CBSE Class X

Question 6:

The vertices of a MABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/AC = 1/4. Calculate the area of the ΔADE and compare it with the area of ΔABC. (Recall Converse of basic proportionality theorem and Theorem 6.6 related to ratio of areas of two similar triangles)

Answer:

Given that, AD/AB = AE/AC = 1/4

Therefore, D and E are two points on side AB and AC respectively such that they divide side AB and AC in a ratio of 1:3.

Clearly, the ratio between the areas of ΔADE and ΔABC is 1:16.

Alternatively,

We know that if a line segment in a triangle divides its two sides in the same ratio, then the line segment is parallel to the third side of the triangle. These two triangles so formed (here ΔADE and ΔABC) will be similar to each other.

Hence, the ratio between the areas of these two triangles will be the square of the ratio between the sides of these two triangles.

Therefore, ratio between the areas of ΔADE and ΔABC = (1/4)² = 1/16

Question 5:

The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the following figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of ΔPQR if C is the origin?

Also calculate the areas of the triangles in these cases. What do you observe?

Answer:

(i) Taking A as origin, we will take AD as x-axis and AB as y-axis. It can be observed that the coordinates of point P, Q, and R are (4, 6), (3, 2), and (6, 5) respectively.

(ii) Taking C as origin, CB as x-axis, and CD as y-axis, the coordinates of vertices P, Q, and R are (12, 2), (13, 6), and (10, 3) respectively.

It can be observed that the area of the triangle is same in both the cases.

Question 4:

The two opposite vertices of a square are (− 1, 2) and (3, 2). Find the coordinates of the other two vertices.

Answer:

Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively. Let (x, y), (x1, y1) be the coordinate of vertex B and D respectively.

We know that the sides of a square are equal to each other.

∴ AB = BC

We know that in a square, all interior angles are of 90°.

In ΔABC,

AB² + BC² = AC²

⇒ 4 + y² + 4 − 4y + 4 + y² − 4y + 4 =16

⇒ 2y² + 16 − 8 y =16

⇒ 2y² − 8 y = 0

⇒ y (y − 4) = 0

⇒ y = 0 or 4

We know that in a square, the diagonals are of equal length and bisect each other at 90°. Let O be the mid-point of AC. Therefore, it will also be the mid-point of BD.

⇒ y + y₁ = 4

If y = 0,

y₁ = 4

If y = 4,

y₁ = 0

Therefore, the required coordinates are (1, 0) and (1, 4).

Question 3:

Find the centre of a circle passing through the points (6, − 6), (3, − 7) and (3, 3).

Answer:

Let O (x, y) be the centre of the circle. And let the points (6, −6), (3, −7), and (3, 3) be representing the points A, B, and C on the circumference of the circle.

 

 

On adding equation (1) and (2), we obtain

10y = −20

y = −2

From equation (1), we obtain

3x − 2 = 7

3x = 9

x = 3

Therefore, the centre of the circle is (3, −2).

Question 2:

Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Answer:

If the given points are collinear, then the area of triangle formed by these points will be 0.

This is the required relation between x and y.

Question 1:

Determine the ratio in which the line 2x + y − 4 = 0 divides the line segment joining the points A(2, − 2) and B(3, 7)

Answer:

Let the given line divide the line segment joining the points A(2, −2) and B(3, 7) in a ratio k : 1.

Coordinates of the point of division

Therefore, the ratio in which the line 2x + y − 4 = 0 divides the line segment joining the points A(2, −2) and B(3, 7) is 2:9.

Question 5:

You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for MABC whose vertices are A (4, − 6), B (3, − 2) and C (5, 2)

Answer:

Let the vertices of the triangle be A (4, −6), B (3, −2), and C (5, 2).

Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.

However, area cannot be negative. Therefore, area of ΔADC is 3 square units.

Clearly, median AD has divided ΔABC in two triangles of equal areas.

Question 4:

Find the area of the quadrilateral whose vertices, taken in order, are (− 4, − 2), (− 3, − 5), (3, − 2) and (2, 3)

Answer:

Let the vertices of the quadrilateral be A (−4, −2), B (−3, −5), C (3, −2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

Question 3:

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, − 1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Answer:

Let the vertices of the triangle be A (0, −1), B (2, 1), C (0, 3).

Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by

Question 2:

In each of the following find the value of ‘k’, for which the points are collinear.

(i) (7, − 2), (5, 1), (3, − k)                            (ii) (8, 1), (k, − 4), (2, − 5)

Answer:

(i) For collinear points, area of triangle formed by them is zero.

Therefore, for points (7, −2) (5, 1), and (3, k), area = 0

(ii) For collinear points, area of triangle formed by them is zero.

Therefore, for points (8, 1), (k, −4), and (2, −5), area = 0

Question 1:

Find the area of the triangle whose vertices are:

(i) (2, 3), (− 1, 0), (2, − 4)                          (ii) (− 5, − 1), (3, − 5), (5, 2)

Answer:

(i) Area of a triangle is given by

Question 10:

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (− 1, 4) and (− 2, −1) taken in order. [Hint: Area of a rhombus = 1/2 (product of its diagonals)]

Answer:

Let (3, 0), (4, 5), (−1, 4) and (−2, −1) are the vertices A, B, C, D of a rhombus ABCD.

Question 9:

Find the coordinates of the points which divide the line segment joining A (− 2, 2) and B (2, 8) into four equal parts.

Answer:

From the figure, it can be observed that points P, Q, R are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.

Question 8:

If A and B are (− 2, − 2) and (2, − 4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.

Answer:

The coordinates of point A and B are (−2, −2) and (2, −4) respectively.

Since AP = 3/7 AB,

Therefore, AP: PB = 3:4

Point P divides the line segment AB in the ratio 3:4.

Question 7:

Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, − 3) and B is (1, 4)

Answer:

Let the coordinates of point A be (x, y).

Mid-point of AB is (2, −3), which is the center of the circle.

Question 6:

If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Answer:

Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a parallelogram ABCD. Intersection point O of diagonal AC and BD also divides these diagonals.

Therefore, O is the mid-point of AC and BD.

If O is the mid-point of AC, then the coordinates of O are

If O is the mid-point of BD, then the coordinates of O are

Since both the coordinates are of the same point O,

Question 5:

Find the ratio in which the line segment joining A (1, − 5) and B (− 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Answer:

Let the ratio in which the line segment joining A (1, −5) and B (−4, 5) is divided by x-axisbe k : 1.

Therefore, the coordinates of the point of division is

We know that y-coordinate of any point on x-axis is 0.

Therefore, x-axis divides it in the ratio 1:1.

Division point =

 

Question 4:

Find the ratio in which the line segment joining the points (− 3, 10) and (6, − 8) is divided by (− 1, 6).

Answer:

Let the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by point (−1, 6) be k : 1.

Therefore, the required ratio is 2 : 7.

Question 3:

To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Answer:

It can be observed that Niharika posted the green flag at 1/4 of the distance AD i.e., (1/4 × 100) m = 25 m from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25).

Similarly, Preet posted red flag at 1/5 of the distance AD i.e., (1/5 × 100) m = 20 m from the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20). Distance between these flags by using distance formula = GR

The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (x, y).

Therefore, Rashmi should post her blue flag at 22.5m on 5th line

Question 2:

Find the coordinates of the points of trisection of the line segment joining (4, − 1) and (− 2, − 3).

Answer:

Let P (x₁, y₁) and Q (x₂, y₂) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

Point Q divides AB internally in the ratio 2:1.