Question

In a right angle triangle, the length of one side is 4.5 cm and the length of the hypotenuse is 20.5 cm. Find the area of the right triangle.​

Answer 1

Step-by-step explanation:

Let, Δ ABC right angle at B.

AB = 4.5 cm

Hypotenuse (AC) = 20.5 cm

BC = x

So,

Area = 1/2 × base × height

= 1/2 × x × 4.5

Therefore, First we have to find the base.

»» AC² = AB² + BC²

BC² = AC² - AB²

x² = (20.5)² - (4.5)²

x² = 420.25 - 20.25

x² = 400 cm

x = √400

x = 20 cm

So, base is 20 cm.

Formula:-

Area = 1/2 × base × height

= 1/2 × 20 × 4.5

= 45 cm²

Hope it helps you.

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

In a right angle triangle, the length of one side is 4.5 cm and the length of the hypotenuse is 20.5 cm.

Let assume that the required triangle be ABC, right-angled at A such that, AB = 4.5 cm and BC = 20.5 cm

Now, In right angle triangle ABC,

Using Pythagoras Theorem, we have

$\rm \: {BC}^{2} = {AB}^{2} + {AC}^{2}$

On substituting the values, we get

$\rm \: {20.5}^{2} = {4.5}^{2} + {AC}^{2}$

$\rm \: {20.5}^{2} - {4.5}^{2} = {AC}^{2}$

$\rm \: {AC}^{2} = (20.5 + 4.5)(20.5 - 4.5) \\ \: \{ \because \: {x}^{2} - {y}^{2} = (x + y)(x - y) \}$

$\rm \: {AC}^{2} = 25 \times 16$

$\rm \: {AC}^{2} = {5}^{2} \times {4}^{2}$

$\rm\implies \:AC \: = \: 20 \: cm$

[ as AC can never be negative ]

Now,

$\rm \: Area_{(\triangle\: ABC)} \: = \: \dfrac{1}{2} \times AB \times AC$

On substituting the value of AB and AC we get

$\rm \: Area_{(\triangle\: ABC)} \: = \: \dfrac{1}{2} \times 4.5 \times 20$

$\rm \: Area_{(\triangle\: ABC)} \: = \: 4.5 \times 10$

$\rm\implies \:Area_{(\triangle\: ABC)} = 45 \: {cm}^{2}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$