Find The Relation Between The Area Of Two Triangles
How do find the relation between the area of the blue triangle and the area of the red triangle?
From figure, MNO is a triangle, MP = 3 cm, PQ = 4 cm, QN = 5 cm, MR = 5 cm & RO = 4 cm. Then find the relation between the area of two triangles
Solution
We can solve this geometry math problem in different ways, here I am presenting two methods
Method 1
Assume angle OMN is θ and area of four triangles are A, B, C and D, so
Area of triangle PMR = ½ × MP × MR × sin θ
⟹ Area of triangle PMR = ½ × 3 × 5 × sin θ
⟹ A = (15 sin θ)/2……………………………………………..eq(1)
Area of triangle MQR = ½ × MQ × MR × sin θ
⟹ Area of triangle MQR = ½ × 7 × 5 × sin θ
⟹ A + B = (35 sin θ)/2……………………………………………..eq(2)
Area of triangle MQO = ½ × MQ × MO × sin θ
⟹ Area of triangle MQO = ½ × 7 × 9 × sin θ
⟹ A + B + C = (63 sin θ)/2……………………………………………..eq(3)
Area of triangle MNO = ½ × MN × MO × sin θ
⟹ Area of triangle MNO = ½ × 12 × 9 × sin θ
⟹ A + B + C + D = (108 sin θ)/2……………………………………………..eq(4)
Subtract equation 3 from equation 4, then
A + B + C + D – (A + B + C)= (108 sin θ)/2 – (63 sin θ)/2
⟹ 2D = 45 sin θ
Subtract equation 1 from equation 2, then
A + B – A = (35 sin θ)/2 – (15 sin θ)/2
⟹ B = (20 sin θ)/2
⟹ sin θ = B / 10
2D = 45 sin θ = 45 (B/10)
⟹ 20D = 45B
⟹ 4D = 9B
So, 4 × Area of the red triangle = 9 × Area of the blue triangle
Method 2
Before moving to 2nd method we need to discuss the equation to find the relation between the area of triangles
Relation between the area of triangle and height
From the figure, area of red triangle = ½ × a × h
area of blue triangle = ½ × b × h
⟹ h/2 = area of red triangle/a = area of blue triangle/b
If the height of the triangle is equal, then the area of the triangle divided with base length then we get a constant
Now solving for the relation between the area of triangles
From the figure triangle MPR and triangle QPR has the same height, so
A/3 = B/4
⟹ B = 4A/3
Triangle MRQ and triangle ORQ has the same height, so
(A + B)/5 = C/4
⟹ A + B = 5C/4
⟹ A + 4A/3 = 5C/4
Thus, 7A/3 = 5C/4
⟹ C = 28A/15
Triangle MQO and triangle QNO has the same height, then
(A + B + C)/7 = D/5
⟹ D/5 = (A + 4A/3 + 28A/15)/7
⟹ D/5 = A(7/3 + 28/15)/7
So, D = 5(1/3 + 4/15)A
⟹ D = 3A
Then, B/D = (4A/3)/(3A)
⟹ B/D = 4/9
⟹ 9B = 4D
So, 4 × Area of the red triangle = 9 × Area of the blue triangle
