Description : A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th
Last Answer : Radius of conical cap, r = 7 cm Height of conical cap, h = 24cm Slant height, l2 = (r2+h2) = (72+242) = (49+576) = (625) Or l = 25 cm CSA of 1 conical cap = πrl = (22/7)×7×24 = 550 CSA of 10 caps = (10×550) cm2 = 5500 cm2 Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.
Description : If the perimeter of an equilateral triangle is 24cm, then find the area of the triangle -Maths 9th
Last Answer : Perimeter of equilateral triangle = 24cm ( there are three equal sides in an e quilateral triangle ) Each side = 24 / 3 = 8cm Area of an equilateral triangle = root 3 / 4 s square = root 3 / 4 × 8 square = root 3 / 4 x 64 = root 3 × 16 = 16 root 3
Description : The base of an isosceles triangle is 24cm and its area is 192cm^2. Find its perimeter -Maths 9th
Last Answer : Given, base of an isosceles triangle =24 cm Area of isosceles triangle =192 sq.cm Area = 21×b×h ∴192=224.h ∴h=16 cm Side=h2+122=256+144=20 cm Perimeter of triangle =2a+b =2(20)+24=64 cm
Description : What is 24cm squared?
Last Answer : (24 cm)² = 576 sq cm, the area of a square with sides 24 cm24 cm² = an area of 24 sq cm
Description : What is the circumference of the circle if the diameter is 24cm?
Last Answer : Circumference of circle: 24*pi = 75.398 cm rounded to 3 decimalplaces
Description : The average height of the students in a group was 360cm. When 10 students whose height is 292.8cm are newly admitted. The average height of the group was reduced by 24cm.How many students are present in the group? A) 29 B) 25 C) 28 D) 14
Last Answer : Answer: C) let the number of students initially in the class be A, then, total height = A * 360------ 1 again the no of students increased = A + 10 then, total height A + 10 student = (A +10) * 336 ... 3 (A+10) *336-2928 =A*360 336A+3360-2928=360A A=18 Number of students in class = 18+10=28