Advertisements
Advertisements
प्रश्न
Show that in a right angled triangle, the hypotenuse is the longest side.
Advertisements
उत्तर
Let us consider a right-angled triangle ABC, right-angled at B.
In ΔABC,
∠A + ∠B + ∠C = 180° ...(Angle sum property of a triangle)
∠A + 90º + ∠C = 180°
∠A + ∠C = 90°
Hence, the other two angles have to be acute i.e., less than 90º.
∴ ∠B is the largest angle in ΔABC.
⇒ ∠B > ∠A and ∠B > ∠C
⇒ AC > BC and AC > AB
In any triangle, the side opposite to the larger (greater) angle is longer.
Therefore, AC is the largest side in ΔABC.
However, AC is the hypotenuse of ΔABC.
Therefore, hypotenuse is the longest side in a right-angled triangle.
APPEARS IN
संबंधित प्रश्न
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
From the following figure, prove that: AB > CD.
In the following figure, write BC, AC, and CD in ascending order of their lengths.
“Caste inequalities are still prevalent in India.” Examine the statement.
"Issues of caste discrimination began to be written about in many printed tracts and essays in India in the late nineteenth century." Support the statement with two suitable examples.
Arrange the sides of the following triangles in an ascending order:
ΔDEF, ∠D = 38°, ∠E = 58°.
In a triangle ABC, BC = AC and ∠ A = 35°. Which is the smallest side of the triangle?
Prove that the perimeter of a triangle is greater than the sum of its three medians.
D is a point on the side of the BC of ΔABC. Prove that the perimeter of ΔABC is greater than twice of AD.
For any quadrilateral, prove that its perimeter is greater than the sum of its diagonals.
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that : PN < RN
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that: SN < SR
In the given figure, T is a point on the side PR of an equilateral triangle PQR. Show that RT < QT
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
In ΔABC, AE is the bisector of ∠BAC. D is a point on AC such that AB = AD. Prove that BE = DE and ∠ABD > ∠C.
