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प्रश्न
Lengths of sides of a triangle are 3 cm, 4 cm and 5 cm. The triangle is ______.
विकल्प
Obtuse angled triangle
Acute-angled triangle
Right-angled triangle
An Isosceles right triangle
उत्तर
Lengths of sides of a triangle are 3 cm, 4 cm and 5 cm. The triangle is right-angled triangle.
Explanation:
Lengths of the sides of a triangle are 3 cm, 4 cm and 5 cm.
Now, 32 + 42 = 9 + 16 = 25 = 52
i.e., Sum of squares of two sides is equal to the square of third side.
Therefore, the triangle is right angled triangle.
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संबंधित प्रश्न
In figure, ∠B of ∆ABC is an acute angle and AD ⊥ BC, prove that AC2 = AB2 + BC2 – 2BC × BD
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM . MR
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(A)\[7 + \sqrt{5}\]
(B) 5
(C) 10
(D) 12
For finding AB and BC with the help of information given in the figure, complete following activity.
AB = BC .......... 
∴ ∠BAC =
∴ AB = BC = × AC
= × `sqrt8`
= × `2sqrt2`
=
M andN are the mid-points of the sides QR and PQ respectively of a PQR, right-angled at Q.
Prove that:
(i) PM2 + RN2 = 5 MN2
(ii) 4 PM2 = 4 PQ2 + QR2
(iii) 4 RN2 = PQ2 + 4 QR2(iv) 4 (PM2 + RN2) = 5 PR2
Prove that `(sin θ + cosec θ)^2 + (cos θ + sec θ)^2 = 7 + tan^2 θ + cot^2 θ`.
A man goes 10 m due east and then 24 m due north. Find the distance from the straight point.
A ladder 15m long reaches a window which is 9m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to other side of the street to reach a window 12m high. Find the width of the street.
Is the triangle with sides 25 cm, 5 cm and 24 cm a right triangle? Give reasons for your answer.
