Area of a triangle
There are so many different approaches for finding the area of a triangle. A suitable result is applicable as per the availability of data.
- General formula
- Heron's formula
- Trigonometric application
A, B and C → measure of three angles
- Application of circum radius (R)
Area = abc/4r
a, b and c → lengths of three sides
R → circum radius
- Application of in-radius (r)
Area = r.S
r → in-radius
S → semi perimeter
- Equilateral Triangle
Area = √3 / 4 * a2
a → length of each side.
Height of an equilateral triangle = √3/4 * a
- Isosceles triangle
Area = b/4 * √(4a2 - b2)
a→ length of equal sides
b → length of third side
Important theorems and useful results
If AB/BC= m/n and AE/ED = p/q
then; Area of ΔABE/Area of ΔACD = pm/(p+q)(m+n)
In the given diagram ΔBDF is inscribed in ΔACE.
AB/BC= 12, CD/DE = 35 and AF/FE = 1/4
Find the ratio of the area of shaded region to the area of non-shaded region.
Area of ΔABF/Area of ΔACE = 1*1/((1+2)(1+4)) = 1/15
Area of ΔBCD/Area of ΔACE= 2*3/((2+1)(3+5)) = 14
Area of ΔDEF/Area of ΔACE= 5*4/((5+3)(4+1)) = 12
Area of (ΔABF + ΔBCD + ΔDEF)/Area of ΔACE= 1/15+ 1/4 + 1/2= 49/60
i.e. Area of non-shaded region = 49/60 * Area of ΔACE
i.e. Area of shaded region = Area of ΔACE - Area of non-shaded region
Area of shaded region = [1- 49/60]Area of ΔACE = 11/60 * Area of ΔACE
Area of shaded region/Area of non-shaded region= 11/49
Hence, the required ratio = 11 : 49
Result 2: Properties of right triangles
ΔABC is a right angled triangle, AD is the altitude drawn from vertex A to hypotenuse BC.
ΔADB ΔCDA ΔCAB (similar triangles)
|AD2 = BD * CD|
|AB2 = BD * BC|
|AC2 = BC * CD|
- In the given diagram AB = 6 cm, AC = 8 cm. Find AD.
AB = 6 cm and AC = 8 cm
BC = 10 cm
Area of ΔABC = 1/2 (AC*AB) = 1/2 (BC *AD)
AC*AB = BC *AD
8 * 6 = 10 * AD
AD = 4.8 cm.
- In right triangle ABC, AB : BC = 1: . Find AD : CD.
Let AB = 1 and BC = √3
AC = √(12 + √32) = 2
Let AD = a
CD = 2 - a
AB2 = AD * AC
12 = a * 2
a = 1/2
i.e. AD = 1/2
CD = 2 - 1/2 = 3/2
AD : CD = 1/2 : 3/2
= 1 : 3
Result 3: Internal angle bisector theorem
In ΔABC, AD is the angle bisector of ∠A, then AD/AC = BD/CD and BD * AC - AB * CD = AD2
Result 4: External angle bisector theorem
BE is the angle bisector of ∠CBD. then BC/AB = CE/AE
Result 5: Apollonius theorem
In triangle ABC, AD is the median from vertex A to side BC.
AB2 + AC2 = 2(AD2 + BD2)
Lengths of the diagonals of a parallelogram are 20cm and 24 cm. if one of its sides is 16 cm, find the length other non-parallel side.
Diagonals of a parallelogram are bisecting each other. Therefore E is the midpoint of BD (and AE).
In ΔABD, AE is a median.
As per Apollonius theorem;
AB2 + AD2 = 2(AE2 + BE2)
162 + x2 = 2(122 + 102)
x2 = 232
x = 2√58
Hence, the length of the other non-parallel side = 2√58 cm