P and O are the mid points of the sides AB and BC respectively of the triangle ABC- right-angled B- then -displaystyle - AQ - 2 - CP - 2 - AC - 2 -displaystyle - AQ - 2 - CP - 2 - frac - 4 - 5 - AC - 2 -displaystyle - AQ - 2 - CP - 2 - frac - 5 - 4 - AC - 2 -displaystyle - AQ - 2 - CP - 2 - frac - 3 - 5 - AC - 2 -

Question

Toppr Admin · Accepted Answer

The correct option is C AQ2-CP2-54AC2Construction- Join PC and AQ as shown in the figure-Consider -x394-ABQ-AQ2-AB2-BQ2-x27F9-AB2-AQ2-x2212-BQ2

$P$ and $O$ are the mid points of the sides AB and BC respectively of the triangle ABC, right-angled at B, then :
${A Q}^{2} + {C P}^{2} = {A C}^{2}$
${A Q}^{2} + {C P}^{2} = {\frac{4}{5} A C}^{2}$
${A Q}^{2} + {C P}^{2} = {\frac{5}{4} A C}^{2}$
${A Q}^{2} + {C P}^{2} = {\frac{3}{5} A C}^{2}$

The correct option is C ${A Q}^{2} + {C P}^{2} = {\frac{5}{4} A C}^{2}$
Construction: Join PC and AQ as shown in the figure.
Consider $Δ A B Q, {A Q}^{2} = A B^{2} + {B Q}^{2} ⟹ {A B}^{2} = A Q^{2} - {B Q}^{2}$

P and O are the mid points of the sides AB and BC respectively of the triangle ABC, right-angled at B, then :AQ2+CP2=AC2AQ2+CP2=45AC2AQ2+CP2=54AC2AQ2+CP2=35AC2

The correct option is C AQ2+CP2=54AC2Construction: Join PC and AQ as shown in the figure.Consider ΔABQ,AQ2=AB2+BQ2⟹AB2=AQ2−BQ2

$P$ and $O$ are the mid points of the sides AB and BC respectively of the triangle ABC, right-angled at B, then :
${A Q}^{2} + {C P}^{2} = {A C}^{2}$
${A Q}^{2} + {C P}^{2} = {\frac{4}{5} A C}^{2}$
${A Q}^{2} + {C P}^{2} = {\frac{5}{4} A C}^{2}$
${A Q}^{2} + {C P}^{2} = {\frac{3}{5} A C}^{2}$

The correct option is C ${A Q}^{2} + {C P}^{2} = {\frac{5}{4} A C}^{2}$
Construction: Join PC and AQ as shown in the figure.
Consider $Δ A B Q, {A Q}^{2} = A B^{2} + {B Q}^{2} ⟹ {A B}^{2} = A Q^{2} - {B Q}^{2}$