The area of a triangle, whose base and the corresponding altitude are

1.	The area of a triangle, whose base and the corresponding altitude are 15 cm and 7 cm, is equal to area of a right triangle whose one of the sides containing the right angle is 10.5 cm. Find the other side of this triangle.
Answer» For the first triangle, given that Base = 15 cm and altitude = 7 cm We know that area of a triangle = ½ (Base x Altitude) = ½ (15 x 7) = 52.5 cm² It is also given that the area of the first triangle and the second triangle are equal. Area of the second triangle = 52.5 c m² One side of the second triangle = 10.5 cm Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle = (2x 52.5)/10.5 =10 cm Hence, the other side of the second triangle will be 10 cm.

The area of a triangle, whose base and the corresponding altitude are 15 cm and 7 cm, is equal to area of a right triangle whose one of the sides containing the right angle is 10.5 cm. Find the other side of this triangle.

Answer»

For the first triangle, given that

Base = 15 cm and altitude = 7 cm

We know that area of a triangle = ½ (Base x Altitude)

= ½ (15 x 7)

= 52.5 cm²

It is also given that the area of the first triangle and the second triangle are equal.

Area of the second triangle = 52.5 c m²

One side of the second triangle = 10.5 cm

Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle

= (2x 52.5)/10.5

=10 cm

Hence, the other side of the second triangle will be 10 cm.