CBSE Class 10 Mathematics Standard 2025: question 34(a)

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. In $△ABQ$ and $△ACR$: $\angle A$ is common, and $\angle ABQ=\angle ACR=90^{\circ }$.

   So $△ABQ\sim △ACR$ [AA similarity].

   1 mark

2. Corresponding sides are proportional:

   $\frac{AB}{AC}=\frac{QB}{RC}=\frac{y}{z}$ …(i)

   ½ mark

3. Similarly, in $△CBQ$ and $△CAP$: $\angle C$ is common, and $\angle CBQ=\angle CAP=90^{\circ }$.

   So $△CBQ\sim △CAP$ [AA similarity].

   ½ mark

4. $\frac{BC}{AC}=\frac{QB}{PA}=\frac{y}{x}$ …(ii)

   ½ mark

5. Adding (i) and (ii):

   $\frac{AB}{AC}+\frac{BC}{AC}=\frac{y}{z}+\frac{y}{x}$

   1 mark

6. $\frac{AB+BC}{AC}=y\left(\frac{1}{x}+\frac{1}{z}\right)$. Since $AB+BC=AC$, this gives $1=y\left(\frac{1}{x}+\frac{1}{z}\right)$.

   1 mark

7. Dividing by $y$: $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$. Hence proved.

   ½ mark

Answer: $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$ (proved)

Revise the chapter

• Chapter: Triangles (notes, formulas, where marks are lost and every past question)
• Topic: Problems on Similar Triangles