The area of an isosceles triangle is $60cm^2$ and the length of equal side is $13cm$. Find height and base.
[Math] To find base and height of an isosceles trangle if sides and area are give
geometrytriangles
Related Solutions
$$a = b - 2 \,c \cos \alpha $$
where $ \alpha = $ angle between the big side and one of equal sides length $c$.
Since, the area is given as 12, and area of a triangle is given by $$\frac{1}{2}\times \text{base}\times \text{height}=12$$
Now, you are also given that $$\text{base}=\text{height}+5$$
So, Use the second information into the formula, generate an equation and solve it!
HINT: $\text{height}=1 \text{cm}$ $\text{base}= 6 \text{cm}$
Best Answer
Hint: Draw a line from the vertex at which the two equal sides meet, and extend it to the base of the triangle such that the line is perpendicular to the base. Then the big triangle consists of two adjacent (and congruent) right-triangles, each of which has height equal to the shared side of the two smaller triangles.
This height is also the height of the large triangle.
Indeed, each equal side becomes the hypotenuse of a right-triangle with height $12$ and base $5$, forming the very nice $5, 12, 13$-right-triangle. Since the bases of the small triangles add up to the base of of the large containing triangle equal to $2\times 5 = 10$.
We can check our solution to find that $\frac 12 bh = \frac 12 (10)(12) = 60$ which is
The other solution is for $h=5,$ and the base of the half-triangles equal to $12$. This gives us $h = 5, b = 2\times 12 = 24$ for the large triangle.