3-a)(a+3-6)}$
                     =  $\sqrt{(a+3)(a-3)\left(9\right)}$
                     = $\sqrt{(a^2-9)\left(9\right)}$
We take square on both sides and get,

63 = $(a^2-9)\times 9$
$\Rightarrow (a^2-9)$ = 7
$\Rightarrow a^2$ = 7+9 = 16

$\Rightarrow$ a= 4cm

Side of the rhombus = 6 cm

One of the diagonal = 8 cm

Semi perimeter of the triangle containing the diagonal = $\frac{(6+6+8)}{2}$ = 10 cm

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

= $\sqrt{10(10-6)(10-6)(10-8)}$

= $\sqrt{10\times 4\times 4\times 2}$
= $\sqrt{2\times 5\times 4\times 4\times 2}$
$\Rightarrow 8\sqrt{5}c{m}^2$

Area of rhombus = 2 (Area of the triangle containing the diagonal)

Area of the rhombus = $2\times 8\sqrt{5}$ = $16\sqrt{5}$${cm}^2$

Area of a rectangle = length $\times$ breadth

$\Rightarrow$ 96 = 12 $\times$ breadth

Breadth = 8 cm

Perimeter of the rectangle = 2(length + breadth) = 2(12+8) cm = 40 cm

Perimeter of the rectangle = Perimeter of the triangle

Therefore, semi perimeter of the triangle $=\frac{40}{2}=20cm$

In a right angled triangle, if the base and the height are given, the area can be calculated directly by using $\frac{1}{2}$ $\times$ base $\times$ height. If the hypotenuse and either the base or the height are given, then the third side can be found using Pythagoras' theorem. Then the area can be calculated using the same formula.

Let the sides of the triangle be 16x, 8x and 10x. 

Then, 16x + 8x + 10x = 340
$\Rightarrow$ 34x = 340
$\Rightarrow$ x = $\frac{340}{34}$
$\Rightarrow$ x =10.

Hence the sides are 160 cm, 80 cm and 100 cm.
Using the Heron's formula, Area A = $\sqrt{s(s-a)(s-b)(s-c)}$

Semi-Perimeter s = $\frac{\text{perimeter}}{2}$
$=\frac{340}{2}$
$=170cm$

$\therefore A=\sqrt{170(170-160)(170-100)(170-80)}$
$=\sqrt{170\times 10\times 70\times 90}$
$=100\sqrt{1071}c{m}^2$

The value of A is 1071

The area of an equilateral triangle is $\frac{\sqrt{3}}{4}{a}^2$ , where a = 2 cm.
Hence the area is $\frac{\sqrt{3}}{4}\times 2^2$
=$\sqrt{3}c{m}^2$
Thus the value of A = 3.

For $△BCD$,
$s=\frac{25+17+12}{2}=27$

Using Heron's formula,
$A=\sqrt{27(27-25)(27-17)(27-12)}$
$=\sqrt{27\times 2\times 10\times 15}$
$=\sqrt{3\times 3\times 3\times 2\times 2\times 5\times 3\times 5}$
$=3\times 3\times 2\times 5$
$=90c{m}^2$

The parallelogram can also be divided into 2 triangles by diagonal AC, as shown below, the area of each triangle being $=90c{m}^2$.

Area of parallelogram $=90\times 2=180c{m}^2$


In, $△ACD$
Area of $△\text{ACD}=\frac{1}{2}\times \text{base}\times \text{Altitude}$
$90=\frac{1}{2}\times 12\times \text{Altitude}$
$\Rightarrow \text{Altitude}=15cm$