Find the altitude of a triangular region whose base is 28 m and the area is 224 metre square.?

Question

Write statements

Answer 1

area = (1/2) * base * altitude

224 = (1/2) * 28 * x
448 = 28x
x = 16 m

Answer 2

Area of a triangular region = 1/2 * base * altitude
1/2 * 28 * h = 224
14 h = 224
h = 224/14 =16
altitudes =16 m. Ans.

Answer 3

Area of a triangle A = 1/2*base*altitude
therefore, Altitude = 2*Area/base
= 2*224/28
= 16
Altitude of the triangle is 16m

Answer 4

area of triangle = 1/2*base*altitude
area = 224 m sq.
base = 28 m
therefore
altitude = 2*area divided by base
=2*224/28 m
=16 m

Answer 5

Area formula

let

a = altitude

a = h

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A = 1/2 b h

224 = 1/2 28 h

2(224) = 28 h

448 = 28 h

448/28 = 28h/28

16 = h

The answer is h = 16

The altitude is 16 meters

insert the h value into the equation

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A = 1/2 b h

224 = 1/2 (28)(16)

224 = 1/2(448)

224 = 224

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Answer 6

1\2 base x alt = area {in a triangle}
1\2 x 28 x alt = 224
alt = 224 x 2/28
alt = 224 x 1/14
alt = 16m

Answer 7

1\2 base x alt = area {in a triangle}
1\2 x 28 x alt = 224
alt = 224 x 2/28
alt = 224 x 1/14
alt = 16m

Answer 8

The region has 4 faces that is it is a pyramid.Each base is 28m long,base is equilateral area is sq. root of(28*28-14*14)*28/2=(let it be called)BA .therefore area of remaining 3 faces=224-BA.area of 3 remainingeach face divide by 3.base is 28 therefore alt will be ( 224-BA)/3=1/2*28*alt.solvin this ans is in negative as value of BA is339.472smt.there ans is(-)2.74933mts

Answer 9

area of a triangle is 1/2*base*height.
area(A)=224m^2
base(b)=28m.
altitude(a)=?
A=1/2ba
224=1/2*28a
224=14a
a=224/14
=16m
the altitude of the triangular region is 16m.

Answer 10

We know that area of a triangle is 1/2XbaseXaltitude.
Therefore altitude=(areaX2)/base
=224X2/28
=16 m