Question

10 . ਉਸ ਸਮਕੋਣ ਤ੍ਰਿਭੁਜ ਦਾ ਖੇਤਰਫਲ ਪਤਾ ਕਰੋ
ਜਿਸ ਦੀਆਂ ਦੋ ਭੁਜਾਵਾਂ 24 ਸੈ.ਮੀ ਅਤੇ 7 ਸੈ.ਮੀ ਹਨ
ਉਸ ਦਾ ਪਰਿਮਾਪ 56 ਸੈ.ਮੀ ਹੈ। Find the area
of a right triangle whose two sides
are 24cm and 7cm and the perimeter
is 56cm. ਧਨ ਜਸਲੀ ਕਿ ਨਾ ਖੇਲਰ ਬਾਰ
ਲਵੇਂ ਕਿਸੀ ਵੀ ਥਾਂ 24 ਜੈਸੀ ਮੀਟ 7 ਸੈਸੀ ਸੈਂ
ਚਲੀ ਧਉਸਾਧ 56 ਸੈਸੀ ਫੈ।
( 4 ਹੈ ਪੀ2 cm2 ਜੈਸੀ
ਹੈ​

Answer 1

Given:

A right triangle whose two sides  are 24cm and 7cm and the perimeter

is 56cm

To find:

Area of the triangle

Solution:

Let the third side of the right-angled triangle be "x".

We know,

$\boxed{\bold{Perimeter\:of\:a \:triangle = sum\:of\:all\:3\:sides\:of\:the \:triangle}}$

But we have,

Perimeter = 56 cm

∴ 24 + 7 + x = 56

⇒ 31 + x = 56

⇒ x = 56 - 31

⇒ x = 25 cm ← the third side of the triangle

So, we can conclude that,

• The side measuring 25 cm will be the hypotenuse
• The other two sides with the measures 24 cm & 7 cm will be the base and height or vice versa.

Also, we know,

$\boxed{\bold{Area\:of\:a\:triangle\:=\:\frac{1}{2}\times base\times height }}$

Now, substituting the values of base and height in the formula above, we get

The area of the right-angled triangle is,

$=\:\frac{1}{2}\times base\times height }}$

= $\frac{1}{2}\times 24\times 7 }}$

= $12 \times 7$

= $\bold{84 \:cm^2}$

Thus, the area of the right-angled triangle is → 84 cm².

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Answer 2

$Let \: a , b \: and \: c \: are \: three$

$sides \:of \: a \: triangle$

$Given \:a = 24 \:cm , b = 7 \:cm \:and$

$perimeter (P) = 56 \:cm$

$\implies a + b + c = 56$

$\implies 24 + 7 + c = 56$

$\implies 31 + c = 56$

$\implies c = 56 - 31$

$\implies c = 25 \: cm$

$c \gt a \: and \: c \gt b$

$\blue{\therefore c \: is \: hypotenuse \: of \: the \: triangle }$

$Now, \red{Area \: of \: the \: triangle }$

$= \frac{1}{2} \times a \times b$

$= \frac{1}{2}\times 24 \times 7$

$= 12 \times 7$

$= 84 \: cm^{2}$

Therefore.,

$\red{ Area \: of \:the \: Triangle } \green { = 84 \:cm^{2} }$

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