SOLUTION: Land in the shape of an isosceles triangle has a base of 130 m. An altitude from the vertex of one of the base angles of the triangle is 120 m. What is the area of this property

Algebra ->  Surface-area -> SOLUTION: Land in the shape of an isosceles triangle has a base of 130 m. An altitude from the vertex of one of the base angles of the triangle is 120 m. What is the area of this property       Log On


   

Question 1150800: Land in the shape of an isosceles triangle has a base of 130 m. An altitude from
the vertex of one of the base angles of the triangle is 120 m. What is the area
of this property
in m²?
Found 3 solutions by MathLover1, Edwin McCravy, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!


We can equate the area as follows

A=%281%2F2%29base+%2A+height+=++65+%2A++height+... (1)
A=120+%2A+%281%2F2%29side++=++60+%2A+side.......... (2)

Set (1) = (2) and solve for the side

60+%2A+side++=++65+%2A+height

side++=+%2865%2F60%29+%2A+height

side++=++%2813%2F12%29%2A+height++=+++%2813%2F12%29%2Ah

The side is the hypotenuse+of a right triangle with the height and 1%2F2+base being the legs

So, using the Pythagorean Theorem, we have that

sqrt++%28+side%5E2++-++height%5E2+%29++=++%281%2F2%29%2A+base}

Substitute

sqrt++%28+%28+%2813%2F12%29h%29%5E2++-+h%5E2%29++=++%281%2F2%29%2A+base+

sqrt++%28%28169%2F+144%29h%5E2+-++%28144%2F144%29h%5E2%29++=+%281%2F2%29%2A+base+


sqrt+%28+%28169+-+144%29+h%5E2++++%2F+144+%29+++=++%281%2F2+%29%2Abase

sqrt+%28+25h%5E2+%2F+144+%29+=++%281%2F2%29%2A+base

%28h+%2F+12%29+%2A+sqrt+%2825%29++=++%281%2F2%29%2A+base

%285%2F12%29h++=++%281%2F2+%29130

%285%2F12%29h++=++65

h+=+%2865%2A12%29+%2F+5+

h++=+13+%2A+12++=+++156+m

So,
Area++=+%281%2F2%29base+%2A+height

Area++=%281%2F2%29130m+%2A+156m

Area++=65m+%2A+156m++

Area++=10140+m%5E2



Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
 

The green line BD is the altitude drawn from vertex of base angle B, so BD is
perpendicular to AC. So ADB is a right triangle, so by the Pythagorean theorem,

AD%5E2%2BBD%5E2=AB%5E2
AD%5E2%2B120%5E2=130%5E2
AD%5E2=130%5E2-120%5E2
AD%5E2=16900-14400
AD%5E2=2500
AD=sqrt%282500%29
AD=50

CDB is also a right triangle, so also by the Pythagorean theorem,

BD%5E2%2BCD%5E2=BC%5E2
120%5E2%2BCD%5E2=BC%5E2

But since BC = AD + CD
          BC = 50 + CD

120%5E2%2BCD%5E2=%2850%2BCD%29%5E2
120%5E2%2BCD%5E2=50%5E2%2B100CD%2BCD%5E2
120%5E2=50%5E2%2B100CD
14400-2500=100CD
11900=100CD
119=CD

AC+=+AD+%2B+CD
AC+=+50+%2B+119
AC+=+169

Area=expr%281%2F2%29base%2Aaltitude
Area=expr%281%2F2%29AC%2ABD
Area=expr%281%2F2%29169%2A120
Area=matrix%281%2C2%2C10140%2Cm%5E2%29


Edwin

Answer by ikleyn(52817) About Me  (Show Source):
You can put this solution on YOUR website!
.

 


Red line  CE  is the altitude in triangle ABC drawn to the base.

Green line BD is the altitude drawn from the base vertex B to the lateral side AC.


Triangle ABD is a right angled triangle with the hypotenuse of 130 meters and the leg BD of 120 meters.

Hence, it is (5,12,13)-triangle and its leg AD is 50 meters.


Thus the area of the triangle ABD is  %281%2F2%29%2A50%2A120 = 3000 square meters.


Triangle ACE is similar to the triangle ABD (since they both are right angled triangles and have common acute angle A).


The similarity coefficient is the ratio of their shorter legs  65%2F50 = 13%2F10.


Therefore, the area of the triangle ACE is  3000%2A%2813%2F10%29%5E2 = 30*169 square meters,

and the area of the triangle ABC is twice this value, i.e.  2*30*169 = 10140 square meters.    ANSWER

Solved.