Question

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

$\Large{\underline{\underline{\tt{\purple{Answer !!}}}}}$

i) 6cm²

ii) 5cm

iii) 1.65 cm

$\Large{\underline{\underline{\tt{\purple{solution!!}}}}}$

$We \: know \: that \: Area \: of \: triangle= \frac{1}{2} ×Base×Height.$

● i) Area of ️△ABC

BASE = AB = 3cm

HEIGHT = AC = 4cm

▪︎Area of ️ABC = 1/2 × 3 × 4 = 6 cm²

● ii) We can apply Pythagoras theorem in △ABC and get

BC² = AC² + AB²

BC² = 4² + 3²

BC² = 16 + 9

BC² = $\sqrt{25}$

BC = 5cm

● iii) We can apply Pythagoras theorem in △ABD we get

AB² = AD² + BD²

AD² = AB² - BD²

AD² = 3²- BD²

AD² = 9 - BD²....eq(1)

▪︎By applying Pythagoras theorem in ️△ACD we get

AC² = AD² + CD²

AD² = AC² - CD²

AD² = AC² - (BC - BD)²...............(BC = BD + CD)

AD² = AC² - (BC² + BD² - 2 BC × BD)

AD² = AC² - BC² - BD² + 2 BC × BD

AD² = 3² - 5² - BD² + 2 × 5 × BD

AD² = 9 - 25 - BD² + 10BD

AD² = -16 - BD² + 10BD

▪︎substituting eq(1) in eq (2)

9 - BD² = - 16 - BD² + 10BD

9 = -16 + 10BD

10BD = 16 + 9

10BD = 25

BD = 2.5cm

▪︎substituting value of BD in eq(1) , we get

AD² = 9 - BD²

AD² = 9 - 2.5²

AD² = 9 - 6.25

AD² = 2.75

▪︎Therefore,

AD = 1.65cm