Question

CLASS 9 MATHEMATICS [ HERON’S FORMULA ]
Q] The perimeter of a right triangle is 24 cm. If its hypotenuse is 10 cm, find the (5) other two sides. Find its area by using the formula area of a right triangle. Verify
your result by using Heron’s formula.
PLEASE REPLY SOON. I NEED IT URGENTLY

Answer 1

$\textbf{Solution:}$

Let the sides of right ∆ be 'a' cm and 'b' cm.

Then,

a + b + c = 24

=> a + b + 10 = 24

=> a + b = 24 - 10

=> a + b = 14.............(1)

Also, $a {}^{2} + b {}^{2} = (10) {}^{2} \\ \\ a {}^{2} + b {}^{2} = 100$

=> $a {}^{2} + b {}^{2} = 100$.......(2)

We know that

$(a + b) {}^{2} = a {}^{2} + 2ab + b {}^{2} \\ \\ (14) {}^{2} = 100 + 2ab \\ \\ - 2ab = 100 - 196 \\ \\ - 2ab = - 96 \\ \\ ab = \frac{96}{2} = 48$
=> ab = 48 .........(3)

Also,

$(a - b) {}^{2} = a {}^{2} - 2ab + b {}^{2} \\ \\ (a - b) {}^{2} = 100 - 2 \times 48 \\ \\ (a - b) {}^{2} = 100 - 96\\ \\ (a - b) {}^{2} = 4 \\ \\ (a - b) = \sqrt{4} = 2$
=> (a - b) = 2 ......(4)

Solving (1) and (4) we get;

•°• a = 8 and b = 6

Now,

s = $\frac{a + b + c}{2} \\ \\ \frac{24}{2} = 12$

Area of ∆ =$\sqrt{s(s - a)(s - b)(s - c)}$

=> $\sqrt{12(12 - 8)(12 - 6)(12 - 10)}$

=> $\sqrt{12 \times 4 \times 6 \times 2}$

=>$\sqrt{2 \times 2 \times 3 \times 2 \times 2 \times 2 \times 3 \times 2}$

=>$\sqrt{2 {}^{2} \times 2 {}^{2} \times 2 {}^{2} \times 3 {}^{2} }$

=>$2 \times 2 \times 2 \times 3$

=>${ 24}$

Hence,

The area of ∆ is 24 cm.

_______________

$\textbf{Thank you!!}$

Answer 2

$\bf \large \it{Hey \: User!!! }$

given the perimeter of the right angle triangle is 24cm and it's hypotenuse is 10cm.

let the sides of the right angle triangle be a, b and c.

therefore a + b + c = 24cm
>> a + b + 10 = 24cm
>> a + b = 24 - 10
>> a + b = 14

therefore a = 14 - b --------(i)

we know that the sum of square of the two sides is the square of the hypotenuse of the right angle triangle.

>> a² + b² = c²
>> a² + b² = 10²
>> a² + b² = 100 --------(ii)

substitute (i) in (ii)

>> ( 14 - b )² + (b)² = 100
>> ( 14² - 28b + b² ) + b² = 100
>> 196 + 2b - 28y = 100
>> 2b² - 28b = 100 - 196
>> 2b² - 28b = -96
>> 2b² - 28b + 96 = 0

>> 2 ( b² - 14b + 48 ) = 0
>> b² - 28b + 48 = 0

by splitting method :-

>> b² - ( 8b + 6b ) + 48 = 0
>> b² - 8b - 6b + 48 = 0
>> b ( b - 8 ) - 6 ( b - 8 ) = 0
>> ( b - 8 ) ( b - 6 )

( b - 8 ) = 0
>> b = 8

( b - 6 ) = 0
>> b = 6

hence, the other two sides of the triangle is 8cm and 6cm.

let us consider 8cm as "a" and 6cm as "b"

area of the right angle triangle = 1/2 × b × h
= 1/2 × 6 × 8
= 3 × 8
= 24cm²

now, by heron's formula :-

s (semi-perimeter) of the triangle = 24/2 = 12cm

$\bf \small { = \sqrt{s(s - a)(s - b)(s - c)} }$
$\bf \small{ = \sqrt{12(12 - 8)(12 - 6)(12 - 10)} }$
$\bf \small{ = \sqrt{12 \times 4 \times 6 \times 2} }$
$\bf \small{ = \sqrt{576} }$
$\bf \small{ = {24cm}^{2} }$

HENCE PROVED...

$\bf \large \it{Cheers!!!}$