Vardaan Learning Institute
Answer Key: Triangles PYQ
Class: 10 (CBSE)
Subject: Mathematics
Chapter: 06
BOARD EXAM 2025 (Set 30/6/3)
Q15. In figure, \(PQ || XY || BC\), \(AP = 2\) cm, \(PX = 1.5\)
cm, \(BX = 4\) cm. If \(QY = 0.75\) cm, then \(AQ + CY =\)
Q16. Given \(\Delta ABC \sim \Delta PQR\), \(\angle A = 30�\)
and \(\angle Q = 90�\). The value of \((\angle R + \angle B)\) is
Q23. In figure, \(AP = 1\) cm, \(BP = 2\) cm, \(AQ = 1.5\) cm,
\(AC = 4.5\) cm. Prove that \(\Delta APQ \sim \Delta ABC\). Find \(PQ\) if \(BC = 3.6\) cm.
Q32 (a). State and prove the converse of BPT: If a line divides
any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Given: \(\frac{AD}{DB} = \frac{AE}{EC}\). To prove: \(DE || BC\).
Assume \(DE\) not parallel to \(BC\). Draw \(DF || BC\). By BPT: \(\frac{AD}{DB} = \frac{AF}{FC}\).
This gives \(AF = AE\), so \(F = E\). Hence \(DE || BC\). Proved.
Assume \(DE\) not parallel to \(BC\). Draw \(DF || BC\). By BPT: \(\frac{AD}{DB} = \frac{AF}{FC}\).
This gives \(AF = AE\), so \(F = E\). Hence \(DE || BC\). Proved.
OR
Q32 (b). In figure, \(\Delta CAB\) is right-angled at A and
\(AD \perp BC\). Prove that \(\Delta ADB \sim \Delta CDA\). If \(BC = 10\) cm and \(CD = 2\) cm,
find \(AD\).
By AA similarity: \(\angle ADB = \angle CDA = 90�\), \(\angle ABD = \angle CAD\). Hence \(\Delta ADB
\sim \Delta CDA\).
\(AD^2 = BD \times CD = 8 \times 2 = 16 \Rightarrow AD = 4\) cm.
\(AD^2 = BD \times CD = 8 \times 2 = 16 \Rightarrow AD = 4\) cm.