Advertisements
Advertisements
Question
If the sides of a triangle are 3 cm, 4 cm, and 6 cm long, determine whether the triangle is a right-angled triangle.
Solution
We have,
Sides of triangle
AB = 3 cm
BC = 4 cm
AC = 6 cm
∴ AB2 = 32 = 9
BC2 = 42 = 16
AC2 = 62 = 36
Since, AB2 + BC2 ≠ AC2
Then, by converse of Pythagoras theorem, triangle is not a right triangle.
APPEARS IN
RELATED QUESTIONS
Construct a triangle ABC with sides BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct a triangle whose sides are `3/4` times the corresponding sides of ∆ABC.
The sides of triangle is given below. Determine it is right triangle or not.
a = 7 cm, b = 24 cm and c = 25 cm
The sides of triangle is given below. Determine it is right triangle or not.
a = 9 cm, b = l6 cm and c = 18 cm
A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.
The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus.
In ∆ABC, ∠A is obtuse, PB ⊥ AC and QC ⊥ AB. Prove that:
(i) AB ✕ AQ = AC ✕ AP
(ii) BC2 = (AC ✕ CP + AB ✕ BQ)
∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that
(i) AB2 = BC x BD
(ii) AC2 = BC x DC
(iii) AD2 = BD x CD
(iv) `"AB"^2/"AC"^2="BD"/"DC"`
State the converse of Pythagoras theorem.
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 side and perimeter of a square whose diagonal is `13sqrt2` cm.
