AMC 8 Geometry Math Olympiad

Radius of a Semicircle | AMC 8, 2016 | Problem 25

Try this beautiful problem from Geometry based on Radius of a semicircle inscribed in an isosceles triangle.

Radius of a Semi circle – AMC-8, 2016 – Problem 25

A semicircle is inscribed in an isoscles triangle with base 16 and height 15 so that the diameter of the semicircle is contained in the base of the triangle as shown .what is the radius of the semicircle?

Semicircle in an isosceles triangle

  • $\frac{110}{19}$
  • $\frac{120}{17}$
  • $\frac{9}{5}$

Key Concepts




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AMC-8, 2016 problem 25

Challenges and Thrills of Pre College Mathematics

Try with Hints

Draw a perpendicular from the point C on base AB

semicircle inscribed in an isosceles triangle

Can you now finish the problem ……….

D be the midpoint of the AB(since $\triangle ABC $ is an isoscles Triangle)

Find AC and area

can you finish the problem……..

radius of a semicircle in an isosceles triangle

Area of the $\triangle ABC= \frac{1}{2} \times AB \times CD$

= $ \frac{1}{2} \times 16 \times 15 $

=120 sq.unit

Using the pythagoras th. $ AC^2= AD^2+CD^2$

i.e $AC^2=(8)^2+(15)^2$

i.e $AC=17$

Let$ ED = x$ be the radius of the semicircle

Therefore Area of $\triangle CAD = \frac{1}{2} \times AC \times ED$=$\frac {1}{2} area of \triangle ABC$

i.e $\frac{1}{2} \times AC \times ED $=60

i.e $\frac{1}{2} \times 17 \times x$ =60

i.e $x=\frac {120}{7}$

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AMC 10 USA Math Olympiad

Area of Triangle – AMC 10A – 2019 – Problem No. – 7

What is Area of Triangle ?

The area of a triangle is defined as the total space that is enclosed by any particular triangle. The basic formula to find the area of a given triangle is A = 1/2 × b × h,  where b is the base and h is the height of the given triangle, whether it is scalene, isosceles or equilateral.

Try This Problem from AMC 10A – 2019 -Problem No.7

Two lines with slopes \(\frac{1}{2}\) and 2 intersect at (2,2) . What is the area of the triangle enclosed by these two lines and the line \(x + y = 10 \) ?

A) 4 B) \(4\sqrt 2\) C) 6 D) 8 E) \(6 \sqrt 2\)

American Mathematics Competition 10 (AMC 10A), 2019, Problem Number – 7

Area of Triangle

6 out of 10

Problems in Plane Geometry by Sharygin

Knowledge Graph

area of triangle - Knowledge Graph

Use some hints

If you need a hint to start this sum use this

Lets try to find the slop – intercept form of all three lines : (x,y) = (2,2) and y =

\(\frac{x}{2}+b\) implies \(2 = \frac{2}{2}+b = 1+b\). So, b = 1 . While y = 2x + c implies 2 = 2.2 + c So, c = -2 And again x+y = 10 implies y = -x + 10.

Thus the lines are \( y = \frac {x}{2} + 1 \) , y = 2x – 2 and y = -x + 10 . Now we find the intersection points between each of the lines with y = -x + 10 , which are (6,4) and (4,6) .

In the last hint we can apply the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle where the base is \(2\sqrt 2\) and the height \(3 \sqrt 2\), whose area is 6 .The answer is 6 (c) .

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