Free Heron's Formula 02 Practice Test - 9th Grade 

$S=\frac{Perimeter}{2}$ $=\frac{(2+5+5)}{2}=6m$

Area of the triangle= $\sqrt{s(s-a)(s-b)(s-c)}$
 $=\sqrt{6(6-2)(6-5)(6-5)}$
 $=\sqrt{6\times 4\times 1\times 1}$
 $=2\sqrt{6}$$m^2$

Since there are 20 rocks of 4 colours, number of rocks of each colour $=\frac{20}{4}=5$
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Therefore the area covered by each type of rock will be $=2\sqrt{6}\times 5$ $=10\sqrt{6}$$m^2$

Third side = perimeter - (sum of other two sides) = 38 - 20 = 18 cm
Semi perimeter of a triangle = $\frac{\text{Perimeter}}{2}=\frac{38}{2}=19\text{cm}$

According to heron's formula,
Area of a triangle $=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{19(19-12)(19-8)(19-18)}=\sqrt{19\times 7\times 11\times 1}=\sqrt{1463}{\text{cm}}^2$

$\text{Area of a triangle}=\frac{1}{2}\times base\times height=40$

$\Rightarrow base=\frac{40\times 2}{height}$

$\Rightarrow base=\frac{80}{10}=8cm$
 

Perimeter = a +b +c (i.e. the sum of length of all the sides)
$\Rightarrow$ Third side = 26 - 8 - 6
                                          = 26 - (8+6)
                                          = 26 - 14
                                          =12

Third side = 12 cm

Area of triangle = $\frac{1}{2}\times base\times height$

The hypotenuse is the longest side of a right-angled triangle. Hence, the base and the height are 6 cm and 8 cm.
$\therefore \text{area}=\frac{1}{2}\times 6\times 8$
$=24c{m}^2$

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Let the triangle be ABC where AC = 13 m and BC = 5 m

Using Pythagoras theorem we can write,

$A{B}^2=A{C}^2-B{C}^2$

              
Therefore, $A{B}^2={13}^2–5^2\Rightarrow AB=12m$
Area of triangle = $\frac{1}{2}\times base\times height=\frac{1}{2}\times 5\times 12=30m^2$

$a=6cm,b=8cm,c=10\text{cm}$
$s=\frac{a+b+c}{2}$
$s=\frac{6+8+10}{2}$
$=12cm$
$A=\sqrt{s(s-a)(s-b)(s-c)}$
$A=\sqrt{12(12-6)(12-8)(12-10)}$
$=24{\text{cm}}^2$
The cost of painting it $=24\times 25=600$ paise
$=\text{Rs.}6$