Question

In the school garden Ajay(A), Brijesh(B), Chinki(C) and Deepak(D) planted their flower plants [4]

of Rose, Sunflower, Champa and Jasmine respectively as shown in the following figure. A fifth

student Eshan wanted to plant her flower in this area. The teacher instructed Eshan to plant his

flower plant at a point E such that CE: EB = 3 : 2.

Answer the following questions:

i. Find the coordinates of point E where Eshan has to plant his flower plant.

Find the area of △ ECD.

a. 9.5 square unit

b. 11.5 square unit

c. 10.5 square unit

d. 12.5 square unit

iii. Find the distance between the plants of Ajay and Deepak.

a. 8.60 unit

b. 6.60 unit

c. 5.60 unit

d. 7.60 unit

iv. The distance between A and B is:

a. 5.5 units

b. 7 units

c. 6 units

d. 5 units

v. The distance between C and D is:

a. 5.5 units

b. 7 units

c. 6 units

d. 5 unit​

Answer 1

Answer:

1. e(6,5)

2.area of triangle ECD= 10.5

3. 8.60 unit

4. 5 unit

5. 7 unit

Answer 2

Step-by-step explanation:

Given: In the school garden Ajay(A), Brijesh(B), Chinki(C) and Deepak(D) planted their flower plants [4] of Rose, Sunflower, Champa and Jasmine respectively as shown in the following figure.

A fifth student Eshan wanted to plant her flower in this area. The teacher instructed Eshan to plant his flower plant at a point E such that CE: EB = 3 : 2.

To find: Answer the following questions:

i. Find the coordinates of point E where Eshan has to plant his flower plant.

Ans: Let coordinates of Eshan plant is E(x,y)

Given that: CE: EB = 3 : 2

3 2

C(3,2)•___________•E(x,y)_________•B(8,7)

Section formula: If point P(x,y) divides the line segment AB in m:n,when coordinates of A(x1,y1) and B(x2,y2)

then

$\boxed{\bf x = \frac{mx_2 + nx_1}{m + n} } \\ \\ \boxed{\bf y = \frac{my_2 + ny_1}{m + n}}$

So,

According to section formula

$x = \frac{3 \times 8 + 2 \times 3}{3 + 2}$

$x = \frac{24 + 6}{5}$

$x = \frac{30}{5}$

$\bf x = 6$

and

$y = \frac{3 \times 7 + 2 \times2 }{3 + 2}$

$y = \frac{21 + 4 }{5}$

$y = \frac{25}{5}$

$\bf y = 5$

Coordinates of E(6,5).

ii) Find the area of △ ECD.

a. 9.5 square unit

b. 11.5 square unit

c. 10.5 square unit

d. 12.5 square unit

Ans:

If A(x1,y1), B(x2,y2) and C(x3,y3)

$\boxed{\bf Ar(∆ABC) = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| }$

Here,

Vertices of triangle are E(6,5),C(3,2) and D(10,2)

$Ar(∆ECD) = \frac{1}{2} |6(2 - 2) + 3(2 - 5) + 10(5 - 2)|$

$Ar(∆ECD) = \frac{1}{2} |0 - 9 +30|$

$Ar(∆ECD) = \frac{1}{2} |21|$

Ar(∆EDC)=10.5 sq-unit

Option C is correct.

iii. Find the distance between the plants of Ajay and Deepak.

a. 8.60 unit

b. 6.60 unit

c. 5.60 unit

d. 7.60 unit

Ans:

Distance formula: If P(x1,y1) and Q(x2,y2), then

$\boxed{\bf PQ = \sqrt{ {(x_2 - x_1)}^{2} + ( {y_2 - y_1)}^{2} }}$

To find distance between Ajay and Deepak apply distance formula.

Coordinates of A(3,7) and D(10,2)

$AD = \sqrt{ {(10 - 3)}^{2} + ( {2 - 7)}^{2} }$

$AD = \sqrt{ 49 + 25}$

$AD = \sqrt{74}$

$\bf AD = 8.6 \:$

Option a is correct.

iv. The distance between A and B is:

a. 5.5 units

b. 7 units

c. 6 units

d. 5 units

Ans: A and B lies on the straight line,

thus,

Distance between A(3,7) and B(8,7) is difference of 8 and 3.

AB= 5 units

Option d is correct.

v. The distance between C and D is:

a. 5.5 units

b. 7 units

c. 6 units

d. 5 unit

Ans: C and D lies on the straight line,

thus,

Distance between C(3,2) and D(10,2) is difference of 10 and 3.

CD= 7 units

Option b is correct.

Note*: Distance AB and CD can be calculated by distance formula also.

Final answer:

I) Coordinates of E(6,5)

ii) Option c is correct.

iii) Option a is correct.

iv) Option d is correct.

v) Option b is correct.

Hope it will help you.

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