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.

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 = $\dfrac{24}{2}$ = 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 = $\sqrt{225}$

⇒ 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 = $\dfrac{1}{2}$ × Base × Height

= $\dfrac{1}{2}$ × BC × AD

= $\dfrac{1}{2}$ × 24 × 9

= 12 × 9

= 108 cm².

Hence, area of triangle ABC = 108 cm².