Question

The hypotenuse of a right angled triangle is 26m and the sum of the other two sides is e4m.Find the leght of its sides.
It a question of Quadratic equation​

Answer 1

Answer:

Correct question:-

The hypotenuse of right-angled triangle is 26 m and the sum of other two sides is 34 m.Find the length of other two sides.

Solution :-

Given :-

• Hypotenuse of right angled triangle = 26 m

• Sum of other 2 sides = 34 m

To find:-

• Length of other 2 sides = ?

As the triangle is right-angled,so we have to apply Pythagoras theorem.

Now,sum of other 2 sides = 34 m

Let base be x m.

$\therefore$ Altitude = (34 - x) m

According to Pythagoras theorem,

$\dashrightarrow\sf (Hypotenuse)^2 = (Base)^2 + (Altitude)^2 \\\\\dashrightarrow\sf (26)^2 = (x)^2 + (34 - x)^2 \\\\\dashrightarrow\sf 676 = x^2 + (34)^2 + x^2 - 2 \times 34 \times x \\\\\dashrightarrow\sf 676 = 2x^2 + 1156 - 68x \\\\\dashrightarrow\sf 2x^2 - 68x + 1156 - 676 = 0 \\\\\dashrightarrow\sf 2x^2 - 68x + 480 = 0 \\\\\dashrightarrow\sf 2(x^2 - 34x + 240) = 0 \\\\\dashrightarrow\sf x^2 - 34x + 240 = 0 \\\\\dashrightarrow\sf x^2 - 24x - 10x + 240 = 0 \\\\\dashrightarrow\sf x(x - 24) - 10(x - 24) = 0 \\\\\dashrightarrow\sf (x - 10)(x - 24) = 0 \\\\\dashrightarrow\sf (x - 10) = 0 \: \: or, \: \: (x - 24) = 0 \\\\\dashrightarrow\large\sf\green{x = 10}\: \: \: or, \: \: \: \large\sf\blue{x = 24}$

$\rule{200}{2}$

$\therefore\sf\blue{x = 10 \:\: \: \: \: \: \: (case:1)}$

$\therefore\sf\red{x = 24 \:\: \: \: \: \: \:(case:2)}$

From (case:1) :-

$\dashrightarrow\sf Base = x =\large{\boxed{\sf\red{ 10 \: m }}}$

$\dashrightarrow\sf Altitude = (34 - x) = (34 - 10) = \large{\boxed{\sf\blue{24 \: m }}}$

$\rule{200}{1}$

From (case:2) :-

$\dashrightarrow\sf Base = x = \large{\boxed{\sf\red{24 \: m }}}$

$\dashrightarrow\sf Altitude = (34 - x) = (34 - 24) = \large{\boxed{\sf\blue{10 \: m }}}$

Answer 2

Solution:-

• Check the question in the above answer
• The sum of two sides =34

Given:-

hypotenuse of right ∆ is 26 cm

The sum of two sides = 34

Let the side be x

and 2nd side =34 - x

Now

• We have to find the Length of its side

A.T.Q

By Pythagoras

$\boxed{\sf{Hypotenuse {}^{2}=Base{}^{2}+Perpendicular {}^{2}}}$

• Now find the values

● Base = x

●perpendicular= 34-x

$\sf\implies (26){}^{2}= x{}^{2}+(34-x){}^{2}\\ \\ \sf\implies 676= x{}^{2}+ 1156-68x+x{}^{2})\\ \\ \sf\implies 676= 2x{}^{2} -68x+1156\\ \\ \sf\implies 0= 2x{}^{2}-68x +1156-676\\ \\ \sf\implies 0= 2x{}^{2} -68x+480\\ \\ \bf\:\:or\:we\:write\:it\:as:-\\ \\ \sf\implies 2(x{}^{2}- 34x+240)=0\\ \\ \sf\implies x{}^{2}-34x+240=\dfrac{0}{2}\\ \\ \sf\implies x{}^{2}-(24+10)x+240=0\\ \\ \sf\implies x{}^{2}-24x-10x+240=0\\ \\ \sf\implies x(x-24)-10(x-24)=0\\ \\ \sf\implies (x-10)(x-24)=0\\ \\ \sf\:\: x= 24 \:\:\:or\:\:\: x= 10$

So now the sides of triangle

$\sf\implies base = x =24\\ \\ \sf\implies height=34- 24= 10$

Or

• Base =10cm
• Height= 24cm

$\boxed{\sf{length\:of\:sides= 10cm\:\:and\:24cm\: respectively}}$