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Let ABC be a right-angled triangle …

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Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.

Posted by Shashikant ? 6 years, 11 months ago

CBSE > Class 10 > Mathematics

6 answers

Shashikant ? 6 years, 11 months ago

Dear Puja Sahoo, area of A(6,3), B(-3,5) and C(4,-2) will be 1/2 * {6(5+2) -3(-2-3)+4(3-5)} = 1/2 * (42 + 15 - 8) = 1/2 * 49 sq units Now area of triangle PBC with (x,y) (-3,5) (4,-2) will be 1/2 * {x(5+2) -3(-2-y)+4(y-5)} = 1/2 * (7x + 7y - 14) = 7/2 * (x+y-2) sq units Hence area ? PBC / area ? ABC = 7/2 * (x + y - 2) / 49/2 So required ratio = (x + y - 2) / 7 ]proved]

Puja Sahoo? 6 years, 11 months ago

Naaa........repeat the question again......

Shashikant ? 6 years, 11 months ago

Nahi...kal ke question ka answer hai. 100m distance..okkk

Puja Sahoo? 6 years, 11 months ago

Agar, solve ho jaye ,to plz ap bata dena........

Puja Sahoo? 6 years, 11 months ago

Apke, kl ke question ka answer hai, ................-200 root 3/3. Correct hai na.....

Puja Sahoo? 6 years, 11 months ago

The coordinates of A, B, C are (6,3),(-3,5)&(4,-2)respectively and P is any point (x, y). show that the ratio of the areas of triangle PBC and ABC is [x+y-2/7]..............solve this.....

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