Question

In triangle abc angle abc is equal to 90 degrees ad is equals to DC ab is equals to 12 CM and BC is equal to 6.5 CM find the area of triangle abd

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

Area of Triangle ABD = 19.5 $cm^{2}$

Step-by-step explanation:

Given,

∠ABC = 90°

AB = 12

BC = 6.5

In triangle ABC, using Pythagoras theorem:

    $AC^{2} =AB^{2} +BC^{2}$

⇒ $AC^{2} = (12)^{2} +(6.5)^{2}$

⇒ $AC^{2} = 144 + 42.25$

⇒ $AC^{2} = 186.25$

∴ AC = 13.6 cm

Now, given

 AD =  DC = $\frac{AC}{2}$  =  $\frac{13.6}{2}$  = 6.8 cm

A perpendicular line from D to AD, such that ED is parallel to BC

 ∴ ∠AED = ∠ABC = 90°

Since,  D is the mid point of AC and ED is parallel to BC

Applying mid-point theorem:

ED = $\frac{BC}{2}$  = $\frac{6.5}{2}$  = 3.25 cm

Now ,

In triangle ABD,

Height (h) = ED

Base (b) = AB

Therefore, Area of Triangle ABD =  $\frac{1}{2}$ × b × h

                                                       = $\frac{1}{2}$ ×  12 ×  3.25

                                                       = 19.5 $cm^{2}$