Question

the areof a right angled triangle is 600sq cm .if the base of the triangle exceeds the altitude by 10 cm .find the dimension of the triangle

Answer 1

let the altitude of the triangle =x cm

base = (x+10) cm

area = 600 sq cm

1/2* base* altitude = 600

x(x+10)= 2*600

x^2+10x -1200=0

x^2 +40x-30x-1200=0

x(x+40)-30(x+40)=0

(x+40)(x-30)=0

x+40=0 or x-30=0

x=-40 or x= 30

x should be negative

x=30

base = x+10 =30+10=40

by Pythagoras theorem

hypotenuse square = base square+ altitude square

= (40)^2+(30)^2
=1600+900
=2500
hypotenuse =50

three sides of a triangle

30cm,40cm and 50cm

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let the altitude of the triangle be x cm.

And its base = (x + 10) cm.

According to the Question,

Area of triangle = 1/2 × Base × height

Area of triangle = 1/2 × x × (x + 10)

⇒ 1/2x(x + 10) = 600

⇒ x(x + 10) = 1200

⇒ x² + 10x - 1200 = 0

⇒ x² + 40x - 30x - 1200 = 0

⇒ x(x + 40) - 30(x + 40) = 0

⇒ (x + 40) (x - 30) = 0

⇒ x = - 40, 30 (As x can't be negative)

⇒ x = 30

Altitude of triangle = x = 30 cm

Base of triangle = x + 10 = 30 + 10 = 40 cm

Hence, the dimension of the triangle is 30 cm and 40 cm.