In Fig. ∆ MNO is a right-angled triangle. Its legs are 6 cm and 8 c

1.	In Fig. ∆ 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
Answer» (a) 4.8 cm From the question it is given that, ∆ MNO is a right-angled triangle, its legs are 6 cm and 8 cm long. By the rule of Pythagoras theorem, Hypotenuse² = perpendicular² + base² In the given figure, MO² = MN² + NO² MO² = 6² + 8² MO² = 36 + 64 MO²= 100 MO = √100 MO = 10 cm Then, consider the triangle MNO, Area of triangle MNO = ½ × MN × NO = ½ × MO × NP = ½ × 6 × 8 = ½ × 10 × NP Therefore, NP = 24/5 NP = 4.8 cm

In Fig. ∆ 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

Answer»

(a) 4.8 cm

From the question it is given that,

∆ MNO is a right-angled triangle, its legs are 6 cm and 8 cm long.

By the rule of Pythagoras theorem,

Hypotenuse² = perpendicular² + base²

In the given figure,

MO² = MN² + NO²

MO² = 6² + 8²

MO² = 36 + 64

MO²= 100

MO = √100

MO = 10 cm

Then, consider the triangle MNO,

Area of triangle MNO = ½ × MN × NO

= ½ × MO × NP

= ½ × 6 × 8

= ½ × 10 × NP

Therefore, NP = 24/5

NP = 4.8 cm