Question

A side of a right angled triangle is
3 cm more than twice the
other side. If the area of
the triangle is 84sq cm then findthe hypotenuse of the triangle​

Answer 1

Answer:

Correction in the question: A side of a right angled triangle is 3 cm more than thrice the  other side. If the area of  the triangle is 84 cm² then find the hypotenuse of the triangle​.

Lets assume that one side of the triangle is 'x' cm. Then, according to the question, the other side of the triangle is '3x + 3' cm.

Also, it is given that the Area of the triangle is 84 cm².

This implies that,

$\text{Area of Triangle} = \dfrac{1}{2} \times x \times (3x+3) = 84\:cm^2$

$\rightarrow \dfrac{x(3x+3)}{2} = 84\:cm^2\\\\\rightarrow x(3x+3) = 84 \times 2 = 168\:cm^2 \\\\\rightarrow 3x^2 + 3x = 168\\\\\rightarrow 3x^2 + 3x - 168 = 0\\\\\rightarrow 3x^2 + 24x - 21x - 168 = 0\\\\\rightarrow 3x ( x + 8 ) - 21 ( x + 8 ) = 0\\\\\rightarrow ( 3x - 21 ) ( x + 8 ) = 0\\\\\rightarrow x = 7, -8$

Therefore one side is 8 cm. Then the other side is calculated as:

→ 3 ( 7 ) + 3

→ 21 + 3 = 24 cm

Therefore the two sides of the triangle are 7 cm and 24 cm.

Applying Pythagoras Theorem in the given right angled triangle we get:

→ 7² + 24² = y²

→ 49 + 576 = y²

→ 625 = y²

→ y = √625 = 25 cm

Therefore the hypotenuse of the given right angled triangle is 25 cm.

Answer 2

Answer:

$\huge{\blue{\boxed{\mathfrak\red{♡Answer\::25\:cm♡}}}}$

Correct question:-

A side of a right angled triangle is 3 cm more than the thrice the other side.If the area of the triangle is 84 sq.cm then find the hypotenuse of the triangle.

Solution:-

Let the other side be Y

So the one side = 3Y + 3

According to question now:-

$\Rightarrow\tt \frac{1}{2} \times Y \times (3Y+3) = 84$

$\Rightarrow\tt \frac{Y(3Y+3)}{2}=84$

$\Rightarrow\tt Y(3Y+3) = 168$

$\Rightarrow\tt {3Y}^{2}+3Y=168$

$\Rightarrow\tt {3Y}^{2} - 168 + 3Y = 0$

$\Rightarrow\tt {3Y}^{2} + 24Y - 21Y - 168 =0$

$\Rightarrow\tt 3Y ( Y + 8) - 21(Y + 8) = 0$

$\Rightarrow\tt (3Y - 21)(Y + 8) = 0$

$\Rightarrow\tt 3(Y - 7)(Y + 8) = 0$

$\Rightarrow\tt (Y - 7)(Y + 8) = 0 \\ \\ \Rightarrow\tt Y - 7 = 0 \: \: \: or\: Y + 8 = 0 \\ \\ \Rightarrow\tt Y = 7 \: \: \: or \:Y = - 8$

$\because \tt We\:will\:avoid\: negative \: values$

$\therefore\tt Y = 7 \:cm$

$\rule{300}{1.5}$

One side of triangle = 7 cm

So, other side =(3 × 7 + 3) cm

$\mapsto\tt Other\:side = (21 + 3) \:cm \\ \\ \mapsto\tt Other\: side = 24 \:cm$

According to theorem of Pythagoras

$\tt {7}^{2} + {24}^{2} = {h}^{2}$

$\Rightarrow\tt 49 + 576 = {h}^{2}$

$\Rightarrow\tt {h}^{2} = 625 \:cm$

$\Rightarrow\tt h =\sqrt{625}\:cm$

$\Rightarrow\tt h = 25 \: cm$

$\therefore\tt{\underline{Hypotenuse\:of\:triangle=25\:cm}}$