Question

pls solve this
class 10 triangles chapter​

Answer 1

Answer:

I am not solve this problem

Step-by-step explanation:

because i am not know

Answer 2

Step-by-step explanation:

Given :-

P and Q are the mid points of CA and CB of ∆ABC ,right angled at C .

Required To Prove :-

Prove that :

i) 4AQ² = 4AC² + BC²

ii) 4BP² = 4BC² +AC²

iii) 4(AQ²+BP²) = 5AB²

Proof :-

Given that

∆ ABC is a right angled triangle, right angle at C.

P and Q are the mid points of CA and CB

AP = PC --------(1)

AC = 2 AP = 2PC -----(2)

and

CQ = QB -------(3)

=> BC = 2CQ = 2QB ----(4)

(i) In ∆ ACQ ,

By Pythagorous Theorem,

AQ² = AC² +CQ²

=> AQ² = AC² +(BC/2)² (From (4))

=> AQ² = AC² + (BC² /4)

=> AQ² = (4AC² + BC²)/4

=> 4AQ² = 4AC² + BC² --------(5)

Hence, Proved.

(ii) In ∆PCB,

By Pythagoras theorem,

=>BP² = PC² + BC²

=> BP² = (AC/2)²+BC² ( from (2))

=> BP² = (AC²/4)+BC²

=> BP² = (AC²+4BC²)/4

=> 4BP² = AC²+4BC²

=> 4BP² = 4BC² + AC² --------(6)

Hence, Proved.

(iii) In ∆ ABC,

By Pythagorous theorem

AB² = AC²+BC² --------------(7)

On adding (5) &(6)

=>4AQ²+4BP² = 4AC² + BC²+ 4BC² +AC²

=> 4(AQ²+BP²) = (4AC²+AC²)+(4BC²+BC²)

=> 4(AQ²+BP²) = 5AC²+5BC²

=> 4(AQ²+BP²) = 5(AC²+BC²)

=> 4(AQ²+BP²) = 5AB² (from (7))

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".