Question 14 - NCERT Exemplar - MCQ - Chapter 6 Class 10 Triangles (Term 1)
Last updated at Oct. 20, 2021 by

If S is a point on side PQ of a ∆ PQR such that PS = QS = RS, then
(A)PR . QR = RS
^{
2
}
(B) QS
^{
2
}
+ RS
^{
2
}
= QR
^{
2
}
(C) PR
^{
2
}
+ QR
^{
2
}
= PQ
^{
2
}
(D) PS
^{
2
}
+ RS
^{
2
}
= PR
^{
2
}

Transcript

Question 14
If S is a point on side PQ of a ∆ PQR such that PS = QS = RS, then
PR . QR = RS2 (B) QS2 + RS2 = QR2
(C) PR2 + QR2 = PQ2 (D) PS2 + RS2 = PR2
Given
PS = QS = RS
Now,
We know that
Angles opposite equal sides are equal
In Δ PSR
∠ 1 = ∠ 2
In Δ QSR
∠ 3 = ∠ 4
In Δ PQR
By Angle sum property
∠ P + ∠ Q + ∠ R = 180°
∠ 2 + ∠ 4 + (∠ 1 + ∠ 3) = 180°
Putting ∠ 1 = ∠ 2 and ∠ 3 = ∠ 4
∠ 1 + ∠ 3 + (∠ 1 + ∠ 3) = 180°
2(∠ 1 + ∠ 3) = 180°
(∠ 1 + ∠ 3) = (180° )/2
(∠ 1 + ∠ 3) = 90°
∠ R = 90°
Thus, Δ PRQ is a right angled triangle
where ∠ R = 90°
By Pythagoras theorem
PQ2 = PR2 + QR2
So, the correct answer is (C)

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