Question

These questions are *Hots* Please answer these questions.​

Answer 1

Answer:

☞QUESTION :

A rectangular garden 40 m ×30 m is surrounded from outside by a path of equal width. if the area of the path is 456 m² , find the width of the path?

☞ANSWER :

FOR RECTANGLE

LENGTH = 40M

BREADTH = 30M

SO AREA OF GARDEN=L×B

=40×30

=1200M²

SO LET THE SUM OF WIDTH BE X FOR THE PATH FROM BOTH SIDES

SO 2(WIDTH)=X

SO OVERALL LENGTH = (40+X)M

AND OVERALL BREADTH = (30+X)M

SO OVERALL AREA = L×B

=(40+X)(30+X)

=1200+40X+30X+X²

= (X²+70X+1200)M

NOW IT IS GIVEN THAT

AREA OF PATH=456M²

SO AREA OF PATH = OVERALL AREA-AREA OF GARDEN

456=X²+70X+1200-1200

X²+70X-456=0

So by splitting the middle term method

X²+76X-6X-456=0

X(X+76)-6(X+76)=0

(X+76)(X-6)=0

SO

X+76=0 OR X-6=0

X=-76 OR X=6

BUT BREADTH CAN NEVER BE NEGATIVE

SO X=-76 IS REJECTED

SO X=6 IS ACCEPTED

SO 2(WIDTH)=X

Width=X/2

Width=3M=6/2

Width=3M

☞HENCE THE WIDTH OF THE PATH IS 3 METRES

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☞Question:

6th term of ap is zero. Prove that it's 21st terms is triple its 11th term

☞Answer:

First term = a

Common diff. = d

6th term = a + 5d

a + 5d = 0

a = -5d

21st term = a + 20d = -5d + 20d = 15d

11th term = a + 10d = -5d + 10d = 5d

Therefore,

☞21st term = 3 * 11th term = 3 * 5d = 15d

(proved)

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Answer 2

Answer:

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