Dear Student, In isosceles triangle ABC, AB = ACand AD is the altitude on side BC The altitude to the base of an isoscelestriangle bisects the base of the triangle Therefore, BD = CD Area of triangle = 12 × Base × HeightAreas of triangle ABC = 12 × BC × AD48 = 12 × BC × 848 = BC × 4BC = 484BC = 12 cmNow, BD = CD = BC2 = 122 = 6cmSo, in right triangle ABD,AB2 = AD2 + BD2 (Using Pythagoras Theorem)AB2 = (8)2 + (6)2AB2 = 64 + 36AB2 = 100AB = 100AB = 10 cmSo, AC = AB = 10cmPerimeter of triangle = Sum of all sides = AB + AC + BC = 10 + 10 + 12 = 32cmThe primeter of triangle ABC is 32 cm Regards

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Dear Student,

I n i s o s c e l e s t r i a n g l e A B C, A B = A C a n d A D i s t h e a l t i t u d e o n s i d e B C . T h e a l t i t u d e t o t h e b a s e o f a n i s o s c e l e s t r i a n g l e b i s e c t s t h e b a s e o f t h e t r i a n g l e . T h e r e f o r e, B D = C D A r e a o f t r i a n g l e = \frac{1}{2} \times B a s e \times H e i g h t A r e a s o f t r i a n g l e A B C = \frac{1}{2} \times B C \times A D 48 = \frac{1}{2} \times B C \times 8 48 = B C \times 4 B C = \frac{48}{4} B C = 12 c m N o w, B D = C D = \frac{B C}{2} = \frac{12}{2} = 6 c m S o, i n r i g h t t r i a n g l e A B D, A B^{2} = A D^{2} + B D^{2} (U \sin g P y t h a g o r a s T h e o r e m) A B^{2} = (8)^{2} + (6)^{2} A B^{2} = 64 + 36 A B^{2} = 100 A B = \sqrt{100} A B = 10 c m S o, A C = A B = 10 c m P e r i m e t e r o f t r i a n g l e = S u m o f a l l s i d e s = A B + A C + B C = 10 + 10 + 12 = 32 c m T h e p r i m e t e r o f t r i a n g l e A B C i s 32 c m .

Regards

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