Question

The base of a triangle is 5cm longer than its altitude.If the area of the triangle is 18sq.Cm then find base and altigufe

Answer 1

Answer:

LET THE HEIGHT BE H.

THEN, ITS BASE= H+5 CM

AREA OF TRIANGLE= BASE* HEIGHT /2

A/Q

1/2*H*(H+5)=18 SQ. CM

H*(H+5)=18*2

H^2+5H=36

H^2+5H-36=0

H^2-4H+9H-36=0

H(H-4) +9(H-4)=0

(H+9)(H-4)=0

H= 4, -9.

WE IGNORE -9 AS LENGTH CANNOT BE NEGATIVE.

THUS, HEIGHT OF THE TRIANGLE=4 CM

AND BASE= 5+4=9  CM

Answer 2

altitude=x

base=y

given that,

y=x+5

1/2*b*h=18

b*h=18*2

(x+5)x=36

x^2+5x=36

x^2+5x-36=0

x^2+9x-4x-36=0

x (x+9)-4(x+9)=0

(x-4)(x+9)=0

x-4=0

x=4

(x+9)=0

x=-9

We take the positive value

altitude=4

base=4+5=9

Hope it helps......