Question

the altitude of a right triangle is 7 cm less than its base . if hypotenuse is 13 cm find other two sides​

Answer 1

Answer:

let altitude be x and base be x+7

By Pythagoras Theorem - x2 + x2+49+14x= 169

2x2+14x= 120

x2+7x-60=0

(x-5)(x+12)=0

x=5

therefore altitude is 5 and base is 12

Answer 2

Answer:

Base = x = 12

Altitude = (x-7)  = 5

Step-by-step explanation:

Let base be 'x' and altitude be (x -7)

$(hyp)^2 = (b)^2 + (alt)^2$

$(13)^2 = (x)^2 + (x - 7 )^2$

$(13)^2 = (x)^2 + (x)^2 -14x + 49$

$169 = (2x)^2 - 14x + 49$

$(2x)^2 -14x - 120 = 0$

Solve the quadratic

-b±√(b²-4ac))/(2a)

14±√((14)² + 4 × 2 × 120))/(2 × 2)

14±√(196 + 960))/(4)

14±√(1156)/(4)

14±(34)/(4)

x = 12 / -5

Base = x = 12

Altitude = (x-7)  = (12 - 7) = 5