triangles _ exercise 7.2
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CBSE Class IX MathsTRANSCRIPT
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Triangles : Exercise 7.2 (Mathematics NCERT Class9th)
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Q.1 In an isosceles triangle ABC, with AB = AC, the bisectors of intersect each other at O. Join A to O. Show that : (i) OB = OC (ii) AO bisects .Sol.
(i) In ABC, we have AB = AC [Since angles opposite to equal sides are equal]
(http://dronstudy.com/wpcontent/uploads/2013/12/102.png)
[Since OB and OC bisect B and C respectively. Therefore ]
OB = OC [Since sides opp. to equal are equal](ii) Now, in ABO and ACO , we have AB = AC [Given]
[From (1)]OB = OC [From (2)]Therefore By SAS criterion of congruence, we have
[Since corresponding parts of congruent triangles are equal] AO bisects .
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Q.2 In ABC, AD is the perpendicular bisector of BC (see figure). Show that ABC is an isosceles triangle in which AB = AC.
(http://dronstudy.com/wpcontent/uploads/2013/12/92.png)Sol.
In ABD and ACD, we have DB = DC [Given]
[since AD BC]AD = AD [Common]Therefore by SAS criterion of congruence, we have.
∠B and ∠C
∠A
Δ
⇒ ∠C = ∠B
⇒ ∠B = ∠C12
12
⇒ ∠OBC = ∠OCB
∠s
∠OBC = ∠B and ∠OCB = ∠C12
12
⇒ ∠s
Δs
∠OBC = ∠OCB
ΔABO ≅Δ ACO
⇒ ∠BAO = ∠CAO
⇒ ∠A
Δ Δ
Δs
∠ADB = ∠ADC ⊥
(http://dronstudy.com/wpcontent/uploads/2013/12/113.png) AB = AC
[Since corresponding parts of congruent triangles are equal]Hence, ABC is isosceles.
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Q.3 ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides ACand AB respectively (see figure). Show that these altitudes are equal.
(http://dronstudy.com/wpcontent/uploads/2013/12/123.png)Sol.
In ABE and ACF, we have [Since Each = 90º] [Common]
and BE = CF [Given]Therefore By AAS criterion of congruence, we have
AB = AC [Since Corresponding parts of congruent triangles are equal]Hence, ABC is isosceles.
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Q.4 ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal(see figure). Show that (i) (ii) AB = AC , i.e., ABC is an isosceles triangle.
Δ ABD ≅ΔACD
⇒
Δ
Δs
∠AEB = ∠AFC
∠BAE = ∠CAF
Δ ABE ≅ Δ ACF
⇒
Δ
Δ ABE ≅ Δ ACF
(http://dronstudy.com/wpcontent/uploads/2013/12/1212.png)Sol.
Let BE AC and CF AB. In ABE and ACF, we have
[Since Each = 90º] [Common]
and , AB = AC [Given]Therefore By AAS criterion of congruence,
BE = CF [Since corresponding parts of congruent triangles are equal]
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Q.5 ABC and DBC are two isosceles triangles on the same base BC (see figure). Showthat .
(http://dronstudy.com/wpcontent/uploads/2013/12/151.png)Sol.
In ABC, we have AB = AC ... (1)
[Since angles opposite to equal sides are equal]In , we have BD = CD ... (2)
[Since angles opposite to equal sides are equal]Adding (1) and (2), we have
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⊥ ⊥Δs
∠AEB = ∠AFC
∠A = ∠A
ΔABE ≅ΔACF
⇒
∠ABD = ∠ACD
Δ
⇒ ∠ABC = ∠ACB
Δ BCD
⇒ ∠DBC = ∠DCB
∠ABC + ∠DBC = ∠ACB + ∠DCB
⇒ ∠ABC = ∠ACD
Δ
Q.6 ABC is an isosceles triangle in which AB = AC. Side BA is produced to Dsuch that AD = AB (see figure). Show that is a right angle.
(http://dronstudy.com/wpcontent/uploads/2013/12/161.png)Sol.
In ABC, we have AB = AC [given] ... (1)
[Since angles opp. to equal sides are equal]Now, AB = AD [Given]Therefore AD = AC [Since AB = AC]Thus , in ADC, we have AD = AC ... (2)
[Since angles opp. to equal sides are equal]Adding (1) and (2) , we get
[Since ] [Adding on both sides] [Angle sum proprty] 90º
Hence , is a right angle.
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Q.7 ABC is a right angled triangle in which 90º and AB = AC. Find .
Sol.
We have, 90º
AB = AC
(http://dronstudy.com/wpcontent/uploads/2013/12/191.png)[Since angles opp. to equal sides ofa triangle are equal]Also 180º [Angle sum property]
Δ∠BCD
Δ
⇒ ∠ACB = ∠ABC
Δ
⇒ ∠ACD = ∠ADC
∠ACB + ∠ACD = ∠ABC + ∠ADC
⇒ ∠BCD = ∠ABC + ∠BDC ∠ADC = ∠BDC
⇒ ∠BCD + ∠BCD = ∠ABC + ∠BDC + ∠BCD ∠BCD
⇒ 2∠BCD = 180∘
⇒ ∠BCD =∠BCD
∠A =∠B and ∠C
∠A =
⇒ ∠C = ∠B
∠A + ∠B + ∠C =
90º + 180º [Since ] 180º – 90º = 90º
Therefore
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Q.8 Show that the angles of an equilateral triangle are 60º each.Sol.
Let ABC be an equilateral triangle so that AB = AC = BC. Now Since AB = AC ... (1) [Since angles opp. to equal sides are equal]
Also since, CB = CA .... (2) [Since angles opp. to equal sides are equal]
From (1) and (2), we have
(http://dronstudy.com/wpcontent/uploads/2013/12/181.png)Also 180º [Angle sum property]Thererfore 180º 180º
60ºTherefore 60ºThus , each angle of an equilateral triangle is 60º.
⇒ 2∠B = ∠C = ∠B
⇒ 2∠B =⇒ ∠B = =90o
245o
∠C = ∠B = 45o
Δ
⇒ ∠C = ∠B
⇒ ∠A = ∠B
∠A = ∠B = ∠C
∠A + ∠B + ∠C =∠A + ∠A + ∠A =
⇒ 3∠A =⇒∠A = 180∘
3⇒∠A =
∠A = ∠B = ∠C =
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