Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# In a right-angled triangle, five times the square on the hypotenuse is equal to four times the sum of the squares on the medians drawn from the acute angles. Prove it.

Last updated date: 13th Jun 2024
Total views: 402k
Views today: 4.02k
Answer
Verified
402k+ views
Hint: In a right-angled triangle, as the Pythagoras theorems say, ${h^2} = {p^2} + {b^2}$, where h is the hypotenuse of a right-angle triangle, p is the perpendicular and b is the base.

Complete step-by-step answer:
Consider a right angle triangle, $\vartriangle ABC$

In $\vartriangle ABC$, using Pythagoras theorem, ${h^2} = {p^2} + {b^2}$, where h is the hypotenuse of a right-angle triangle, p is the perpendicular and b is the base.
$A{C^2} = A{B^2} + B{C^2}$
Let AB=x and BC=y
$\Rightarrow A{C^2} = {x^2} + {y^2}$

AD and CE are the median drawn from A and B respectively.
Now, further applying Pythagoras theorem in $\vartriangle ABD$, we get
$A{D^2} = {x^2} + {\left( {\dfrac{y}{2}} \right)^2} \\ \Rightarrow AD = \sqrt {{x^2} + {{\left( {\dfrac{y}{2}} \right)}^2}} \\$
Similarly, In $\vartriangle ECB$,
$C{E^2} = {y^2} + {(\dfrac{x}{2})^2} \\ \Rightarrow CE = \sqrt {{y^2} + {{(\dfrac{x}{2})}^2}} \\$
Now, the sum of the squares of the medians is
$\Rightarrow A{D^2} + C{E^2} = {x^2} + {\left( {\dfrac{y}{2}} \right)^2} + {y^2} + {\left( {\dfrac{x}{2}} \right)^2} \\ \Rightarrow A{D^2} + C{E^2} = \dfrac{5}{4}({x^2} + {y^2}) \\ \Rightarrow 4\left( {A{D^2} + C{E^2}} \right) = 5({x^2} + {y^2}) \\$
As, we know $A{C^2} = {x^2} + {y^2}$
$\Rightarrow 4(A{D^2} + C{E^2}) = 5A{C^2}$
Hence, five times the square on the hypotenuse is equal to four times the sum of the squares on the medians drawn from the acute angles

Note: The Median joins the vertex to the midpoint of the opposite side. The properties of the median are as follows:-
The median divides the triangle into two parts of equal area.
The point of concurrency of medians is called Centroid.
The centroid divides the median in the ratio 2:1 with the larger parts toward the vertex.