![]() ![]() In the right angled isosceles triangle, the altitude on the hypotenuse is half the length of the hypotenuse. In the right angled isosceles triangle, one angle is a right angle (90 degrees) and the other two angles are both 45 degrees. Two isosceles triangles are always similar. The medians drawn from vertex B and vertex C will not bisect the opposite sides AB and AC. The median drawn from vertex A will bisect BC at right angles. In the above figure, triangle ADB and triangle ADC are congruent right-angled triangles. The altitude from the vertex divides an isosceles triangle into two congruent right-angled triangles. The altitude from vertex A to the base BC is the angle bisector of the vertex angle ∠ A. The altitude from vertex A to the base BC is the perpendicular bisector of the base BC. In the above figure, ∠ B and ∠C are of equal measure. The angles opposite to equal sides are equal in measure. ![]() In the above figure, sides AB and AC are of equal length ‘a’ unit. Now, we will discuss the properties of an isosceles triangle.Īn Isosceles Triangle has the Following Properties: Obtuse angled triangle: A triangle whose one interior angle is more than 90 0. Right angled triangle: A triangle whose one interior angle is 90 0. Scalene triangle: A triangle whose all three sides are unequal.Ĭlassification of Triangles on the Basis of their Angles is as FollowsĪcute angled triangle: A triangle whose all interior angles are less than 90 0. Isosceles triangle: A triangle whose two sides are equal. Each of them has their own individual properties.Ĭlassification of Triangles on the Basis of their Sides is as Follows:Įquilateral triangle: A triangle whose all the three sides are equal. For this, either use the inscribed angle theorem (had to look up the name) or just note that the isosceles triangle on the right has angles of $\alpha,\alpha$ and $180-2\alpha$, making the supplementary angle $2\alpha$.Triangles are classified into different types on the basis of their sides and angles. Once that's calculated, double everything to account for the congruent left half of the big triangle. That just leaves the area of the lower right triangle. So the isosceles portion at the right has area of If you draw an altitude from the center of the circle to the base of the isosceles triangle right of center, you can calculate the lengths of the base and height through trigonometric formulae given $\alpha$ and the length of the radii. The only thing I can think of for part b is to cut the larger triangle into several smaller ones. ![]() That makes the angles of the large triangle $60$ degrees each. The supplementary angle is then $120$ and the other $2$ angles of the isosceles triangle on the right are $30$ each. Therefore, the angle in between is $60$ degrees. One of the legs of that lower right triangle is $h$ which is equal to $3$. It is possible, however, to verify your guess from what you've already done. Part b looks to be the tough part and c would likely quickly follow from it. Therefore, the equilateral triangle has the maximum area. Theorem: Among all triangles inscribed in a given circle, the equilateral one has the largest area. $$b=\sqrt.$$įor $(c)$, By the Isoperimetric Theorem, it states So, we can say that the total height of the triangle is $9$ units. For solving $(b)$, we can find the total height $(h_T)$ of the isosceles triangle by adding the value for $h$ found in $(a)$ to the other height $6$: ![]()
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