Advertisements
Advertisements
प्रश्न
The sides of the triangle are given below. Find out which one is the right-angled triangle?
40, 20, 30
Advertisements
उत्तर
It is known that, if in a triplet of natural numbers, the square of the biggest number is equal to the sum of the squares of the other two numbers, then the three numbers form a Pythagorean triplet. If the lengths of the sides of a triangle form such a triplet, then the triangle is a right-angled triangle.
The sides of the given triangle are 40, 20, and 30.
Let us check whether the given set (40, 20, 30) forms a Pythagorean triplet or not.
The biggest number among the given set is 40.
(40)2 = 1600; (20)2 = 400; (30)2 = 900
Now, 400 + 900 = 1300 ≠ 1600
∴ (20)2 + (30)2 ≠ (40)2
Thus, (40, 20, 30) does not form a Pythagorean triplet.
Hence, the given triangle with sides 40, 20, and 30 is not a right-angled triangle.
संबंधित प्रश्न
From a point O in the interior of a ∆ABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove
that :
`(i) AF^2 + BD^2 + CE^2 = OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2`
`(ii) AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2`
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

`"AC"^2 = "AD"^2 + "BC"."DM" + (("BC")/2)^2`
In the given figure, point T is in the interior of rectangle PQRS, Prove that, TS2 + TQ2 = TP2 + TR2 (As shown in the figure, draw seg AB || side SR and A-T-B)

Digonals of parallelogram WXYZ intersect at point O. If OY =5, find WY.
A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.
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 a rectangle ABCD,
prove that: AC2 + BD2 = AB2 + BC2 + CD2 + DA2.
In ΔABC, AD is perpendicular to BC. Prove that: AB2 + CD2 = AC2 + BD2
Find the distance between the helicopter and the ship
