Advertisements
Advertisements
प्रश्न
In a quadrilateral ABCD, ∠B = 90°, AD2 = AB2 + BC2 + CD2, prove that ∠ACD = 90°.
उत्तर
We have, ∠B = 90° and AD2 = AB2 + BC2 + CD2
∴ AD2 = AB2 + BC2 + CD2 [Given]
But AB2 + BC2 = AC2 [By pythagoras theorem]
Then, AD2 = AC2 + CD2
By converse of by pythagoras theorem
∠ACD = 90°
APPEARS IN
संबंधित प्रश्न
If the sides of a triangle are 3 cm, 4 cm, and 6 cm long, determine whether the triangle is a right-angled triangle.
In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC.
Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Using Pythagoras theorem determine the length of AD in terms of b and c shown in Figure.
The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus.
Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm find the length of the other diagonal.
In right-angled triangle ABC in which ∠C = 90°, if D is the mid-point of BC, prove that AB2 = 4AD2 − 3AC2.
If D, E, F are the respectively the midpoints of sides BC, CA and AB of ΔABC. Find the ratio of the areas of ΔDEF and ΔABC.
The co-ordinates of the points A, B and C are (6, 3), (−3, 5) and (4, −2) respectively. P(x, y) is any point in the plane. Show that \[\frac{ar\left( ∆ PBC \right)}{ar\left( ∆ ABC \right)} = \left| \frac{x + y - 2}{7} \right|\]
Find the altitude of an equilateral triangle of side 8 cm.
