If area of isosceles triangle os 60 and lemgth of its equal side is 13 cm find its base
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Construct triangle ABC with the vertex B common to the two congruent sides upward and the unknown side as the base. Construct a perpendicular to the base through the vertex at the top of the triangle. Label the point of intersection of the perpendicular and the base with D.
Let x represent the measure of segement AD. Let y represent segment BD. Since we know from a common geometric proof that the perpendicular from the apex to the base of an isosceles triangle bisects the base, the apex angle, and the triangle itself. Therefore the area of triangle ABD is equal to the area of triangle BCD and the area of each is one-half of the area of the entire triangle ABC.
From the formula for the area of a triangle, we have:
x
Hence,
y
And further,
x=5
From which we can derive:
y=6
From the fact that segment BD is perpendicular to segment AD, we can deduce that triangle ABD is a right triangle, for which we are given the measure of the hypotenuse is 13 and we have identified the measure of the legs as  and . Using Pythagoras:
y
And using what we learned earlier about the relationship between x and y, we can write:
x^2
Multiply both sides by y^2:
x
Let y, substitute, and put the resulting quadratic into standard form:
x
Factor:
y
Then
y
or
x
Discarding the negative roots because we are attempting to calculate length, we have
Having previously determined that segment BD bisects segment AC, and by construction, AD plus DC is equal to AC, we can deduce that AC is equal to two times AD. Therefore the base measurement that we seek as an answer to this problem is two times the value of x
Let x represent the measure of segement AD. Let y represent segment BD. Since we know from a common geometric proof that the perpendicular from the apex to the base of an isosceles triangle bisects the base, the apex angle, and the triangle itself. Therefore the area of triangle ABD is equal to the area of triangle BCD and the area of each is one-half of the area of the entire triangle ABC.
From the formula for the area of a triangle, we have:
x
Hence,
y
And further,
x=5
From which we can derive:
y=6
From the fact that segment BD is perpendicular to segment AD, we can deduce that triangle ABD is a right triangle, for which we are given the measure of the hypotenuse is 13 and we have identified the measure of the legs as  and . Using Pythagoras:
y
And using what we learned earlier about the relationship between x and y, we can write:
x^2
Multiply both sides by y^2:
x
Let y, substitute, and put the resulting quadratic into standard form:
x
Factor:
y
Then
y
or
x
Discarding the negative roots because we are attempting to calculate length, we have
Having previously determined that segment BD bisects segment AC, and by construction, AD plus DC is equal to AC, we can deduce that AC is equal to two times AD. Therefore the base measurement that we seek as an answer to this problem is two times the value of x
ababab:
hello bro
this is what happens when u dont read the question and copy paste the ans from google !! (lol)
how to copy paste the Answer??? from Google
Answered by
1
Divide the isosceles triangle in two halves by dividing the base side from middle .
Now we have 2 right angle triangles and the area of each of two resultant triangles is 30cm^2.
Using prognathous theorem where hypotenuse is fixed as 13 and also 1/2*base*height= 30
We get two equations:
b^2 + h^2 = 13^2
1/2 *b*h= 30
=> b*h = 60
Solving we get 5, 12. As we have divided the base in two equal parts we need to multiply by 2
Hence answer is 5*2, 12*2 = 24, 10
hope this helps you
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