In triangle ABC in which AB=BC=4cm and angle B= 90 degrees, what is the area of the triangle?
If AB=BC=4 cm and b is right angled,So either AB is perpendicular or BC,If you take AB as base,then Bc would be perpendicular and vice versa,Applying formula for area of triangle,(1/2) *base*perpendicular,It is (1/2)*4*4=8cm^2Hope u understood, feel free to ask if u have doubts.
An equilateral triangle has a height of 8, what is the side length?
sin60=8/side side=8/sin60=8/(sqrt3/2)=16/sqrt3
ABC is a right-angled triangle, having the right angle at B, such that BC = 6 cm and AB = 8 cm. How do I find the radius of its incircle?
In a right angle triangle, in-circle touches all the inner sides, means the small two sides are the tangents to the circle.Source: GoogleIn above image, O is centre of in-circle.BO = √2*rOM = r. BO and OM are collinear or forms a single line. Think about it?So BM = (√2 + 1) * rNow BM is the altitude of the triangle.Area if the triangle, since it is right angled triangle,Area = (1/2)* (altitude)*(hypotenuse)hypotenuse = 10cm(1/2)*(√2+1)*r * 10 = (1/2)*(6)*(8)(√2 +1)*r = 4.8r = (4.8)*(√2–1)For any right angled triangle , the radius of in-circler = {(a*b)/c}*(√2 - 1)where a and b are the side lengths and c is hypotenuse length.
In triangle ABC, AC= AB. Angle A= 80 °. What is Angle B(do not use inequalities in triangle)?
A triangle with the two equal side is an isoceles triangle.In isoceles triangle two angles are equalAs give Angle A is 80°And as we know A+B+C=180→ B+C=180–80As angle B and C are equal2B=100→ Angle B =50° ans.
In an isosceles triangle, its two equal sides are 20 cm each and the angle between them is 45 degrees. What is the area of the triangle?
Consider any triangle [math]\Delta ABC.[/math]In right triangle [math]\Delta ABD,[/math][math]\sin \theta=\dfrac{BD}{AB}\tag*{}[/math][math]BD=AB\sin\theta\tag*{}[/math]The area of this triangle is easy right?[math]\operatorname{ar}{(\Delta ABC)}=\frac{1}{2}BD\cdot AC\tag*{}[/math][math]\operatorname{ar}{(\Delta ABC)}=\frac{1}{2}AB\cdot AC\sin\theta\tag*{}[/math]We can generalise this as follows, the area of a triangle with consecutive sides [math]a,b[/math] and included angle [math]\theta[/math] is [math]\frac{1}{2}ab\sin\theta.[/math]Apllying this to your question,[math]A=\frac{1}{2} 20\times 20\times\dfrac{1}{\sqrt{2}}\tag*{}[/math][math]A=100\sqrt{2}\text{ cm}^2\tag*{}[/math]
How do I construct a triangle with a base, median on the base and vertical angle?
How do i construct a triangle with a base, median on the base and vertical angle?Suppose you are required to construct a triangle with a base, BC (8 cm), median on the base, AD (12 cm) and vertical angle,
An isosceles trapezoid has a base of 10, an angle of 30º and the upper base has double the size the non parallel sides, (a). What is the value of (a)?
Let ABCD be the trapezium with AB = 10, BC = AD = a and CD = 2a.Produce AD and BC to meet at O. Let OD = OC = r, so that OA = OB = r+a.Now, OA = OB = r+a = 5 sec 30 = 5.773502692.r = a sec 30 = 1.154700538 aThus r+a = 1.154700538 a + a = 2.154700538 a = 5.773502692, ora = 5.773502692/2.154700538 = 2.679271118, henceCD the upper base = 2a = 2*2.679271118 = 5.358542236
Triangle ABC is similar to triangle ADE. DE = 6 cm and BC = 12 cm. If the area of triangle ABC is 90cm^2, what is the area of triangle ADE?
Similar as in it has the same angles but different lengths. The ratio of the sides is the same. So the length BC=12cm corresponds to length DE=6cm. When a triangle’s lengths grow by a multiplier its area increases by the multiplier squared. The multiplier in this case is 1/2 (6/12) so its area increases by [math]\frac{1}{2}^2 = \frac{1}{4}[/math].So the answer is [math]\frac{90}{4} cm^2[/math] =[math]22.5 cm^2[/math]