In - PQR- if ST -QR- then RT-PQ-PS- equals -1- PR-SQ-PT -2- SQ-PR-PT- -3- 2-PR-PT- -4- PR-SQ-2PT-

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Standard X

Mathematics

Pythagoras Theorem

In △ PQR, if ...

Question

In △ PQR, if ST \parallel QR, then RT(PQ+PS) equals (1) PR+SQ+PT (2) SQ(PR+PT) (3) 2(PR+PT) (4) PR(SQ+2PT)

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Q. In

△ P Q R, S T ∥ Q R

and

\frac{P S}{S Q} = \frac{3}{5}

P R = 6

cm find

P T

Q. In a triangle PQR, ST is parallel to QR Show that RT(PQ + PS) = SQ (PR + PT)

Q. In the given figure, in a triangle PQR, ST

| |

QR and

\frac{P S}{S Q} = \frac{3}{5}

and

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cm, find PT.

Q. In the given figure seg PS is the median of ∆PQR and PT ⊥ QR. Prove that,

(1)

{PR}^{2} = {PS}^{2} + QR \times ST + {(\frac{QR}{2})}^{2}

(2)

{PQ}^{2} = {PS}^{2} - QR \times ST + {(\frac{QR}{2})}^{2}

In the figure below, if the line segment ST is parallel to line segment QR such that $\frac{P S}{S Q} = \frac{P T}{T R}$ . This provided data is not sufficient to prove that
$\frac{S T}{Q R} = \frac{P T}{T R}$

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