Math, asked by Anonymous, 5 months ago

the altitude of a triangle is 3/4 the length of its base. if the altitude were increased by 3 feet the base decreased by 3 feet. the area would be unchanged. find the length of the altitude and base.

Answers

Answered by janiyabrooks
1

Answer:

Original triangle:

Let b = length of base

then

(3/4)b = altitude

Area = (1/2)b(3/4)b = (3/8)b^2

.

Changed triangle:

b-3 = length of base

(3/4)b+3 = altitude

Area = (1/2)(b-3)(3/4)b+3 = (3/8)(b-3)b+3

.

Set area equal to each other:

(3/8)b^2 = (3/8)(b-3)b+3

b^2 = (b-3)b+3

b^2 = b^2-3b+3

0 = -3b+3

-3 = -3b

1 feet = b (base of triangle)

.

Altitude:

(3/4)b = (3/4)1 = 3/4 feet (altitude)

Step-by-step explanation:

Answered by 170388shyammsundargb
0

Answer:

Orginial Triangle  :

Step-by-step explanation:

Let b = length of base

then

(3/4)b = altitude

Area = (1/2)b(3/4)b = (3/8)b^2

.

Changed triangle:

b-3 = length of base

(3/4)b+3 = altitude

Area = (1/2)(b-3)(3/4)b+3 = (3/8)(b-3)b+3

 

.

Set area equal to each other:

(3/8)b^2 = (3/8)(b-3)b+3

b^2 = (b-3)b+3

b^2 = b^2-3b+3

0 = -3b+3

-3 = -3b

1 feet = b (base of triangle)

.

Altitude:

(3/4)b = (3/4)1 = 3/4 feet (altitude)

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