which of the following can be sides of a right triangle -2.5cm,6.5cm,6cm ans in solution
Answers
Answer:
: In a right triangle, the longest side is the hypotenuse. We also know that according to Pythagoras theorem:
Hypotenuse2 = Base2 + Perpendicular2
Let us check if the given three sides fulfill the criterion of Pythagoras theorem.
25
2
=
24
2
+
7
2
Or,
625
=
576
+
49
Or,
625
=
625
Here; LHS = RHS
Hence; this is a right triangle.
(b) 3 cm, 8 cm, 6 cm
Solution: Let us check if the given sides fulfill the criterion of Pythagoras theorem.
8
2
=
6
2
+
3
2
Or,
64
=
36
+
9
Or,
64
≠
45
Here; LHS is not equal to RHS
Hence; this is not a right triangle.
(c) 50 cm, 80 cm, 100 cm
Solution: Let us check if the given sides fulfill the criterion of Pythagoras theorem.
100
2
=
80
2
+
50
2
Or,
10000
=
6400
+
2500
Or,
10000
≠
8900
Here; LHS is not equal to RHS
Hence; this is not a right triangle.
(d) 13 cm, 12 cm, 5 cm
Solution: Let us check if the given sides fulfill the criterion of Pythagoras theorem.
13
2
=
12
2
+
5
2
Or,
169
=
144
+
25
Or,
169
=
169
Here; LHS = RHS
Hence; this is a right triangle.
Question 2: PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM. MR.
similar triangles exercise solution
Solution: In triangles PMQ and RMP
∠ PMQ = ∠ RMP (Right angle)
∠ PQM = ∠ RPM (90 – MRP)
Hence; PMQ ∼ RMP (AAA criterion)
So,
P
M
Q
M
=
M
R
P
M
Or,
P
M
2
=
Q
M
.
M
R
proved
Question 3: In the given figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
similar triangles exercise solution
(a)
(
A
B
)
2
=
(
B
C
)
×
(
B
D
)
Solution: In triangles ACB and DAB
∠ ACB = ∠ DAB (Right angle)
∠ CBA = ∠ ABD (common angle)
Hence; ACB ∼ DAB
So,
A
B
B
C
=
B
D
A
B
Or,
A
B
2
=
B
D
.
B
C
proved
(b) AC2 = BC. DC
Solution: In triangles ACB and DCA
∠ ACB = ∠ DCA (right angle)
∠ CBA = ∠ CAD
Hence; ACB ~ DCA
So,
A
C
B
C
=
D
C
A
C
Or,
A
C
2
=
B
D
.
C
D
proved
(c) AD2 = BD. CD
Solution: In triangles DAB and DCA
∠ DAB = ∠ DCA (right angle)
∠ ABD = ∠ CAD
Hence; DAB ∼ DCA
So,
A
D
B
D
=
C
D
A
D
Or,
A
D
2
=
B
D
.
C
D
proved