Question

a right angled triangle has the largest aide as 13 cm and one of the sides containing the right angle 12 cm its area in cm sq is​

Answer 1

Answer ::

The area of triangle is 30cm².

Step-by-step explanation ::

To Find :-

• The area of triangle

Solution :-

Given that,

• A right-angled triangle has the largest side as 13cm and one of the sides containing the right angle 12cm

Figure :-

• Refer the attachment.

Assumption :-

Let us assume the right-angled triangle as ∆ABC,

ACCORDING THE QUESTION,

To get the area, first we need to find out the measure of AB [ the base side of ∆ ],

By applying Pythagoras theorem to ∆ABC,

AB² + BC² = AC²,

➟ AB² + 12² = 13²

➟ AB² = 13² - 12²

➟ AB² = 169 - 144

➟ AB² = 25

➟ AB = √25

➟ AB = 5

The base side of ∆ABC is 5cm.

Now, the area of triangle is,

As we know that,

Area of ∆ = 1/2 × base × height,

➟ Area = 1/2 × 5 × 12

➟ Area = 1 × 5 × 6

➟ Area = 5 × 6

➟ Area = 30

The area of ∆ is 30cm².

Answer 2

$\bigstar \underline{ \sf Diagram} \bigstar$

• $\small \sf \:In \: the \: Attachment\:$

$\bigstar \underline{ \sf Given} \bigstar$

• $\small \sf Hypotenuse \: of \: the \: right \: angled \: traingle : \longrightarrow \boxed{ \sf 13 \: cm}$
• $\small \sf Perpendicular \: of \: the \: right \: angled \: traingle : \longrightarrow \boxed{ \sf 12 \: cm}$

$\bigstar \underline{ \sf To \: Find} \bigstar$

• $\small \sf Base \: of \: the \: right \: angled \: traingle : \longrightarrow \boxed{ \sf?}$
• $\small \sf Area \: \: of \: the \: right \: angled \: traingle : \longrightarrow \boxed{ \sf ?}$

$\bigstar \underline{ \sf Formula \: Used} \bigstar$

• $\small\bigstar {\underline {\boxed { \bf Hypotenuse {}^{2} \:= \: Base {}^{2} \: + \: Perpendicular {}^{2} }}} \bigstar$
• $\small\bigstar {\underline {\boxed { \bf Area \: of \: traingle\:= \: \frac{1}{2} \times base \times height\: }}} \bigstar$

$\bigstar \underline{ \sf Solution} \bigstar$

$\small \sf We \: know,$

$\small\bigstar {\underline {\boxed { \bf Hypotenuse {}^{2} \:= \: Base {}^{2} \: + \: Perpendicular {}^{2} }}} \bigstar$

$\small \sf : \implies \: AC {}^{2} =AB {}^{2} + BC {}^{2}$

$\small \sf : \implies \: (13) {}^{2} =AB {}^{2} + (12) {}^{2}$

$\small \sf : \implies \: 169 =AB {}^{2} + 144$

$\small \sf : \implies \: AB {}^{2} + 144 = 169$

$\small \sf : \implies \: AB {}^{2} =169 - 144$

$\small \sf : \implies \: AB {}^{2} =25$

$\small \sf : \implies \: AB = \sqrt{25}$

$\small \sf : \implies \: AB = 5$

$\small \sf \: \therefore \: Hence, the \: base \: of \: the \: right \: angled \: traingle \: is \: \underline{ 5 \: cm}$

$\small \sf We \: also \: know,$

$\small\bigstar {\underline {\boxed { \bf Area \: of \: traingle\:= \: \frac{1}{2} \times base \times height\: }}} \bigstar$

$\small \sf : \implies Area \: of\: the \: traingle = \frac{1}{ \cancel2} \times 5 \times \cancel{ 12} \:$

$\small \sf : \implies Area \: of\: the \: traingle = 5 \times 6\:$

$\small \sf : \implies Area \: of\: the \: traingle = 30 \: {cm}^{2} \:$

$\small \sf \therefore \: Hence,we \: got \: the \: area \: of \: the \: right \: angled \: traingle \: as \: \underline{30 \: cm {}^{2} }$

$\bigstar{ \underline{ \underline \pink{ \sf★@iTzShInNy☆}}} \bigstar$