CBSE Class 10 Mathematics Basic 2025: question 33(a)

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Step-by-step solution

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1. Let $AE=x$ cm. Tangents drawn from an external point to a circle are equal, so $AF=AE=x$, $CE=CD=10$ cm, $BF=BD=6$ cm.
   Hence $BC=16$, $CA=x+10$, $AB=x+6$.

   1 mark

2. Semi-perimeter $s=\frac{16+(x+10)+(x+6)}{2}=16+x$.

   ½ mark

3. By Heron's formula, $\text{ar}(△ABC)=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(16+x)(x)(6)(10)}$ …(i)

   1 mark

4. Also, joining O to A, B, C (OD ⟂ BC, OE ⟂ CA, OF ⟂ AB, radius 4 cm) [tangent ⟂ radius]:
   $\text{ar}(△ABC)=\text{ar}(OBC)+\text{ar}(OCA)+\text{ar}(OAB)=\frac{1}{2}[16\times 4+(10+x)\times 4+(6+x)\times 4]=4(16+x)$ …(ii)

   1 mark

5. Equating (i) and (ii) and squaring: $60x(16+x)=16(16+x{)}^2$.
   Since $16+x\ne 0$ ($x=-16$ is rejected), $60x=16(16+x)\Rightarrow 44x=256\Rightarrow x=\frac{64}{11}$.

   So $AE=\frac{64}{11}$ cm $\approx 5.8$ cm.

   1½ marks

Answer: $AE=\frac{64}{11}$ cm ≈ 5.8 cm

Revise the chapter

• Chapter: Circles (notes, formulas, where marks are lost and every past question)
• Topic: Lengths of Tangents from an External Point