Question 17 - CBSE Class 10 Sample Paper for 2018 Boards
Last updated at Nov. 27, 2018 by Teachoo

In given figure ∠1 = ∠2 and ∆NSQ ≅ ∆MTR , then prove that ∆PTS ~ ∆ PRQ .
OR
In an equilateral triangle ABC, D is a point on the side BC such that BD = 1/3 BC. Prove that 9AD
^{
2
}
= 7AB
^{
2
}
This is a question of CBSE Sample Paper - Class 10 - 2017/18.

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Question 17
In an equilateral triangle ABC, D is a point on the side BC such that BD = 1/3 BC. Prove that 9AD
^{
2
}
= 7AB
^{
2
}

This is exactly same as Ex 6.5, 15.

Check answer here

Transcript

Question 17
In given figure ∠1 = ∠2 and ∆NSQ ≅ ∆MTR , then prove that ∆PTS ~ ∆ PRQ .
Given: ∠ 1 = ∠ 2
& ΔNSQ ≅ ΔMTR
To Prove: ΔPTS ∼ PRQ
Proof:
Given ΔNSQ ≅ ΔMTR
∴ ∠ NQS = ∠ MRT
i.e. ∠ PQR = ∠ PRQ
Now,
In Δ PST
By angle sum property
∠ P + ∠ 1 + ∠ 2 = 180°
Since ∠ 1 = ∠ 2 given
∠ P + ∠ 1 + ∠ 1 = 180°
∠ P + 2 ∠ 1 = 180°
In Δ PQR
By angle sum property
∠ P + ∠ PQR + ∠ PRQ = 180°
From (1), ∠ PQR = ∠ PRQ
∠ P + ∠ PQR + ∠ PQR = 180°
∠ P + 2 ∠ PQR = 180°
Right side of (2) and (3) are same
So, from (2) and (3)
∠ P + 2 ∠ 1 = ∠ P + 2 ∠ PQR
2 ∠ 1 = 2 ∠ PQR
∠ 1 = ∠ PQR
In ΔPTS & Δ PRQ
∠P = ∠P (Common angle)
∠ PST = ∠ PQR (From (4))
∴ ΔPTS ∼ PRQ (AA similarity)
Hence proved

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