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In right triangle APR, we have AP = 6cm., AR = 8 cm., And, ∠A = 90° ∴ By using Converse of Pythagoras Theorem, We have PR 2 = AR 2 + AP 2 ⇒ PR 2 = 8 2 + 6 2 = 64 + 36 = 100 PR = 10 cm. Now, in triangle PQR, we have PQ = 24 cm., QR = 26 cm. and, PR = 10 cm. Here, we have, PQ 2 + PR 2 = (24) 2 + (10) 2 = 576 + 100 = 676 = QR 2 Hence, By Converse of Phythagoras Theorem ∠QPR = 90°

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Triangles

∆ABC and ∆DBC lie on the same side of base BC. From a point P on base BC, PQ || AB and PR || BD are drawn. Prove that QR || AD.

In ∆ABC
PQ || AB
Therefore, by using basic proportionality theorem,
$CP over PB equals CQ over QA space space space space space space space space space space space... left parenthesis straight i right parenthesis$
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Therefore, by using basic proportionality theorem,
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Using (i) and (ii), we get
$CP over PB equals CR over RD$

115 Views

If D and E are the points on the sides AB and AC of a triangle ABC respectively, such that AD = 6 cm, BD = 9 cm, AE = 8 cm and EC = 12 cm. Show that DE is parallel to BC.

Solution not provided.

101 Views

In the given Fig, PQ = 24 cm, QR = 26 cm, ∠PAR = 90°, PA = 6 cm, and AR = 8, cm. Find ∠QPR.

In right triangle APR, we have
AP = 6cm., AR = 8 cm., And, ∠A = 90°
∴ By using Converse of Pythagoras Theorem, We have
PR² = AR² + AP²⇒ PR² = 8² + 6²= 64 + 36 = 100
PR = 10 cm.
Now, in triangle PQR, we have
PQ = 24 cm., QR = 26 cm. and, PR = 10 cm.
Here, we have,
PQ² + PR² = (24)² + (10)²= 576 + 100 = 676 = QR²Hence,
By Converse of Phythagoras Theorem ∠QPR = 90°