Question

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Answer 1

Answer:

Here is your answer mate ❤️

According to Pythagoras theorem

BC = (4^2+3^2)^1/2

BC = (16+9)^1/2

BC = 5

As D is mid point

CD = 2.5 BD =2.5

IN TRIANGLE ABD

WE KNOW AD=(AB^2-BD^2)^1/2

AD= (16-6.25)^1/2

AD =9.75^1/2

AB= 3.12

Area = 1/2 × base × height

Area =1/2×BD×AD

=1/2 × 3.12×2.5

=1.46×2.5

=3.65 cm^2

Have a good day ❤️

Area is 3.65 sqcm

Answer 2

Given :-

• AC = 3 cm
• AB = 4 cm
• D is the midpoint of CB

To Find :-

• Area of right triangle ∆ ABD

Explanation :-

First of all we need a base for triangle ∆ABC so, the base of this triangle = CB

Now, By Pythagoras theorem we can Find CB

Hence,

CB² = AC² + AB²

CB² = (3)² + (4)²

CB² = 9 + 16

CB² = 25

CB = √25

CB = 5 cm

Now For Area of traingle ∆ABD

We need base of traingle ∆ABD that's BD

we know that D is the midpoint of CB

ADB = 90°

CD = DB

so,

CB = CD + DB

5 = 2 CB

5/2 = CB

CB = 2.5 cm = Base

hypotenuse = 4 cm = AB

$\tt Area \: \: of \: \: traingle = \frac{1}{2} \times base \times hypotenuse \\ \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2} \times CB × AB\\ \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2} \times 2.5 \times 4 \\ \\ \tt \: \: \: \: \: \: \: = 2.5 \times 2 \\ \\ \tt \: \fbox{ \red{ = 5 \: cm²}}$

I hope it helps you ❤️✔️