In the figure : ∠PSQ = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

In right angled triangle PQS, By pythagoras theorem, ⇒ (Hypotenuse) 2 = (Perpendicular) 2 + (Base) 2 ⇒ PQ 2 = PS 2 + QS 2 ⇒ 10 2 = PS 2 + 6 2 ⇒ PS 2 = 10 2 - 6 2 ⇒ PS 2 = 100 - 36 ⇒ PS 2 = 64 ⇒ PS = = 8 cm. From figure, RS = RQ + QS = 9 + 6 = 15 cm. In right angled triangle PRS, By pythagoras theorem, ⇒ (Hypotenuse) 2 = (Perpendicular) 2 + (Base) 2 ⇒ PR 2 = PS 2 + RS 2 ⇒ PR 2 = 8 2 + 15 2 ⇒ PR 2 = 64 + 225 ⇒ PR 2 = 289 ⇒ PR = = 17 cm. Hence, PR = 17 cm.

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Mathematics

In the figure : ∠PSQ = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

Pythagoras Theorem

Related Questions

A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.

A man goes 40 m due north and then 50 m due west. Find his distance from the starting point.

In a quadrilateral PQRS, ∠Q = ∠S = 90° then prove that 2PR² - QR² = PQ² + PS² + SR².

AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.

Mathematics

In the figure : ∠PSQ = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

Pythagoras Theorem

Related Questions

A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.

A man goes 40 m due north and then 50 m due west. Find his distance from the starting point.

In a quadrilateral PQRS, ∠Q = ∠S = 90° then prove that 2PR2 - QR2 = PQ2 + PS2 + SR2.

AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.

In a quadrilateral PQRS, ∠Q = ∠S = 90° then prove that 2PR² - QR² = PQ² + PS² + SR².