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# Ex 7.2 NCERT Solutions- Triangles Class 9 Notes | EduRev

## Class 9 Mathematics by Full Circle

Created by: Full Circle

## Class 9 : Ex 7.2 NCERT Solutions- Triangles Class 9 Notes | EduRev

The document Ex 7.2 NCERT Solutions- Triangles Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.
All you need of Class 9 at this link: Class 9

Question 1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:
(i) OB = OC (ii) AO bisects ∠ A
Solution:
(i) In ΔABC, we have AB = AC                          [Given]
∴ ∠C= ∠B
[Angle opposite to equal sides are equal] ⇒ (1/2)∠C= (1/2)∠B
or ∠OCB = ∠OBC
⇒ OB = OC                          [Sides opposite to equal angles are equal]

(ii) In ΔABO and ΔACO, we have
AB = AC                          [Given]
OB = OC                          [Proved]
∠OBA = ∠OCA ∴ Using SAS criteria,
ΔABO ≌ ΔACO
⇒ ∠OAB = ∠OAC [c.p.c.t.]
⇒ AO bisects ∠A.

Question 2. In ΔABC, AD is the perpendicular bisector of BC (see figure). Show that ΔABC is an isosceles triangle in which AB = AC.
Solution:
∵ AD is bisector of BC.
∴ BD = CD
Now, in ΔABD and ΔACD, we have:
BD = CD                          [Proved]
∴ ΔABD ≌ ΔACD [SAS criteria] ∴ Their corresponding parts are equal.
⇒ AB = AC Thus, ΔABC is an isosceles triangle.

Question 3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Figure) Show that these altitudes are equal.
OR
ABC is an isosceles triangle with AB = AC. Prove that the altitudes BE and CF of the triangle are equal. Solution: DABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ACB = ∠ABC                     [∵ Angles opposite to equal sides are equal]
Now, in ΔBEC and ΔCFB, we have
∠EBC = ∠FCB                          [Proved]
BC = CB                          [Common]
and ∠BEC = ∠CFB                          [Each = 90°]
∴ ΔBEC ≌ ΔCFB                          [Using ASA criteria]
⇒ Their corresponding parts are equal. i.e. BE = CF

Question 4. ABC is a triangle is which altitudes BE and CF to sides AC and AB are equal (see figure). Show that
(i)
ΔABE ≌ ΔACF
(ii) AB = AC, i.e. ABC is an isosceles triangle.
Solution:
(i) In ΔABE and ΔACF, we have ∠AEB = ∠AFC
[each = 90° ∵ BE ⊥ AC and CF ⊥ AB]
∠A= ∠A                         [Common]
BE = CF                          [Given]
∴ ΔABE ≌ ΔACF                          [Using AAS criterion]

(ii) Since, ΔABE ≌ ΔACF
∴ Their corresponding parts are equal.
⇒ AB = AC

Question 5. ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD.
Solution:
In ΔABC, we have AB = AC
[∵ DABC is an isosceles triangle]

But angles opposite to equal sides are equal.
∴ ∠ABC = ∠ACB                          ...(1)
Again, in ΔBDC, we have BD = CD
[∵ ΔBDC is an isosceles triangle.]
∴ ∠CBD = ∠BCD                         ...(2)
[Angles opposite to equal sides are equal] Adding (1) and (2), we have
∠ABC + ∠CBD = ∠ACB + ∠BCD
⇒ ∠ABD = ∠ACD

Question 6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.
Solution:
∵ In ΔABC, AB = AC                          [Given]                  …(1)
From (1) and (2), we have AC = AD Now, in ΔABC, we have ∠B + ∠ACB + ∠BAC = 180°
⇒ 2∠ACB + ∠BAC = 180°                          …(3)
[∵ ∠B = ∠ACB (Angles opposite to equal sides)]
In ΔACD, ∠D + ∠ACD + ∠CAD = 180°
⇒ 2∠ACD + ∠CAD = 180°                          ...(4)
[∵ ∠D = ∠ACD (angles opposite to equal sides)]
Adding (3) and (4), we have
2∠ACB + ∠BAC + 2∠ACD + ∠CAD = 180° + 180°
⇒ 2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°
⇒ 2[∠BCD] + [180°] = 360°                          [∠BAC and ∠CAD form a linear pair]
⇒ 2∠BCD = 360° ∠ 180° = 180º
⇒ ∠BCD = (1800/2)= 90°
Thus,         ∠BCD = 90°.

Question 7. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution
: In ΔABC, we have AB = AC                          [Given] ∴ Their opposite angles are equal.
⇒ ∠ACB = ∠ABC
Now, ∠A + ∠B + ∠C = 180°
⇒ 90° + ∠B + ∠C = 180°                          [∵ ∠A = 90° (Given)]
⇒ ∠B + ∠C = 180°
But ∠ABC = ∠ACB,
i.e. ∠B = ∠C
∴ ∠B= ∠C = (900/2) = 45°
Thus, ∠B = 45° and ∠C = 45°

Question 8. Show that the angles of an equilateral triangle are 60° each.
Solution:
In ΔABC, we have
AB = BC = CA           [∵ ABC is an equilateral triangle]
∴ AB = BC ⇒ ∠A = ∠C                          …(1)
[∵ Angle opposite to equal sides are equal.]
Similarly, AC = BC ⇒ ∠A = ∠B From (1) and (2), we have ∠A= ∠B = ∠C
Let ∠A= ∠B = ∠C = x
Since, ∠A + ∠B + ∠C = 180°
∴ x + x + x = 180°
⇒ 3x = 180° or x =(180o/3) = 60°
∴ ∠A= ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each.

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