Question

the altitude of a triangle is 7 cm less than its base. and its hypotenuse is 13 cm , find the base of the triangle​

Answer 1

Solution:-

According to the Given Statement

Let the base of triangle be = x cm

therefore, the altitude of triangle = x -7 cm

and hypotenuse of triangle is 13 cm

Using Pythagoras theorem

• H² = P² + B²

where,

H is the hypotenuse of triangle

P is the altitude/Perpendicular

B is the base of triangle

Substitute the value we get

→ 13² = x² + (x-7)2

→ 169 = x² + x² -14x + 49

→ 169 = 2x² - 14x + 49

→ 169 -49 = 2x² -14x

→ 120 = 2x² -14x

→ 2x² - 14x -120 = 0

Taking Common 2 we get

→ x² -7x -60 = 0

Splitting the middle term

→ x² -12x +5x -60 = 0

Taking Common

→ x (x-12) + 5(x-12) =0

→ (x-12) (x+5) = 0

Therefore, x = 12 or x = -5

Length can't be negative So ignore negative

Therefore, the base of the triangle is 12 cm

Altitude = x-7 = 12-7 = 5 cm

Answer 2

Diagram:-

$\setlength{\unitlength}{1.1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf (x+7)}\put(2.8,.3){\large\bf x}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf \Theta }\end{picture}$

$\huge \bf \: Solution$

Altitude is 7 cm less than base

Hypotenuse = 13 cm

Let

Base = x

Altitude = x - 7

Hypotenuse = 13 cm

Therefore

$\sf \: 13² = {x}^{2} + {x - 7}^{2}$

$\sf \: 169 {x}^{2} + {x}^{2} - {14x}^{} + {49}^{2}$

$\sf \: 169 - 49 = {x}^{2} - {14x}$

$\sf \: 120 = {x}^{2} - {14x}$

$\sf \: 2 {x}^{2} - {14x} - 120 = 0$

$\sf {x}^{2} - 7x - 60 = 0$

$\sf {x}^{2} - 12x + 5x - 60 = 0$

$\sf \: x(x - 12) + 5(x - 12) = 0$

$\bf \: x \: - 12 \: or \: x + 5$

$\sf \: x \: = 12 \: or \: x \: = - 5$

Length can't be negative

Therefore base = 12 cm and altitude = 12 - 7 = 5 cm