Advertisements
Advertisements
प्रश्न
Find the Pythagorean triplet from among the following set of numbers.
9, 40, 41
उत्तर
The given set of numbers is (9, 40, 41).
We use the Pythagorean theorem:
Hypotenuse2 = Base2 + Height2
∴ 412 = 92 + 402
1681 = 81 + 1600
∴ 412 = 1681
Since 412= 1681, the Pythagorean theorem holds true.
Thus, (9, 40, 41) forms a Pythagorean triplet.
संबंधित प्रश्न
Prove that the diagonals of a rectangle ABCD, with vertices A(2, -1), B(5, -1), C(5, 6) and D(2, 6), are equal and bisect each other.
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
In the following figure, OP, OQ, and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC.
Prove that: AR2 + BP2 + CQ2 = AQ2 + CP2 + BR2
In triangle PQR, angle Q = 90°, find: PR, if PQ = 8 cm and QR = 6 cm
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels at a speed of `(20 "km")/"hr"` and the second train travels at `(30 "km")/"hr"`. After 2 hours, what is the distance between them?
An isosceles triangle has equal sides each 13 cm and a base 24 cm in length. Find its height
From given figure, In ∆ABC, If AC = 12 cm. then AB =?
Activity: From given figure, In ∆ABC, ∠ABC = 90°, ∠ACB = 30°
∴ ∠BAC = `square`
∴ ∆ABC is 30° – 60° – 90° triangle
∴ In ∆ABC by property of 30° – 60° – 90° triangle.
∴ AB = `1/2` AC and `square` = `sqrt(3)/2` AC
∴ `square` = `1/2 xx 12` and BC = `sqrt(3)/2 xx 12`
∴ `square` = 6 and BC = `6sqrt(3)`
In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
(i) `"AC"^2 = "AD"^2 + "BC"."DM" + (("BC")/2)^2`
(ii) `"AB"^2 = "AD"^2 - "BC"."DM" + (("BC")/2)^2`
(iii) `"AC"^2 + "AB"^2 = 2"AD"^2 + 1/2"BC"^2`
In a triangle, sum of squares of two sides is equal to the square of the third side.
If the hypotenuse of one right triangle is equal to the hypotenuse of another right triangle, then the triangles are congruent.
