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Home / Geometry / Solution of exercise 9.2(Triangle)| Class seven

# Solution of exercise 9.2(Triangle)| Class seven

Problem-9: In the triangle ABC, AB>AC and the bisectors of the ∠B and ∠C intersect at the point P. Prove that PB>PC.

Particular enunciation: Given that, in the triangle ABC, AB>AC and the bisectors of the ∠B and ∠C intersect at the point P. We have to prove that PB>PC.

Proof: Since, BP is the bisector of ∠B.

∴∠PBC=½∠ABC

CP is the bisector of ∠C.

∴∠PCB = ½ ∠ACB

We know that, in a triangle the angle opposite the greater side is greater than the angle opposite the smaller side.

Since AB>AC

∴ ∠ACB>∠ABC

⟹½∠ACB> ½∠ABC

⟹∠PCB>∠PBC

⟹PB>PC

∴ PB>PC.(Proved)

Problem -10.ABC is an isosceles triangle and AB=AC. The side BC is extend up to D. Prove that AD>AB.

General enunciation: ABC is an isosceles triangle and AB=AC. The side BC is extend up to D. Prove that AD>AB.

Particular enunciation: Given that, the ∆ABC is an isosceles triangle and AB=AC. The side BC is extended up to D. We have to prove that AD>AB.

Proof: Given that in ∆ ABC,

AB=AC

∴ ∠ACB=∠ABC [∵ opposite angles of equal arms are also equal]

Again, ∠ACB is the external angle of ∆ ACD

Problem – 11: In the quadrilateral ABCD, AB = AD, BC = CD and CD> AD. Prove that ∠DAB > ∠BCD.

General enunciation: In the quadrilateral ABCD, AB = AD, BC = CD and CD> AD. Prove that ∠DAB > ∠BCD.

Particular enunciation: In the quadrilateral ABCD, AB = AD, BC = CD and CD> AD. We have to prove that ∠DAB > ∠BCD.

Construction: Join A and C.

Proof: We know that, in a triangle the angle opposite the greater side is greater than the angle opposite the smaller side.

∴ ∠DAC > ∠ACD ———————————— (1)

Again, in the triangle ∆ ABC, BC > AB

∴ ∠BAC > ∠ACB ————————————- (2)

Adding equation (1) and (2) we get,

∠DAC + ∠BAC > ∠ACD + ∠ACB

⟹ ∠DAB > ∠BCD (Proved)

Problem-12: In the triangle ∆ABC, AB = AC and D is any point on BC. Prove that AB > AD.

General enunciation: In the triangle ∆ABC, AB = AC and D is any point on BC. Prove that AB > AD.

Particular enunciation: Given that, in the triangle ∆ABC, AB = AC and D is any point on BC. We have to prove that AB > AD.

Proof: In the ∆ ABC, AB = AC

∴ ∠ACB = ∠ABC   [∵ opposite angles of equal arms are also equal]

⟹ ∠ACD= ∠ABD

Problem – 13: In the triangle ∆ABC, AB ⊥ AC and D is any point on AC. Prove that BC > BD.

General enunciation: In the triangle ∆ABC, AB ⊥ AC and D is any point on AC. Prove that BC > BD.

Particular enunciation: In the triangle ∆ABC, AB ⊥ AC and D is any point on AC. We have to prove that BC > BD.

Construction: Join B and D.

Proof: In ∆ABC, AB ⊥ AC

∴ ∠BAC = 900, BC be the hypotenuse.

∴ ∠BCA is an acute angle

∴ ∠BAC > ∠BCA

Again, ∠BDC is an exterior angle of ∆ABD.

⟹ ∠BDC > ∠BAC

⟹ ∠BDC > ∠BCA

⟹ ∠BDC > ∠BCD

⟹ BC > BD (Proved)

Problem-14: Prove that the hypotenuse of a right angled triangle is the greatest side.

General enunciation: We have to prove that the hypotenuse of a right angled triangle is the greatest side.

Particular enunciation: Let ∆ABC be right angled triangle in which ∠ABC = right angle or 900 and AC be the hypotenuse. It is required to prove that AC be the greatest side.

Proof: We know that, in a right angled triangle one angle is right angle and the other two angles are acute angles.

i.e., ∠ABC = 1 right angle = 900

∴ ∠ABC> ∠ACB

∴ AC>AB [∵the side opposite to the greater angle is greater.]

Again, ∠ABC> ∠BAC

⟹ AC>BC

∴ Side AC be greater than the sides AB and BC.

Therefore, AC be the greatest side. (Proved)

Problem-15: Prove that the angle opposite the greatest side of a triangle is also the greatest angle of that triangle.

General enunciation: We have to prove that the angle opposite the greatest side of a triangle is also the greatest angle of that triangle.

Particular enunciation: Let the greatest side of ∆ABC is AC. We have to prove that ∠ABC is the greatest angle of the triangle.

Proof: In triangle ABC, AC>AB

∴ ∠ABC> ∠ACB [∵the angle opposite to the greater side is greater.]

Again, AC>BC

⟹∠ABC> ∠BAC

Hence ∠ABC be greater than ∠ACB and ∠BAC.

Therefore, ∠ABC be the greatest angle of the triangle.

Solution exercise 9.1| Class seven| Geometry

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...