KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

Karnataka State Syllabus Class 10 Maths Chapter 2 Triangles Ex 2.6

Question 1.
In Fig. 2.56, PS is the bisector of ∠QPR of ∆ PQR, Prove that QS/SR=PQ/PR

Answer:
Given: In figure, ABC is a triangle in which ∠ ABC > 90° and AD ⊥ BC.
To prove: AC² = AB² + BC² + 2BC. BD.
Proof: In right triangle ABC
∠D = 90°
AC² = AD² + CD² [Pythagoras theorem]
AC² = AD² + (BD + CB)² [DC = BD + BC]
AC² = AD² + BD²+ BC2 +2.BD. BC
AC² = AB² + BC² + 2BD.BC [In right angle ∆^le ADB AB² = AD² + BD²]

Question 4.
In Fig. 2.59, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² – 2BC. BD.

Answer:
Given: In figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC
To prove: AC² = AB² + BC² – 2BC.BD
Proof: In right triangle ABC ∠ D = 90°
AC² = AD² + DC² [By pythagoras theorem]
= AD² + (BC – BD)² [CD = BC – BD]
= AD² ± BC² + BD² – 2BC. BD
AC² = AB² + BC² – 2BC.BD
[In right angle ADB with
∠D = 90° AB² = AD² + BD² By Pythagoras theorem]

Question 5.
In Fig. 2.60,AD is a median of a triangle ABC and AM I BC. Prove that:

Answer:
Given: In figure, AD is a median of a triangle ABC and AM ⊥ BC.
∴ D is mid-point of BC (∵ AD is median)

Proof:
i) In right triangle AMC ∠M = 90°
∴ AC² = AM² + MC² [By Pythagoras theorem]
= AM² +(MD + DC)²
= AM² + MD² + DC² + 2MD. DC

[In right angle ∆^le AMD with ∠M = 90°
AM² + MD² = AD² (By pythagoras theorem)]

ii) In right triangle AMB ∠M = 90°
AB2 = AM2 + MB2 [Pythagoras
theorem]
= AM² + (BD – MD)²
= AM² + BD² + MD² – 2BD . MD
= AM² + MD² + BD² – (2 BD)MD

Question 6.
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Answer:

Given: In parallelogram ABCD
AB = CD & AD = BC
Construction: convert parallelogram into a rectangle and Draw AG ⊥ CD
To prove:
AC² + BD² = AB² + BC²2 + CD² – AD²
Proof: consider ∆^le BDF ∠D = 90°
BD² = BF² + FD² = h² + (x+d)² — (1)
Consider ∆^le AGC ∠G = 90°
AC² = AG2² + GC² = h² + (x – d)² — (2)
Adding (I) and (2)
AC²+ BD² = h² + (x + d)² + h² + (x – d)²
2h² + x² + 2xd + d² + x² – 2xd + d2
2h² + 2x² + 2d²
= 2x² + 2(h2 + d2)
= 2x² + 2y²
= x² + y² + x² + AC² + BD² =AB² +BC² i-CD² + AD².

Question 7.
In Fig. 2.61, two chords AB and CD intersect each other at the point P.
Prove that:
i) ∆ APC ~ ∆ DPB
ii) AP. PB = CP.DP

Answer:
Given: In figure, two chords AB and CD intersect each other at the point P.
To prove: i) ∆ APC ~ ∆ DPB
ii) AP. PB = CP . DP
Proof: i) In ∆ APC and ∆ DPB

Question 8.
In Fig. 2.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

Answer:
Given: In figure, two chords AB and CD of a circle intersect each other at the point P. (when produced) outside the circle.
To prove: i) ∆ PAC ~ ∆ PDB
ii )PA. PB = PC . PD
Proof: i) we know that in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.
Therefore, for cyclic quadrilateral ABCD.
consider ABCD ∆ PAC and ∆ PBD.
∠PAC = ∠PDB → (i)
∠PCA = ∠PBD → (2)
∆ PAC ~ ∆ PDB [A A similarity criterion]

AC = 3m
Hence, she has 3 m string out.
Length of the string pulled in 12 seconds
at the rate of5 cm/ sec 5 × 12cm = 60 cm = O.6 m.
∴ Length of remaining string left out
= AD = 3.0 – 0.6 = 2.4m

In right angled ∆ ABD ∠B = 90°
AD² = AB² + BD²
BD² = AD² – AB²
[pythagoras theorem]
= (2.4)² – (1.8)²
= 5.76 – 3.24 = 2.52
BD = 12.52 = 1.59 m (approx)
Hence, the horizontal distance of the fly form Nazirna after 12 seconds = 1.2 + 1.59 = 2.79 m.