Question

ANS AS SOON AS POSSIBLE..​

Answer 1

$\underline{\bigstar{\it\ Question-1}}$

★ Figure for this question Refer to attachment!

Given :-

• ABCD is a trapezium and AB||CD and diagonal intersect at O

To Prove :-

$\underline{\boxed{\sf \dfrac{AO}{OC}=\dfrac{BO}{OD}}}$

Construction :-

• Through O draw PO || AB ,meeting AD at P

Proof :-

In ∆ABD , AB||PO

$\longrightarrow\sf \ \dfrac{AP}{DP}=\dfrac{BO}{OD} -----eq.(i) \ \ \ \Big[\therefore By\ Thales \ theorm \ (B.P.T)\Big]\\ \\ \\ \bullet\sf \ In\ \triangle ADC, \ DC||PO \ \ (\therefore \ AB||DC||OP)\\ \\ \\ \longrightarrow\sf \dfrac{AP}{DP}=\dfrac{AO}{OC}-----eq.(ii) \ \ \ \Big[\therefore (B.P.T)\Big]\\ \\ \\ \bullet\sf From\ (i) \ and \ (ii)\ we \ get \\ \\ \\ \longrightarrow{\boxed{\sf \dfrac{AO}{OC}=\dfrac{BO}{OD}}}\ \ \ \ \ (Hence \ Proved ! )$

$\underline{\bigstar{\it\ Question-2}}$

Given :-

$\bullet\sf \ AO= (3x-1)cm \ \ , \ \ \bullet\ OC=(5x-3)cm\\ \\ \bullet\sf \ BO=(2x+1)cm \ \ , \ \bullet\sf \ OD=(6x-5)$

• We have to find out the value of x

So ,

$\longrightarrow\sf \dfrac{AO}{OC}=\dfrac{BO}{OD}\ \ \ \ \ \ \Big[above \ Proved\Big]\\ \\ \\ \longrightarrow\sf \dfrac{(3x-1)}{(5x-3)}=\dfrac{(2x+1)}{(6x-5)}\\ \\ \\ \longrightarrow\sf (3x-1)(6x-5)=(2x+1)(5x-3)\\ \\ \\ \longrightarrow\sf 18x^2-15x-6x+5=10x^2-6x+5x-3\\ \\ \\ \longrightarrow\sf 18x^2-21x+5=10x^2-x-3\\ \\ \\ \longrightarrow\sf 18x^2-21x+5-10x^2+x+3=0\\ \\ \\ \longrightarrow\sf 8x^2-20x+8=0\\ \\ \\ \longrightarrow\sf 4(2x^2-5x+2)=0\\ \\ \\ \longrightarrow\sf 2x^2-4x-x+2=0\\ \\ \\ \longrightarrow\sf 2x(x-2)-1(x-2)=0\\ \\ \\ \longrightarrow\sd (2x-1)(x-2)=0\\ \\ \\ \longrightarrow\sf x= \dfrac{1}{2}\ \ or \ \ x= 2$

$\underline{\boxed{\sf x= 2\ or \ 1/2 }}$

• If x= 1/2

$\longrightarrow\sf OC= 5x-3\\ \\ \longrightarrow\sf \dfrac{5}{2}-3\\ \\ \longrightarrow\sf \dfrac{5-6}{2}\\ \\ \longrightarrow\sf OC = \dfrac{-1}{2}$

And ,distance can never be negative . So x≠ 1/2

$\underline{\boxed{\sf Hence , \ x= 2}}$

Answer 2

Answer:

Hello Kitty and opposite reaction to finding the right