2. In the Figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS).
In the Figure, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
6. From a circular sheet of radius 4 cm, a circle of radius 3 cm is cut out. Calculate the area of the remaining sheet after the smaller circle is removed.
The area of the remaining sheet after the smaller circle is removed will be = Area of the entire circle with radius 4 cm – Area of the circle with radius 3 cm
We know,
Area of circle = pr²
So,
Area of the entire circle = p(4)² = 16p cm2
And,
Area of the circle with radius 3 cm which is cut out = p(3)² = 9p cm2
Thus, the remaining area = 16p – 9p = 7p cm2
4. The area of a rhombus is 16 cm2 and the length of one of its diagonal is 4 cm. Calculate the length of other diagonal.
Area of rhombus = ½ × d1 × d2
? 16 = ½ × 4 × d2
So, d2 = 32/4 = 8 cm
3. A lawnmower takes 750 complete revolutions to cut grass on a field. Calculate the area of the field if the diameter of the lawnmower is 84 cm and the length is 1 m.
Given, length of lawnmower = 1m = 100cm
Its circumference = p × D = 22/7 × 84 = 264 cm
Length of field will be = 264 × 750 = 198000 cm
Here, the width of field = length of the lawnmower i.e. 100 cm
So, area of field = 198000 × 100 = 19,800,000 cm²
Or, 1980 m²
2. Calculate the height of a cuboid which has a base area of 180 cm2 and volume is 900 cm3.
Volume of cuboid = base area × height
900 = 180 × height
So, height = 900/180 = 5 cm
1. The parallel sides of a trapezium measure 12 cm and 20 cm. Calculate its area if the distance between the parallel lines is 15 cm.
Area of trapezium = ½ × perpendicular distance between parallel sides × sum of parallel sides
= ½ × 15 × (12 + 20)
= 1/2 × 15 × 32
= 15 × 16
= 240 cm2
3. In a certain school, there are 456 girls. Calculate the total number of students if 24% of the total students are boys.
Assume the total number of students to be 100.
It is given that 24% are boys.
So, total number of boys = 24% of 100 = 24
Thus, the total number of girls will be = (100-24)% i.e. 76%.
So, the total number of girls = 76% of 100 = 76
But, it is given that there are 456 girls.
Now, for 76 girls, total students are 100
For, 1 girl, total students will be 100/76
Therefore, for 456 girls, the total number of students will be = 100/76 × 456 = 45600/76 = 600
Thus, the total number of students = 600.
2. A shopkeeper bought two phones for Rs. 8,000 each. After selling the phones, there was a loss of 4% on the 1st phone while a profit of 8% on the 2nd phone. Calculate the overall gain or loss per cent on the whole transaction.
As the shopkeeper bought both phones at Rs. 8000 each.
Total cost price = Rs. 16,000
Assume that the cost price of the 1st phone is Rs. 100
Consider the deal of phone 1,
As it is given, there is 4% loss, the selling price will be = Rs. 96
For CP = 100, SP = 96
So, for CP = 1, SP = 96/100
Now, given, CP = 8000
Hence, SP = 96/100 × 8000 = 7680
Thus, the selling price of 1st phone = Rs. 7680
Consider the deal of phone 2, there is an 8% profit.
Hence, the selling price will be = Rs. 108
For CP = 100, SP = 108
So, for CP = 1, SP = 108/100
Now, given CP = 8000
hence, SP = 108/100 × 8000 = 8640
Thus, the selling price of 2nd phone = Rs. 8640
Here, the total selling price will be = Rs. 7680 + Rs. 8640 = 16320
Now, it can be seen that,
Total selling price > total cost price i.e. Rs. 16320 > Rs. 16000
So, there is a profit of Rs. (16320 – 16000) = Rs. 320
Now, the overall profit percentage will be-
Profit% = (Profit/Total Cost Price) × 100 = (320/16000) × 100 = 2
Therefore, there is a total of 2% profit for the whole transaction.
1.If 72% of 25 students like maths, find out the number of students who do not like mathematics?
Given, 72% of 25 students like maths.
Hence, 72% of 25 = (72/100) × 25 = 18 students
Now, from 25 students, 18 students like maths
Thus, the number of students who do not like maths = 25 – 18 = 7 students
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