Heron’s Formula - Class 9 Mathematics - Chapter 10 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Heron’s Formula | NCERT | Mathematics | Class 9
Cube root of the perimeter of a triangular field is $7$ $\mathrm{m}$ and its sides are in the ratio $14:15:20$. Find the area of the triangle.
The sides of the triangular field are given in the ratio 14:15:20, and we can denote the sides as $14x$, $15x$, and $20x$ respectively.
Given that the cube root of the perimeter is $7$ $\mathrm{m}$, the perimeter itself is: $$ (7)^3 = 343 \text{ m} $$
The sum of the sides, which is the perimeter, is represented by: $$ 14x + 15x + 20x = 343 $$ Adding the sides, we get: $$ 49x = 343 $$ From the above, solving for $x$ yields: $$ x = \frac{343}{49} = 7 $$
Thus, the sides of the triangle are:
First side $(a) = 14x = 14 \times 7 = 98 \text{ m}$
Second side $(b) = 15x = 15 \times 7 = 105 \text{ m}$
Third side $(c) = 20x = 20 \times 7 = 140 \text{ m}$
To find the area of the triangle, we use Heron's Formula, where: $$ s = \frac{a + b + c}{2} $$ Plugging the sides into the formula gives us the semi-perimeter $s$: $$ s = \frac{98 + 105 + 140}{2} = 171.5 \text{ m} $$
The area $A$ of the triangle using Heron's formula is: $$ A = \sqrt{s(s-a)(s-b)(s-c)} $$ Substituting the values we get: $$ A = \sqrt{171.5(171.5-98)(171.5-105)(171.5-140)} $$ Calculating further: $$ A = \sqrt{171.5 \times 73.5 \times 66.5 \times 31.5} \approx \sqrt{2016283} \text{ m}^2 $$
Thus, the area of the triangle is approximately: $$ A = 2016283 \text{ m}^2 $$ This calculation provides the required area based on the perimeter and given side ratios.
If 8, 15, and $c$ form a Pythagorean triplet, then the value of $c$ is
A) 14
B) 18
C) 17
D) $\mathbf{16}$
Solution:
The correct option is C) 17.
To find the value of $c$ such that 8, 15, and $c$ form a Pythagorean triplet, we can use a specific property of Pythagorean triplets.
The relation: $$ (2m)^2 + (m^2 -1)^2 = (m^2 + 1)^2 $$ indicates that $(2m), (m^2 - 1),$ and $(m^2 + 1)$ are the sides of a Pythagorean triplet.
Given $2m = 8$, we can solve for $m$: $$ m = \frac{8}{2} = 4 $$
Substituting $m = 4$ into the expressions:
-
The second side is: $$ m^2 - 1 = 16 - 1 = 15 $$
-
Calculating the value of $c$, which corresponds to $(m^2 + 1)$: $$ c = 16 + 1 = 17 $$
Therefore, when 8, 15, and $c$ form a Pythagorean triplet and $2m = 8$, the value of $c$ is 17.
87^2 - 13^2
To solve the expression $87^2 - 13^2$, we can employ the difference of squares formula, which is given by:
$$ a^2 - b^2 = (a+b)(a-b) $$
Here, $a = 87$ and $b = 13$. Plugging in these values, we get:
$$ 87^2 - 13^2 = (87 + 13)(87 - 13) $$
Now, compute the sum and difference:
$$ 87 + 13 = 100 \ 87 - 13 = 74 $$
Therefore, the expression simplifies to:
$$ 100 \times 74 = 7400 $$
Thus, the answer is 7400. This technique offers a quick and efficient way to compute differences of squares.
The value of $\prod_{n=3}^{24} \frac{n^{2}}{(n-1)^{2}}$ is
To find the value of the product $$ \prod_{n=3}^{24} \frac{n^2}{(n-1)^2}, $$ we can rewrite each term in the product to see a pattern, simplify it, and ultimately calculate the final value:
The product expands as follows: $$ \prod_{n=3}^{24} \frac{n^2}{(n-1)^2} = \frac{3^2}{(3-1)^2} \times \frac{4^2}{(4-1)^2} \times \cdots \times \frac{24^2}{(24-1)^2}. $$
Almost every term in the sequence has a factor in the numerator that cancels with a factor in the denominator of the adjacent term. Specifically,
- $\frac{3^2}{2^2}$ in the first term will have $3^2$ cancelling with the $3^2$ in the denominator of the next term,
- this pattern continues until the last term $\frac{24^2}{23^2}$.
After cancelling out all intermediate terms, what remains are the factors from the first and last expressions in the product: $$ \frac{24^2}{2^2}. $$ Calculating this gives: $$ \frac{576}{4} = 144. $$
Thus, the value of the product is $\boldsymbol{144}$.
What is Heron's formula?
Heron's formula is a well-known method used to determine the area of a triangle when the height is not provided, but the lengths of all three sides are known. This formula is applicable to any type of triangle as long as the side lengths are given.
Heron's Formula
$$ A = \sqrt{s(s - a)(s - b)(s - c)} $$
where $ s $ is the semi-perimeter of the triangle, calculated as:
$$ s = \frac{a + b + c}{2} $$
In this context:
( a, b, ) and ( c ) represent the lengths of the sides of the triangle.
( s ) stands for the semi-perimeter.
Using Heron's formula, one can efficiently find the area of the triangle without the need to know its height.
If $\Delta$ denotes the area of any triangle and $s$ its semi-perimeter, then
A $\quad \Delta < \frac{s^{2}}{2}$
B $\Delta > \frac{s^{2}}{4}$
C $\Delta < \frac{s^{2}}{4}$
D $\Delta > \frac{s^{2}}{2}$
The correct option is C: $$ \Delta < \frac{s^2}{4} $$
Let ( a, b, c ) be the sides of the triangle. Clearly, the semi-perimeter $ s $ is given by: $$ s = \frac{a + b + c}{2} $$
Thus, $ s $, $ s - a $, $ s - b $, and $ s - c $ will all be positive.
Applying the Arithmetic Mean-Geometric Mean (AM-GM) inequality for positive quantities: $$ AM \geq GM $$
Since $ s $, $ s - a $, $ s - b $, and $ s - c $ are not all equal, we have:
$$ \frac{s + (s - a) + (s - b) + (s - c)}{4} > [s(s - a)(s - b)(s - c)]^{1/4} $$
Simplifying the left-hand side, we obtain: $$ \frac{4s - (a + b + c)}{4} > [s(s - a)(s - b)(s - c)]^{1/4} $$
Given that $ a + b + c = 2s $, the expression further simplifies: $$ \frac{4s - 2s}{4} > [\Delta^2]^{1/4} $$
Which simplifies to: $$ \frac{2s}{4} > \Delta^{1/2} $$
Squaring both sides of the inequality, and bearing in mind that both $ s $ and $ \Delta $ are positive, we get: $$ \left( \frac{s}{2} \right)^2 > \Delta $$
Therefore: $$ \Delta < \frac{s^2}{4} $$
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