Math, asked by Anonymous, 1 month ago

→ AB = 9cm AC = 9cm BC = 10cm
(a) Calculate the area, in cm2
to one decimal place, of triangle ABC.

Answers

Answered by HolyGirl
1

Given:

AB = 9 cm, AC = 9 cm, BC = 10 cm

To Find:

Area of ∆ ABC

Solution:

 \boxed{Area of ∆ ABC = \large \frac{ height × base}{2}}

So, we will draw a perpendicularly line from the vertex A of ∆ ABC, namely AM.

Since, AB = AC, AM will divide BC into 2 equal parts, i.e., BM = CM.

→ BM = CM = BC/2 = 10/2 cm

→ BM = CM = 5 cm

So, in ∆ ABM,

 \angle AMB = 90°

AB = 9 cm, BM = 5 cm

 \\

Using Pythagoras Theorem,

AB² = AM² + BM²

9² = AM² + 5²

81 = AM² + 25

AM² = 81 - 25

AM² = 56

AM =  \sqrt{56} \: cm

AM = 7.48 cm

 \\ \\

Therefore, height (h) = AM = 7.48 cm

and, base (b) = BC = 10 cm

Area of ∆ ABC = \Large \frac{height × base}{2}

 = \Large \frac{7.48×10}{2}

 = \Large \frac {\: ^{^{3.74}}\: \cancel {7.48} × 10}{\cancel 2}

 = \boxed{37.4 \: cm^2} \\

Hence, Area of \triangle ABC is 37.4 cm².

Attachments:
Answered by vikashpatnaik2009
0

Answer:

Given:

AB = 9 cm, AC = 9 cm, BC = 10 cm

To Find:

Area of ∆ ABC

Solution:

 

So, we will draw a perpendicularly line from the vertex A of ∆ ABC, namely AM.

Since, AB = AC, AM will divide BC into 2 equal parts, i.e., BM = CM.

→ BM = CM = BC/2 = 10/2 cm

→ BM = CM = 5 cm

So, in ∆ ABM,

AB = 9 cm, BM = 5 cm

Using Pythagoras Theorem,

AB² = AM² + BM²

9² = AM² + 5²

81 = AM² + 25

AM² = 81 - 25

AM² = 56

AM =  cm

AM = 7.48 cm

Therefore, height (h) = AM = 7.48 cm

and, base (b) = BC = 10 cm

Area of ∆ ABC =  

Hence, Area of ABC is 37.4 cm².

Step-by-step explanation:

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