Question

9. From the adjoining figure, find
(i) the area of AABC
(ii) length of BC
(iii) the length of altitude from A to BC.​

Answer 1

Answer:

1. Area of ABC= 1/2×base×height

= 1/2×3×4

= 12cm²

2. (AB)²+(AC)²=(BC)²

3²+4²=(BC)²

9+16=BC²

Root 25=BC

5cm =BC

3. AREA=1/2×base×height

12 =1/2×5×height

12×2/5= height

4.8cm =height

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Answer 2

Step-by-step explanation:

$(i) area \: of \: triangle \: abc \: = \frac{1}{2} \times b \times h \\ = \frac{1}{2} \times 3 \times 4 \\ 6 {cm}^{2}$

$(ii) {h}^{2} = {b}^{2} + {p}^{2} \\ = {3}^{2} + {4}^{2} \\ = 9 + 16 \\ = 25 \\ {h}^{2} = 25 \\ h = 5cm$

$(iii)area \: of \: triangle = \frac{1}{2} \times b \times h \\ 6 = \frac{1}{2} \times 3 \times h \\ h = 6 \times \frac{2}{3} \\ h = 4cm$