In Fig. 9.22 ∆ MNO is a right-angled triangle. Its legs are 6 cm and 8 cm long. Length of perpendicular NP on the side MO is
(a) 4.8 cm
(b) 3.6 cm
(c) 2.4 cm
(d) 1.2 cm
Solution:
Given, MNO is a right-angled triangle
The legs are 6 cm and 8 cm long
We have to find the length of perpendicular NP on the side MO.
By using Pythagoras theorem,
MO² = MN² + NO²
MO² = (6)² + (8)²
MO² = 36 + 64
MO² = 100
Taking square root,
MO = 10 cm
Area of triangle = 1/2 × base × height
1/2 × MN × NO = 1/2 × MO × NP
6 × 8 = 10 × NP
NP = 48/10
NP = 4.8 cm
Therefore, the value of NP is 4.8 cm
✦ Try This: ∆ ABC is a right-angled triangle. Its legs are 3 cm and 5 cm long. Length of perpendicular BD on the side AC is
☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 11
NCERT Exemplar Class 7 Maths Chapter 9 Problem 17
In Fig. 9.22 ∆ MNO is a right-angled triangle. Its legs are 6 cm and 8 cm long. Length of perpendicular NP on the side MO is (a) 4.8 cm, (b) 3.6 cm, (c) 2.4 cm, (d) 1.2 cm
Summary:
In Fig. 9.22 ∆ MNO is a right-angled triangle. Its legs are 6 cm and 8 cm long. Length of perpendicular NP on the side MO is 4.8 cm
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