The volume of a frustum cone is 1,600 $m m^{3}$ , whose base and upper area of a circle is 16 and 100 $m m^{2}$ . Find the height of the cone.

30.76 mm

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

11.25 mm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

12.25 mm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

13.25 mm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 30.76 mm
Volume of a frustum cone = $\frac{1}{3} (A_{1} + A_{2} + \sqrt{A_{1} A_{2}}) h$
1,600 = $\frac{1}{3} (16 + 100 + \sqrt{16 \times 100}) \times h$
1,600 = $\frac{1}{3} (116 + \sqrt{1600}) \times h$
1,600 = $\frac{1}{3} (116 + 40) \times h$
1,600 = $\frac{1}{3} \times 156 \times h$
h = $(1, 600 \times 3) / 156 = 30.76 m m$

Suggest Corrections

Similar questions

Find the volume of a frustum cone, whose base and upper area of a circle is 36 and 100 $m m^{2}$ . The height of the cone is 32.5 mm.

4, 200 c m^{3}

10 c m^{2}

90 c m^{2}

m^{3}

m^{2}

m m^{3}

Join BYJU'S Learning Program