Question

A right angled triangle PQR where angle Q is 90degree.if QR =16 cm and PR=20cm. compare the curved surface areas of the right circular cones so formed by the triangle​

Answer 1

Answer:

C.S.A(i) : C.S.A(ii) = 4:3

Step-by-step explanation:

Formula used:

Curved surface area of cone $=\pi\:r\:l$ square units

$l^2=h^2+r^2$

In right ΔPQR,

$PR^2=PQ^2+QR^2$

$20^2=PQ^2+16^2$

$400=PQ^2+256$

$400-256=PQ^2$

$PQ^2=144$

$PQ=\sqrt{144}$

$PQ=12\:cm$

case(i): when the triangle is revolving about the side PQ

Curved surface area of cone $=\pi\:r\:l$

Curved surface area$=\pi\:(16)\:(20)$

case(ii): when the triangle is revolving about the side QR

Curved surface area of cone $=\pi\:r\:l$

Curved surface area$=\pi\:(12)\:(20)$

Now,

$\frac{C.S.A(i)}{C.S.A(ii)}$

$=\frac{\pi\:(16)\:(20)}{\pi\:(12)\:(20)}$

$=\frac{16}{12}$

$=\frac{4}{3}$

C.S.A(i) : C.S.A(ii) = 4:3

Answer 2

3:4 is  the ratio of curved surface areas of the right circular cones formed by rotaing about QR and PQ  where QR = 16 cm & PR = 20 cm Q = 90°

Step-by-step explanation:

PR=20cm = Hypotenuse

QR = 16 cm

PQ² = PR² - QR²

=> PQ² = 20² - 16²

=> PQ = 12

PQ = 12 cm Base  when the triangle is revolving about the side QR

Radius of cone = base = 12cm

Slant height = Hypotenuse = 20 cm

Curved suface area = π R L

= π  12 * 20

= 240 π

PR=20cm = Hypotenuse

QR = 16 cm = Base  when the triangle is revolving about the side PQ

Radius of cone = base = 16cm

Slant height = Hypotenuse = 20 cm

Curved suface area = π R L

= π 16 * 20

= 320  π

Ratio of Curved surface area  rotated about QR and PQ  =  240π / 320 π

= 3 /4