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Question
In Fig. below, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D,
E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.
Solution
In right ΔABC, ∠B = 90°
By using Pythagoras theorem
`AC^2 = AB^2+ BC^2`
⇒ `15^2 = 9^2 +BC^2`
⇒ BC =`sqrt(15^2 - 9^2)`
⇒ BC =`sqrt(225-81)`
⇒ BC =`sqrt144`
= 12cm
In ΔABC
D and E are midpoints of AB and AC
∴ DE || BC, DE = `1/2` BC [By midpoint theorem]
AD = OB = `(AB)/ 2= 9/2` = 4 . 5cm [ ∵ D is the midpoint of AB]
DE = `(BC)/2 = 12/2` = 6cm
Area of ΔADE = `1/2 xxAD xx DE `
= `1/2× 4 .5 × 6 = 13.5cm^2`
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