Quadrilaterals and ...
 
Notifications
Clear all

Quadrilaterals and Polygons

Page 1 / 3
   RSS

0
Topic starter

 Hello sir,

Please solve the following question

 

Hello Ananya 

Please find the solution

43 Answers
0

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

0

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. 

0
Topic starter

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 = 45o, BC = 2 cm and CD = 3 cm then AB = ?

(A) 5 cm (B) 4 cm

(C) 3 cm (D) 2 cm

Hello Ananya,

 

Please find the attachment.

Draw DE perpendicular on AB,

CD = BE= 3 cm,

AE = ED = 2 cm.

AB = AE + EB = 2 + 3 = 5 cm

0

Find the value of x

∆DOC ~ ∆BOA 

 

DO / OC= BO/ OA 

 

3/6 =( x - 4)/(x -3) 

 

3x - 9 = 6x - 24 

 

x = 5 

0

Could you please solve this question?

Hello Vernica , 

PFA the solution. 

0

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?

Hello Richa ! 

The area of the region outside the house that the dog can reach = Shaded area shown in the figure =  π(12^2 ) x 240/360 + π ( 6^2) x 120/360 = 108π 

0

Both the questions, please.

3. 

 

AB = OP ( Diagonals of a rectangle ) 

 

OP = r = 36 

 

Diameter = 2r = 72in

 

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

 

0

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 . 

0

Hello Sir, please help.

Hello Richa, 

PFA the solution. 

and please do not post/discuss the questions ( from copy CAT -1)  till the window of the test (Copy CAT -1)  is not closed.

Ok, Sir. I'll keep that in mind. 

0

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

Hello Richa 

 

PFA the approch 

 

0

 

Sir please help

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 

0

Unable to solve 

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

Solution : 

0

Hey Sir, help me solve this problem.

0

Both questions.

This post was modified 6 years ago by Richa

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 ..

Approach for third question : 

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

Page 1 / 3
Share: