Question

The altitude of a right triangle is 7cm less than its base if the hypotenuse is 13cm please find the other two sides class 10

Answer 1

Diagram :-

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\put(7.7,2.9){\large{A}}\put(7.7,1){\large{B}}\put(10.6,1){\large{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\put(10.5,1){\line(-4,3){2.5}}\put(7.3,2){\sf{\large{(x-7)}}}\put(9,0.7){\sf{\large{x cm}}}\put(9.4,1.9){\sf{\large{13 cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

Solution :-

Let base be x cm

and altitude be (x - 7) cm

In right ∆ ABC

(AC)² = (AB)² + (BC)²

➝ (13)² = (x - 7)² + (x)²

➝ 169 = (x)² - 2 × x × 7 + (7)² + x²

➝ 169 = x² - 14x + 49 + x²

➝ 169 - 49 = 2x² - 14x

➝ 120 = 2x² - 14x

➝ 2x² - 14x - 120 = 0

➝ 2(x² - 7x - 60) = 0

➝ x² - 7x - 60 = 0

➝ x² - 12x + 5x - 60 = 0

➝ x(x -12) + 5(x - 12) = 0

➝ (x - 12) (x + 5) = 0

Now,

➝ x - 12 = 0

➝ x = 0 + 12

➝ x = 12

Then,

➝ x + 5 = 0

➝ x = 0 - 5

➝ x = - 5 (not possible because sides cannot be negative)

Hence,the base of given triangle will be 12 cm and altitude of given triangle will be 12 - 7 = 5 cm.