Approach please .

ABC is right angled Traingle. A semicircle with centre O is inscribed inside the triangle shown in the figure. Find area of ABOD.

Hello Kinshuk !

PQR is an Isosceles triangle so PE bisects QR

QE = ER

In ∆PQE => QE² = PQ² - PE² = 7² - PE²

In ∆PDE , DE² = PD² - PE² = 5² - PE²

Now,

QD × DR = (QE + DE) (QE - DE)

=> QD × DR = ( QE)²- ( DE)²

=> QD × DR = 7² -5² = 24

So they can only be 3 × 8 or 4 × 6.

Hence 3+8 = 11,

4+6 =10

Option ( B )

ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED = FD. Find the area of the shaded figure.

Choose one answer.

a. 1/2

b. 13/16

c. 5/9

d. 1/3

Hello Abhi,

Please find the solution

A(ABC) = 1

A(ACF) = 1/3 and A(BCF) = 2/3.

A(BFE) = (1/3)*(2/3) = 2/9. So A(AFEC) = 7/9.

A(FBD = A(EBD) = 1/9.

A(EDC) = 2*A(EBD) = 2/9.

A(FDA) = (1/2)*A(FBD) = 1/18.

**A(ADC) = A(AFEC) - A(EDC) - A(FDA) = (7/9) - (2/9) - (1/18) = 1/2**

Alternate Approach

**In triangle ABC, BC = 4 units, AC = 2 units and sin A = 1. What is angle ABC?**

**30****60****45****90**

Hello sir,

Please solve

3^{2} – (1+x)^{2} = 3^{3}-(2-x)^{2}

1+x^{2}+2x=4+x^{2}-4x

6x = 3

x =1/2

AE = 3(sqrt3)/2

AD^{2} =(3(sqrt3)/2)^{2} + (1/2)^{2} =27/4 + 1/4 = 7

AD = sqrt7

Hello sir,

please solve this question

Hello Tina,

Please find the solutions,

In triangle ABC ,Produce a line from B to meet AC,meeting at D and From C to AB,meeting at E.let BD and CE meeet at X.let triangle BXE have area a,triangle BXC have area b and triangle CXD have area c.find the area of quadrilateral AEXD in terms of a,b,and c

a) bc(2a+b+c)/b^2-ac

b)ac(a+2b +c)/b^2-ac

c)abc/b^2-ac

d)ab(a+b +2c)/b^2-ac

ake a special case. Assume ABC is an equilateral triangle.

Let's assume D is the mid point on AC and E is the mid point on AB.

Now Area(BXE) = Area(CXD) = a = c

Then for this case Area(BXC) = b = 2a

And area of AEXD = 2a (it's very simple to calculate now)

Now, check all options:

- a) 2a*a(2a+2a+a)/4a^2-a^2 = 10a/3
- b) a*a(6a)/4a^2-a^2 = 2a

c)a*2a*a/4a^2-a^2 = 2a/3

- d) a*a(a+2a+2a)/4a^2-a^2 = 5a/3

Only b) matches. So, it is the answer.

Three altitudes of a triangle are 12cm, 20cm and 25 cm. Find the perimeter of the triangle. How to solve this one?

Hello Hemant!

Ratio of sides :

1/12 : 1/20 : 1/15 => 5x : 3x : 4x

Area => 1/2 • 3x•4x = 1/2 • 12 • 5x

x = 5

Perimeter => 5x + 4x + 3x = 5 × ( 5+3+4) = 60 🙂

The perimeter of a right triangle is 12+8√3. The sum of the squares of the three sides is 294. Find the length of the hypotenuse.

Given that P^{2}+ B^{2} + H^{2} = 294

and P^{2} + B^{2} = H^{2}

So, H^{2} + H^{2} = 294

2H^{2} = 294

H^{2} = 147 = 49 x 3

H = 7 sqrt(3)

The sides of a right triangle are 3 4 5

A point is taken on the hypotenuse at a distance of 2 cm from the vertex adjacent to the 4 side. Find the distance from this point to the vertex of the right triangle.

Question 28 sir

The alternate approach for the question

The sides of a right triangle are 3 4 5

A point is taken on the hypotenuse at a distance of 2 cm from the vertex adjacent to the 4 side. Find the distance from this point to the vertex of the right triangle.

Let's say MN intersect AX at O ,

MO = 1/2 BX

∆AMO and ∆ABX are similar triangles so

Area∆ AMO/ Area∆ABX = MO²/BX² = 1/4

So if Area of ∆AMO = x ,then Area ABX = 4x and Area MOXB = 4x - x = 3x ( shaded area )

Area of ∆ABC = 2 × Area ∆ABX = 8x

Hence required ratio = 3x/ 8x = 3 : 8