Mrs. Sharmila, a mathematics teacher of class 9 started a very important geometry chapter on Congruency in triangles and decided to test the previous concept by setting a geometry problem on the blackboard. In △ ABC, AD is drawn as perpendicular bisector of BC. It is given that AD = 9 cm and BC = 24 cm.
Based on the above information answer the following:
(i) Prove △ ABD ≅ △ ACD
(ii) Assign a special name to △ ABC
(iii) Calculate perimeter of △ABC.
(iv) Calculate Area of △ABC.

Question

Mrs. Sharmila, a mathematics teacher of class 9 started a very important geometry chapter on Congruency in triangles and decided to test the previous concept by setting a geometry problem on the blackboard. In △ ABC, AD is drawn as perpendicular bisector of BC. It is given that AD = 9 cm and BC = 24 cm.

Based on the above information answer the following:

(i) Prove △ ABD ≅ △ ACD

(ii) Assign a special name to △ ABC

(iii) Calculate perimeter of △ABC.

(iv) Calculate Area of △ABC.

KnowledgeBoat · Accepted Answer

Given, In △ABC, AD is the perpendicular bisector of BC. This means ∠ADB = ∠ADC = 90 . D is the mid-point of BC, so BD = DC. AD = 9 cm BC = 24 cm. (i) In △ABD and △ACD: ⇒ BD = CD (Since AD is the perpendicular bisector of BC; BD = DC = = 12 cm) ⇒ ∠ADB = ∠ADC = 90 (Given AD ⊥ BC) ⇒ AD = AD (Common side). ∴ △ABD ≅ △ACD (By S.A.S axiom) Hence proved that △ABD ≅ △ACD. (ii) Given, BD = DC Since △ABD ≅ △ACD, their corresponding parts are equal. (C.P.C.T) ∴ AB = AC. So, two sides of a triangle are equal, ∴ It is an isosceles triangle. Hence, △ABC is an isosceles triangle. (iii) In △ABD using Pythagoras theorem, ⇒ AB2 = AD2 + BD2 ⇒ AB2 = 92 + 122 ⇒ AB2 = 81 + 144 ⇒ AB2 = 225 ⇒ AB = ⇒ AB = 15 cm. ∴ AB = AC = 15 cm. Perimeter = Sum of all sides of a triangle. = AB + BC + CA = 15 + 24 + 15 = 54 cm. Hence, perimeter of triangle ABC = 54 cm. (iv) Area of triangle = × Base × Height = × BC × AD = × 24 × 9 = 12 × 9 = 108 cm2. Hence, area of triangle ABC = 108 cm2.

Mathematics

Triangles

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