Question

The sides of a triangle are 11 m, 60 m and 61 m. The altitude to the smallest side is
A. 11 m
B. 66 m
C. 50 m
D. 60 m

Answer 1

Answer:

your question answer is A

Answer 2

The altitude to the smallest side is is 60 m

Solution:

Given that:

The sides of a triangle are 11m , 60m, 61m

We have to find the altitude of smallest side of triangle

The figure is attached below

In figure, the smallest side is BC which is 11 m

We have to find the length of AD

Semiperimeter = sum of all sides divided by 2

$s=\frac{a+b+c}{2}=\frac{60+61+11}{2}=\frac{132}{2}=66cm$

The area of triangle is:

$Area = \frac{1}{2} \times base \times height$

By heron's formula, find the area of triangle

$\text{Area of ABC}=\sqrt{s(s-a)(s-b)(s-c)}$

$\frac{1}{2} \times BC \times AD = \sqrt{66(66-60)(66-61)(66-11)}\\\\ \frac{1}{2}\times 11\times AD=\sqrt{66(6)(5)(55)}\\\\\frac{1}{2} \times 11 \times AD = 330\\\\AD = 330 \times 2 \times \frac{1}{11}\\\\AD = 60$

Thus altitude to the smallest side is is 60 m

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