CBSE Class 10 Mathematics Standard 2024: question 28(b)

Hints

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Step-by-step solution

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1. Let the circle touch $AB,BC,CD,DA$ at $P,Q,R,S$. Join $OP,OQ,OR,OS$. Name $\angle POB=\angle 1$, $\angle BOQ=\angle 2$, $\angle QOC=\angle 3$, $\angle COR=\angle 4$, $\angle ROD=\angle 5$, $\angle DOS=\angle 6$, $\angle SOA=\angle 7$, $\angle AOP=\angle 8$.

   ½ mark

2. In $△OPB$ and $△OQB$: $OP=OQ$ [radii], $OB$ common, $\angle OPB=\angle OQB=90^{\circ }$ [tangent ⟂ radius]. So $△OPB\cong △OQB$ (RHS) $\Rightarrow \angle 1=\angle 2$ (CPCT).

   1 mark

3. Similarly $\angle 3=\angle 4$, $\angle 5=\angle 6$, $\angle 7=\angle 8$.

   ½ mark

4. Angles round point $O$: $\angle 1+\angle 2+\cdots +\angle 8=360^{\circ }\Rightarrow 2(\angle 1+\angle 8+\angle 4+\angle 5)=360^{\circ }$.

   ½ mark

5. $\Rightarrow (\angle 1+\angle 8)+(\angle 4+\angle 5)=180^{\circ }$, i.e. $\angle AOB+\angle COD=180^{\circ }$. Hence proved.

   ½ mark

Answer: $\angle AOB+\angle COD=180^{\circ }$ (proved)

Revise the chapter

• Chapter: Circles (notes, formulas, where marks are lost and every past question)
• Topic: Solving Problems Using Tangent Properties