**ABCD is a trapezium, in which AD and BC are parallel. If the four sides AB, BC, CD and DA are respectively 9cm, 12cm, 15cm and 20cm then the magnitude of the sum of the squares of the two diagonals is:**

(a) 638 (b) 786 (c) 838 (d) 648

Let the height be h. Base be 20 top be 12.

Draw perpendiculars from the edges of the top to the base.

We have left overs of the base on both the sides. call them x and y.

Now X+Y =8

Now sum of the diagonals is nothing but h^2 + (20-y)^2 + h^2 + (20-x)^2

it comes out to be 800-40(x+y)+2 h^2 + X^2 + Y^2

We have h^2+X^2 = 9^2 and h^2+ Y^2 = 15^2 and X + Y = 8

Ans = 786.

Hello sir,

Please solve this question:

ABCD is a trapezium, such that AB, DC are parallel and BC is perpendicular to them. If angle DAB = 45^{o}, BC = 2 cm and CD = 3 cm then AB = ?

(A) 5 cm (B) 4 cm

(C) 3 cm (D) 2 cm

∆DOC ~ ∆BOA

DO / OC= BO/ OA

3/6 =( x - 4)/(x -3)

3x - 9 = 6x - 24

x = 5

A dog is tied to the corner of a house with a regular hexagonal base that measures 6 ft on each side. If the rope is 12 ft in length, what is the area in square feet of the region outside the house that the dog can reach?

Thankyou, Sir. Can you also share the solution for Q4 in the same image?

Hey sir, posting a question below. Though I've solved the question with the approach mentioned in the image (solution with blue pen), but is there any way to solve it faster? If somehow we could prove triangle PBC as a right angled triangle, it's just a 10sec question. But my question is, how could we do that?

Hello Richa !

Rotate P to P' 90 degrees with centre B .

We get PP' = 6√2

So BPP' = 45 degrees

Now cosP'PC = √2/2

cosP'PC = 45 degrees too.

So , Area = BC^2 = 14^2 + 6^2 = 232 .

Ok, Sir. I'll keep that in mind.

In regular octagon ABCDEFGH

Find the area of ACEG, if side of octagon is 2 cm

A. 13.24

B.12.97

C.14.18

D 13.66

Triangle MEC ~ Triangle BEA

Area MEC / Area BEA = MC²/AB² = 1²/2²

If , Area MEC = x then Area AEB = 4x

EC : AE = 1 : 2 hence area BEC = 2x

Area [Square ABCD] = 2 [ Area ABC] = 12x

Area AEMD = 12x - 2x - x - 4x = 5x

Required ratio : x : 5x = 1: 5

Hint : Area Base Ratio - Ratio of area of triangles having base on the same line and third vertex common is equal to the ratio of length of bases of the triangles

In 2nd question.

Just square the sides given ,add them and see the sum will be square of 73

For example u can consider a triangle or sqaure or hexagon u will get same result ..

Sir, in third question they should tell that assuming there is no slag . shouldn't they ?