Mix Examples - Heron’s Formula Questions in English

(D) In Heron's formula,the area of a triangle with sides $a$,$b$,and $c$ is given by $\sqrt{s(s-a)(s-b)(s-c)}$.
Here,$s$ is the semiperimeter of the triangle,which is calculated as $s = \frac{a+b+c}{2}$.

(A) Heron's formula is used to calculate the area of a triangle when the lengths of all three sides are known.
Let the sides of the triangle be $a$,$b$,and $c$.
First,calculate the semi-perimeter $s$ of the triangle,which is given by $s = \frac{a+b+c}{2}$.
Then,the area of the triangle is given by the formula $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$.

(B) The area of a triangle is given by the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
In a right-angled triangle,the two sides that form the right angle act as the base and the height.
Therefore,the area of a right triangle is $\frac{1}{2} \times \text{product of the sides forming the right angle}$.

(C) The area of a trapezium is calculated by the formula: $\text{Area} = \frac{1}{2} \times (a + b) \times h$,where $a$ and $b$ are the lengths of the parallel sides and $h$ is the perpendicular distance (height) between them.
Therefore,the correct formula is $\frac{1}{2} \times \text{sum of parallel sides} \times \text{distance between parallel sides}$.

(D) The area of an equilateral triangle with side $a$ is given by the formula: $Area = \frac{\sqrt{3}}{4} a^2$.
For a triangle with side $a_1 = 16\,cm$,the area $A_1 = \frac{\sqrt{3}}{4} (16)^2 = \frac{\sqrt{3}}{4} \times 256$.
For a triangle with side $a_2 = 8\,cm$,the area $A_2 = \frac{\sqrt{3}}{4} (8)^2 = \frac{\sqrt{3}}{4} \times 64$.
To find how many times $A_1$ is of $A_2$,we calculate the ratio: $\frac{A_1}{A_2} = \frac{\frac{\sqrt{3}}{4} \times 256}{\frac{\sqrt{3}}{4} \times 64} = \frac{256}{64} = 4$.
Therefore,the area of the triangle with side $16\,cm$ is $4$ times the area of the triangle with side $8\,cm$.

(A) The side length of the equilateral triangle is $a = 10\,cm$.
The perimeter of the triangle is $P = a + a + a = 3a$.
$P = 3 \times 10 = 30\,cm$.
The semi-perimeter $s$ is defined as half of the perimeter:
$s = \frac{P}{2} = \frac{30}{2} = 15\,cm$.

(B) First,observe that the side lengths $12 \, cm$,$16 \, cm$,and $20 \, cm$ form a Pythagorean triple because $12^2 + 16^2 = 144 + 256 = 400 = 20^2$.
Since the square of the longest side equals the sum of the squares of the other two sides,this is a right-angled triangle.
The area of a right-angled triangle is given by the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Taking the base as $12 \, cm$ and the height as $16 \, cm$,we get: $\text{Area} = \frac{1}{2} \times 12 \times 16 = 6 \times 16 = 96 \, cm^2$.

(C) Given sides of the triangle are $a = 13 \, cm$, $b = 14 \, cm$, and $c = 15 \, cm$.
First, calculate the semi-perimeter $(s)$:
$s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21 \, cm$.
Now, use Heron's formula for the area of the triangle:
$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$
$\text{Area} = \sqrt{21(21 - 13)(21 - 14)(21 - 15)}$
$\text{Area} = \sqrt{21 \times 8 \times 7 \times 6}$
$\text{Area} = \sqrt{(3 \times 7) \times (2^3) \times 7 \times (2 \times 3)}$
$\text{Area} = \sqrt{2^4 \times 3^2 \times 7^2}$
$\text{Area} = 2^2 \times 3 \times 7 = 4 \times 21 = 84 \, cm^2$.

(D) The area of a rhombus can be calculated using the formula: $\text{Area} = \frac{1}{2} \times d_1 \times d_2$,where $d_1$ and $d_2$ are the lengths of the diagonals.
Given: $d_1 = 40 \, cm$ and $d_2 = 42 \, cm$.
Substituting the values into the formula:
$\text{Area} = \frac{1}{2} \times 40 \times 42$
$\text{Area} = 20 \times 42 = 840 \, cm^2$.
Thus,the correct option is $D$.

(A) Let the diagonals of the rhombus be $d_1 = 16\,cm$ and $d_2 = 30\,cm$.
In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the side of the rhombus be $a$. The diagonals divide the rhombus into four congruent right-angled triangles.
The legs of these triangles are half the lengths of the diagonals: $d_1/2 = 8\,cm$ and $d_2/2 = 15\,cm$.
Using the Pythagorean theorem: $a^2 = (8)^2 + (15)^2$.
$a^2 = 64 + 225 = 289$.
$a = \sqrt{289} = 17\,cm$.
The perimeter of a rhombus is $4 \times \text{side} = 4 \times 17 = 68\,cm$.

(B) First,check if the triangle is a right-angled triangle by verifying the Pythagorean theorem: $AB^2 + BC^2 = 20^2 + 48^2 = 400 + 2304 = 2704$.
Since $AC^2 = 52^2 = 2704$,we have $AB^2 + BC^2 = AC^2$.
Thus,$\Delta ABC$ is a right-angled triangle with base $BC = 48 \, cm$ and height $AB = 20 \, cm$.
The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 48 \times 20 = 24 \times 20 = 480 \, cm^2$.

(C) First,check if the triangle is a right-angled triangle by verifying the Pythagorean theorem: $12^2 + 35^2 = 144 + 1225 = 1369$. Since $37^2 = 1369$,the triangle is a right-angled triangle.
For a right-angled triangle,the area is given by $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Taking base as $12 \ cm$ and height as $35 \ cm$,we get:
$\text{Area} = \frac{1}{2} \times 12 \times 35 = 6 \times 35 = 210 \ cm^2$.

(D) The semi-perimeter $s$ of a triangle is given by the formula:
$s = \frac{a + b + c}{2}$
Given values are $a = 15 \, cm$,$b = 20 \, cm$,and $s = 31 \, cm$.
Substituting these values into the formula:
$31 = \frac{15 + 20 + c}{2}$
$31 \times 2 = 35 + c$
$62 = 35 + c$
$c = 62 - 35$
$c = 27 \, cm$
Therefore,the value of $c$ is $27 \, cm$.

(A) The sides of the triangle are $a = 20 \, cm$,$b = 24 \, cm$,and $c = 32 \, cm$.
The semi-perimeter $s$ of a triangle is given by the formula:
$s = \frac{a + b + c}{2}$
Substituting the given values:
$s = \frac{20 + 24 + 32}{2}$
$s = \frac{76}{2}$
$s = 38 \, cm$
Therefore,the semi-perimeter of the triangle is $38 \, cm$.