Question

ch7 triangles answer of class9 exeecise 7.2 all answers If any answer i will mark as brainliest

Answer 1

Q.1     In an isosceles triangle ABC, with AB = AC, the bisectors of ∠Band∠C intersect each other at O. Join A to O. Show that : 
              (i) OB = OC               (ii) AO bisects ∠A.
Sol.

(i) In Δ ABC, we have 
AB = AC
⇒ ∠C=∠B [Since angles opposite to equal sides are equal]

⇒ 12∠B=12∠C
⇒ ∠OBC=∠OCB 
[Since OB and OC bisect ∠s B and C respectively. Therefore ∠OBC=12∠Band∠OCB=12∠C]
⇒ OB = OC [Since sides opp. to equal ∠s are equal]
(ii) Now, in Δs ABO and ACO , we have 
AB = AC [Given]
∠OBC=∠OCB [From (1)]
OB = OC [From (2)]
Therefore By SAS criterion of congruence, we have 
ΔABO≅ΔACO
⇒ ∠BAO=∠CAO
[Since corresponding parts of congruent triangles are equal]
⇒ AO bisects ∠A.H

Q.2      In Δ ABC, AD is the perpendicular bisector of BC (see figure). Show that Δ ABC is an isosceles triangle in which AB = AC.

Sol.

In Δs ABD and ACD, we have 
DB = DC [Given]
∠ADB=∠ADC [since AD ⊥BC]
AD = AD [Common]
Therefore by SAS criterion of congruence, we have.

ΔABD≅ΔACD
⇒ AB = AC
[Since corresponding parts of congruent triangles are equal]
Hence, Δ ABC is isosceles.

Q.3      ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively (see figure). Show that these altitudes are equal.

Sol.

In Δs ABE and ACF, we have 
∠AEB=∠AFC [Since Each = 90º]
∠BAE=∠CAF [Common]
and BE = CF [Given]
Therefore By AAS criterion of congruence, we have 
ΔABE≅ΔACF
⇒ AB = AC 
[Since Corresponding parts of congruent triangles are equal]
Hence, Δ ABC is isosceles.

Q.4      ABC is a triangle in 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.

Sol.

Let BE ⊥ AC and CF ⊥ AB. 
In Δs ABE and ACF, we have 
∠AEB=∠AFC [Since Each = 90º]
∠A=∠A [Common]
and , AB = AC [Given]
Therefore By AAS criterion of congruence, 
ΔABE≅ΔACF
⇒ BE = CF 
[Since corresponding parts of congruent triangles are equal]

Q.5    ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that ∠ABD=∠ACD.

Sol.

In Δ ABC, we have 
AB = AC 
⇒ ∠ABC=∠ACB ... (1) 
[Since angles opposite to equal sides are equal]
In ΔBCD, we have 
BD = CD 
⇒ ∠DBC=∠DCB ... (2) 
[Since angles opposite to equal sides are equal]
Adding (1) and (2), we have 
∠ABC+∠DBC=∠ACB+∠DCB
⇒ ∠ABC=∠ACD


Q.6      Δ ABC 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.

Sol.

In Δ ABC, we have 
AB = AC    [given]
⇒ ∠ACB=∠ABC ... (1) 
[Since angles opp. to equal sides are equal]
Now, AB = AD [Given]
Therefore AD = AC [Since AB = AC]
Thus , in Δ ADC, we have 
AD = AC 
⇒ ∠ACD=∠ADC ... (2) 
[Since angles opp. to equal sides are equal]
Adding (1) and (2) , we get 
∠ACB+∠ACD=∠ABC+∠ADC
⇒ ∠BCD=∠ABC+∠BDC[Since ∠ADC=∠BDC]
⇒ ∠BCD+∠BCD=∠ABC+∠BDC+∠BCD  [Adding ∠BCD on both sides]
⇒ 2∠BCD=180∘ [Angle sum proprty]
⇒ ∠BCD= 90º
Hence , ∠BCD is a right angle.

Q.7     ABC is a right angled triangle in which ∠A= 90º and AB = AC. Find ∠Band∠C.
Sol.

We have, 
∠A= 90º
AB = AC 
⇒ ∠C=∠B
[Since angles opp. to equal sides of a triangle are equal]
Also ∠A+∠B+∠C= 180º [Angle - sum property]
⇒ 90º + 2∠B= 180º [Since ∠C=∠B]
⇒ 2∠B= 180º – 90º = 90º
⇒ ∠B=90o2=45o
Therefore ∠C=∠B=45o

Q.8      Show that the angles of an equilateral triangle are 60º each.
Sol.

Let Δ ABC be an equilateral triangle so that AB = AC = BC. 
Now Since AB = AC 
⇒ ∠C=∠B ... (1)  [Since angles opp. to equal sides are equal]
Also since, CB = CA 
⇒ ∠A=∠B .... (2)  [Since angles opp. to equal sides are equal]
From (1) and (2), we have

∠A=∠B=∠C
Also ∠A+∠B+∠C= 180º [Angle - sum property]
Thererfore ∠A+∠A+∠A=180º
⇒ 3∠A= 180º
⇒∠A=180∘3
⇒∠A= 60º
Therefore ∠A=∠B=∠C=60º
Thus , each angle of an equilateral triangle is 60º.

Answer 2

Step-by-step explanation:

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