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If the median to a side of a triangle is also an altitude to that side then the triangle is isosceles How do you write this Proof?
Let the triangle be ABC and suppose the median AD is also an altitude.AD is a median, therefore BD = CDAD is an altitude, therefore angle ADB = angle ADC = 90 degreesThen, in triangles ABD and ACD,AD is common,angle ADB = angle ADCand BD = CDTherefore the two triangles are congruent (SAS).And therefore AB = AC, that is, the triangle is isosceles.
What is the area of the quadrilateral ABCD when AB is 0.38cm BC is 0.69cm and AD is 0.42cm with angles ABC 109 degrees and BAD 123 degrees?
Form two triangles from diagonal A to C and by finding the areas of each triangle they will add up to 0.305 square cm rounded to three decimal places.
What is the proof that equiangular triangle is also called equilateral triangle?
Label the triangle ABC. Draw the bisector of angle A to meet BC at D. Then in triangles ABD and ACD, angle ABD = angle ACD (equiangular triangle) angle BAD = angle CAD (AD is angle bisector) so angle ADB = angle ACD (third angle of triangles). Also AD is common. So, by ASA, triangle ABD is congruent to triangle ACD and therefore AB = AC. By drawing the bisector of angle B, it can be shown that AB = BC. Therefore, AB = BC = AC ie the triangle is equilateral.
How many medians are needed to find the centroid of a triangle?
Two. However, you can actually do it with just one. Consider the median AD of triangle ABC. Then the point G, 2/3 of the way from A to D, is the centroid. This process (2/3 of the way from the vertex to the opposite side) can be applied to any median.
Her original design for the ad is 6 inches wide and 4.5 inches tall but the space available is only 4 18 inches wide. If she shrinks the design to fit how tall will the ad be?
It will be 4.18 by 3.14 inches, approx.
In triangle ABC seg AD perpendicular to side BC and B-D-CIf BC 8cm area of triangle ABC is 12 cm square what is the length of AD?
Not too sure of the question but if one side is 8 cm then its height must be 3 cm because:- Area of the triangle is: 0.5*8*3 = 12 square cm
What is the lenght of the hypotenuse of the right triangle ABC in if AC 6 and AD 5?
61
Triangle abc is an equilateral triangle bc equals 10 what is the length of ad give your answer to the nearest whole number?
5
Triangle ABC is an equilateral triangle. BC 10 What is the length of AD Give you answer to the nearest whole number?
That depends on what is meant by length of AD but the 3 sides of an equilateral triangle are all equal in lengths.
Given an equilateral triangle abc 'd' is the midpoint of bc if ad x then what is the perimeter of the triangle in terms of x?
P = x*2*sqrt(3)
What are the dimensions of an isosceles triangle of least area that can be circumscribed about a circle of radius r?
The isosceles triangle of least area that can be circumscribed about a circle of radius r turns out to be not just isosceles, but also equilateral. Each side has length 2r x ( 3 )0.5 . The area is r2 x (27)0.5 . Thanks are due to litotes for pointing out that the original answer did not actually answer the question ! tpm Since the equilateral triangle is also an isosceles triangle, we can say that at least area that can be circumscribed to a circle is the area of an equilateral triangle.If we are talking only for isosceles triangle where base has different length than two congruent sides, we can say that at least area circumscribed to a circle with radius r, is the area of an isosceles triangle whose base angles are very close to 60 degrees. Solution: Let say that the isosceles triangle ABC is circumscribed to a circle with radius r, where BA = BC. We know that the center of the circle inscribed to a triangle is the point of the intersection of the three angle bisectors of the triangle. Let draw these angle bisectors, and denote with D the point where the bisector drawn from the vertex, B, of the triangle, intersects the base AC. Since the triangle is an isosceles triangle, then BD bisects the base and it is perpendicular to the base. So that AD = DC, OD = r, and the triangles ADB and AOD are right triangles (O is the center of the circle). In the triangle ADB, we have:tan A = BD/AD, so that AD = BD/tan A In the triangle AOD, we have:tan A/2 = OD/AD, so that AD = r/tan A/2, and AC = 2r/tan A/2 Therefore,BD/tan A = r/tan A/2, andBD = (r tan A)/tan A/2 Area of triangle ABC = (1/2)(AC)(BD) = (1/2)(2r/ tan A/2)[(r tan A)/tan A/2] = (r2 tan A)/tan2 A/2 After we try different acute angles measure, we see that the smallest area would be: If the angle A= 60⁰,then the Area of the triangle ABC = r2 tan 60⁰/tan2 30⁰ ≈ 5.1961r2 If the angle A= 59.8⁰,then the Area of the triangle ABC = (r2 tan 59.8⁰)/tan2 29.9⁰ ≈ 5.1962r2
How do you find base of triangle?
let abc be the triangle with base bc. Consider it is equilateral triangle.. Now draw ad perpendicular to bc. Now ad equalls dc. Now tan 60 degree equalls ad/dc. With value of tan 60 and ad we can find dc. There fore bc equalls 2 * dc
How can you prove a triangle ABC is isosceles if angle BAD is congruent to angle CAD and line AD is perpendicular to line Bc?
Given: AD perpendicular to BC; angle BAD congruent to CAD Prove: ABC is isosceles Plan: Principle a.s.a Proof: 1. angle BAD congruent to angle CAD (given) 2. Since AD is perpendicular to BC, then the angle BDA is congruent to the angle CDA (all right angles are congruent). 3. AD is congruent to AD (reflexive property) 4. triangle BAD congruent to triangle CAD (principle a.s.a) 5. AB is congruent to AC (corresponding parts of congruent triangles are congruent) 6. triangle ABC is isosceles (it has two congruent sides)
Isosceles trapezoid abcd has an area of 276 in squared if ad equals 13 inches and de equals 12 inches find ab?
Isosceles trapezoid ABCD has an area of 276 If AD = 13 inches and DE = 12 inches, find AB.
How do i prove if the base angles of a triangle are congruent then the triangle is isosceles?
Suppose you have triangle ABC with base BC, and angle B = angle C. Draw the altitude AD.Considers triangles ABD and ACDangle ABD = angle ACD (given)angle ADB = 90 deg = angle ACDtherefore angle BAD = angle CADAlso the side AD is common to the two triangles.Therefore triangle ABD is congruent to triangle ACD (ASA) and so AB = AC.That is, triangle ABC is isosceles.
If the median to a side of a triangle is also an altitude to that side then the triangle is isosceles How do you write this Proof?
Let the triangle be ABC and suppose the median AD is also an altitude.AD is a median, therefore BD = CDAD is an altitude, therefore angle ADB = angle ADC = 90 degreesThen, in triangles ABD and ACD,AD is common,angle ADB = angle ADCand BD = CDTherefore the two triangles are congruent (SAS).And therefore AB = AC, that is, the triangle is isosceles.
What is the area of the quadrilateral ABCD when AB is 0.38cm BC is 0.69cm and AD is 0.42cm with angles ABC 109 degrees and BAD 123 degrees?
Form two triangles from diagonal A to C and by finding the areas of each triangle they will add up to 0.305 square cm rounded to three decimal places.
