Question

anyone can help me
solve this question​

Answer 1

In ∆ALB, angle ALB = 90°

using Pythagoras' theorem;

AB² = AL² + BL²

BL² = AB² - AL²

BL² = (6.5)² - (2.5)²

BL² = 42.25 - 6.25

BL² = 36

BL = √36

BL = 6 cm

now,

In ∆ALC, angle ALC = 90°

using Pythagoras' theorem;

AC² = AL² + LC²

LC² = AC² - AL²

LC² = (3.4)² - (2.5)²

LC² = 11.56 - 6.25

LC² = 5.31

LC = √5.31

BC = BL + LC = (6 + √5.31) cm

ar(||gm ABCD)

=> ½ x AB x CM = ½ x BC x AL

$cm \\ =\frac{bc \times al}{ab} \\ = \frac{(2.5)(6 + \sqrt{5.31} )}{6.5} \\ multiplying \: both \: numerator \: and \\ denominator \: by \: 10 \\ = \frac{25(6 + \sqrt{5.31} )}{65} \\ = \frac{5(6 + \sqrt{5.31} )}{13} \\ = \frac{30 + 5 \sqrt{5.31} }{13}$

Hence, length of the altitude from C is

$\frac{30 + \sqrt{5.31} }{13} \: cm$