In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC.
Answers
SOLUTION :
GIVEN : ∆ABC is an isosceles ∆ in which AB = AC = 13cm and altitude AD = 5 cm.
In ∆ADB, by Pythagoras theorem
AB² = AD² + BD²
13² = 5² + BD²
169 = 25 + BD²
169 − 25 = BD²
BD² = 144
BD = √144
BD = 12 cm
In ∆ ADB & ∆ADC,
∠ADB = ∠ADC [Each 90°]
AB = AC (GIVEN)
AD = AD [Common]
∆ADB ≅ ∆ADC [By RHS criterion]
∴ BD = CD [By c.p.c.t]
BC = BD + CD
BC = 12 + 12
BC = 24 cm
Hence, the length of BC is 24 cm.
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Answer:
Step-by-step explanation:
Given :-
∆ABC is an isosceles.
AB = AC = 13 cm and altitude AD = 5 cm.
To Find :-
BC
Solution :-
In ∆ADB, By applying Pythagoras theorem , we get
AB² = AD² + BD²
⇒ 13² = 5² + BD²
⇒ 169 = 25 + BD²
⇒ 169 − 25 = BD²
⇒ BD² = 144
⇒ BD = √144
⇒ BD = 12 cm
Now, In ∆ ADB & ∆ADC,
∠ADB = ∠ADC [Each 90°]
AD = AD [Common]
AB = AC (GIVEN)
∆ADB ≅ ∆ADC [By RHS criterion]
∴ BD = CD [By c.p.c.t]
BC = BD + CD = 12 + 12 = 24 cm
Hence, the length of BC is 24 cm.
