Explain these questions from que 12-15 with colour diagrams and images or figure if possible in the video and write all step by step solution of each question in simplest way for class 8 child and write all formulas or rules whereas required..read image uploaded carefully of class 8 india,delhi.explain in best way.also keep in mind that I will be also copying the solutions done by you in my school register,so write the solutions in good presentable format showing all working n steps of each question.write the steps in easiest understandable way to grasp concept in once.. solution should be complete with all answer statements.---**Question 12**
**Question Stem:**
12. In the figure given below, O is the centre of the three circles, I ,II and III . The length of PQ is 20 cm and the diameter of the circle II is 12 cm.
What will be the radius of the largest circle ?
**Options:**
A) 8 cm
B) 10 cm
C) 14cm
D) 16cm
**Chart/Diagram Description:**
* **Type:** Geometric figure, showing three concentric circles.
* **Main Elements:**
* **Circles:** Three concentric circles labeled I (innermost), II (middle), and III (outermost).
* **Point:** O is the center of all three circles.
* **Line Segment:** A horizontal line segment passes through the center O. Points P and Q are on this line.
* **Points on Line Segment:** P is on the outermost circle (III), and Q is on the middle circle (II). O is between P and Q.
* **Labels:** Circles are labeled I, II, III. The center is labeled O. Points on the diameter are labeled P and Q.
**Question 13**
**Description:**
13. Rahul used 2 identical white squares and 8 identical grey squares to make the shapes below.
**Question Stem:**
What can we say about the perimeter of the two shapes?
**Options:**
A. The perimeter of Shape 2 is greater than that of Shape 1.
B. The perimeter of Shape 1 is greater than that of Shape 2.
C. The perimeters of both Shape 1 and Shape 2 are equal.
D. (cannot be concluded without knowing the actual measures of the white and grey squares)
**Chart/Diagram Description:**
* **Type:** Arrangement of squares forming two shapes.
* **Main Elements:**
* Two shapes are shown side-by-side, labeled "Shape 1" and "Shape 2" below them.
* **Shape 1:** Composed of one white square and four grey squares. The white square is in the center. Two grey squares are attached to the top edge of the white square, one on the left and one on the right, adjacent to each other. Two grey squares are attached to the bottom edge of the white square, one on the left and one on the right, adjacent to each other. All squares appear to be the same size.
* **Shape 2:** Composed of one white square and four grey squares. The white square is in the center. One grey square is attached to the middle of the top edge of the white square. One grey square is attached to the middle of the bottom edge of the white square. One grey square is attached to the middle of the left edge of the white square. One grey square is attached to the middle of the right edge of the white square. All squares appear to be the same size.
**Question 14**
**Description:**
14. Shown below are three shapes. Shape 2 can be formed by arranging 2 triangles shown in shape 1. Shape 3 can be formed by arranging 3 triangles shown in shape 1.
**Chart/Diagram Description (Initial Shapes):**
* **Type:** Geometric shapes.
* **Main Elements:**
* Three shapes shown side-by-side, labeled "Shape 1", "Shape 2", and "Shape 3" below them.
* **Shape 1:** A right-angled triangle. Appears to be an isosceles right-angled triangle.
* **Shape 2:** A square. Formed by arranging two triangles like Shape 1.
* **Shape 3:** A trapezoid/quadrilateral. Formed by arranging three triangles like Shape 1.
**Description (New Shape):**
A new shape is formed by arranging the above three shapes.
**Chart/Diagram Description (New Shape):**
* **Type:** Composite geometric shape.
* **Main Elements:**
* A larger shape is shown below the initial three shapes.
* This shape is formed by combining Shape 1, Shape 2, and Shape 3.
* Shape 3 (the trapezoid) forms the base.
* Shape 1 (the triangle) is placed on the left side of Shape 3, with its hypotenuse along the slanted edge of the trapezoid.
* Shape 2 (the square) is placed on the right side of Shape 3, attached to its vertical edge.
**Question Stem:**
The area of the new shape is _________.
**Options:**
A. thrice the area of shape 1 + twice the area of shape 2
B. twice the area of shape 1 + twice the area of shape 2
C. twice the area of shape 2 + thrice the area of shape 3
D. once the area of shape 1 + once the area of shape 2 + once the area of shape 3
**Question 15**
**Question Stem:**
15. Simple interest on a certain amount is ( 9 / 16 ) of the principal . If the number representing the rate of interest in percent and time in years be equal, then time, for which the principal is lent out, is
**Options:**
A) 5 and ½ years
B) 6 and ½ years
C) 7 years
D) 7 and ½ years
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Let's solve Question 12 step by step. We have three concentric circles with center O.
Circle I is the smallest, Circle II is in the middle, and Circle III is the largest.
Point P is on the largest circle, and point Q is on the middle circle.
The distance PQ is 20 centimeters, and the diameter of circle II is 12 centimeters.
We need to find the radius of the largest circle.
Now let's solve this step by step. First, we find the radius of Circle II.
Since the diameter is 12 centimeters, the radius is 12 divided by 2, which equals 6 centimeters.
This means OQ equals 6 centimeters. Next, we use the relationship that PQ equals PO plus OQ.
Since PQ is 20 centimeters and OQ is 6 centimeters, we can calculate PO as 20 minus 6,
which equals 14 centimeters. Therefore, the radius of the largest circle is 14 centimeters.
Now let's solve Question 13 about comparing perimeters. We have two shapes made from identical squares.
Shape 1 has a white square in the center with grey squares attached to each of its four sides, forming a plus or cross pattern.
Shape 2 has a different arrangement. To find the perimeter, we count the exposed outer edges.
For Shape 1, counting all the outer edges gives us 12 unit lengths. For Shape 2, the outer boundary gives us 10 unit lengths.
Therefore, Shape 1 has a greater perimeter than Shape 2.
Now let's solve Question 14 about the area of a composite shape. We have three basic shapes:
Shape 1 is a triangle, Shape 2 is a square made from 2 triangles, and Shape 3 is a trapezoid made from 3 triangles.
A new shape is formed by combining all three shapes together. Since the shapes don't overlap when combined,
the total area of the new shape is simply the sum of the areas of the individual shapes.
Therefore, the area equals once the area of Shape 1, plus once the area of Shape 2, plus once the area of Shape 3.
Finally, let's solve Question 15 about simple interest. We're given that the simple interest is nine-sixteenths of the principal,
and the rate of interest equals the time in years. Using the simple interest formula, we substitute the given values.
After simplification, we get T squared equals 900 divided by 16. Taking the square root of both sides,
we find T equals 30 divided by 4, which equals 7.5 years. Therefore, the time for which the principal is lent out is 7 and a half years.