class 10 triangles chapter
pls solve
Answers
Step-by-step explanation:
Given :-
ABC is a right angled triangle,right angle at C. Let BC = a , CA = b and AB = c and let p be the length of perpendicular from C on AB.
Required To Prove:-
Prove that :
(i) cp = ab
(ii) 1/p² = (1/a²) + (1/b²)
Proof :-
Given that
ABC is a right angled triangle,right angle at C.
BC = a
CA = b
AB = c
The length of perpendicular from C on AB = p
=> CD = p
p divides the ∆ ABC into two triangles
They are ∆ ACD and ∆BCD
(i) In ∆ ABC and ∆ACD
<C = <D = 90°
<A = <A ( Common angle of ∆ ABC and ∆ACD )
By A.A criterion of similarity
∆ ABC and ∆ACD are similar triangles.
∆ ABC ~ ∆ ACD
We know that
In similar triangles, The corresponding sides are in the same ratio.
=> AB/AC = BC/CD
=> c/b = a/p
On applying cross multiplication then
=> cp = ab -----(1)
Hence, Proved.
from (1) , c = (ab)/p --------(2)
(ii)In ∆ ABC, By Pythagoras Theorem,
AB² = BC² + AC²
=> c² = a² + b²
=> (ab/p)² = a² + b²
=>a²b²/p² = a² + b²
=> On dividing by a²b² both sides then
=> (a²b²/p²)/(a²b²) = (a² + b²)/(a²b²)
=> (a²b²)/(p²a²b²) = (a²/a²b²) + (b²/a²b²)
=> 1/p² = (1/b²) + (1/a²)
=> 1/p² = (1/a²) + (1/b²)
Hence, Proved.
Used formulae:-
Pythagoras Theorem:-
" In a right angled triangle, The square of the hypotenuse is equal to the sum of the squares of the other two sides".
Properties of Similar triangles :-
If two triangles are said to be similar if,
→ The corresponding angles are equal.
→ The corresponding sides are in the same ratio or in the proportion.
