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P-2 Pythagoras Theorem Maths - P-2 Pythagoras Theorem

Maharashtra Board - STD 10 - Maths (MSBSHSE - English Medium). 4 questions in this chapter.

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P-2 Pythagoras Theorem question types

4 questions across 4 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.

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Sample Questions

P-2 Pythagoras Theorem questions

One sample from each question group in this chapter. Select any group above to see the full set with answer keys.

Q 33 Marks Question3 Marks
Complete the following activity to prove that the sum of squares of diagonals of a rhombus is equal to the sum of the squares of the sides.
Image
Given :
$\square PQRS$ is a rhombus. Diagonals PR and SQ intersect each other at point T.
To prove : $PS ^2+ SR ^2+ QR ^2+ PQ ^2= PR ^2+ QS ^2$
Activity :
Diagonals of a rhombus bisect each other.
In $\Delta PQS$, PT is the median and in $\Delta QRS$, RT is the median.
$\therefore$ By Apollonius theorem,
$PQ^{2} + PS^{2} = ⬜ + 2QT^{2}$ .......... (I)
$QR^{2} + SR^{2} = ⬜ + 2QT^{2}$ .......... (II)
adding (I) and (II),
$PQ^{2}+PS^{2}+QR^{2}+SR^{2} = 2(PT^{2}+ ⬜) + 4QT^{2}$
$= 2(PT^{2}+⬜) + 4QT^{2}$ .......... ($RT=PT$)
$= 4PT^{2} + 4QT^{2}$
= (⬜)2 + (2QT)2
$\therefore$ $PQ^{2}+PS^{2}+QR^{2}+SR^{2} = PR^{2} + ⬜.$
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