# [SOLVED] Math homework help...Algebra ... PLS help! As soon as possible...?

• Nov 20th 2007, 04:18 AM
waterqueen7
[SOLVED] Math homework help...Algebra ... PLS help! As soon as possible...?
Math homework help...Algebra ... PLS help! As soon as possible...?

Q. 1. Rajan saw a wooden piece near his bed. He measured it and found it to be 32 cm. long. He wanted to divide it into two parts. Help him divide the wooden piece such that if the larger part of the wood is divided by the smaller part the quotient is 2 and the remainder is 5.

Q 2. Rajan calculated the length and the breadth of his rectangular math notebook and found the length exceeds the breadth by 2 cm. If both the length and the breadth are increased by 1 cm, then the area of the book increased by 11 sq cm. Find the length and breadth of the book.

Q 3. The sum of the digits of a two digit number is 12. If the digits are reversed the new number is 4/7 times the original number. Find the original number.

Q 4. Rajan constructed rhombus ABCD. He then drew a straight line RABS such that RA = AB = BS. Help him prove that RD and SC when produced meet at right angles.
• Nov 20th 2007, 05:19 AM
earboth
Quote:

Originally Posted by waterqueen7
Math homework help...Algebra ... PLS help! As soon as possible...?

Q. 1. Rajan saw a wooden piece near his bed. He measured it and found it to be 32 cm. long. He wanted to divide it into two parts. Help him divide the wooden piece such that if the larger part of the wood is divided by the smaller part the quotient is 2 and the remainder is 5....

Hello,

let x be the larger part, then (32-x) is the smaller part:

$\frac{x}{32-x}=2+\frac{5}{32-x}$ . Multiply both sides of the equation by (32-x):

$x=2(32-x)+5$ . Solve for x. (I've got: The larger part is 23 cm, the smaller part is 9 cm).
• Nov 20th 2007, 05:24 AM
earboth
Quote:

Originally Posted by waterqueen7
Math homework help...Algebra ... PLS help! As soon as possible...?
...
Q 2. Rajan calculated the length and the breadth of his rectangular math notebook and found the length exceeds the breadth by 2 cm. If both the length and the breadth are increased by 1 cm, then the area of the book increased by 11 sq cm. Find the length and breadth of the book.
...

Hi,

let the width be x, then the length is (x+2) and the area is A=x(x+2)

Now both increased by 1 cm and the area increases by 11 cm²:

$(x+1)(x+3)=x(x+2)+11$ . Solve for x. I've got x = 4 cm. (That seems to be a very small math book!)
• Nov 20th 2007, 05:31 AM
earboth
Quote:

Originally Posted by waterqueen7
Math homework help...Algebra ... PLS help! As soon as possible...?

...
Q 3. The sum of the digits of a two digit number is 12. If the digits are reversed the new number is 4/7 times the original number. Find the original number.
...

Hi,

a 2 digit number has the form $10x + y\ ,\ x \ \in \ \{1, 2, 3, ..., 9\},\ y\ \in \ \{0,1,2,,...,9\}$

With your problem x + y = 12 ==> y = 12 - x.

$10y + x = \frac47 \cdot (10x + y)~\implies~10(12-x)+x=\frac47\cdot (10x+12-x)$ . Solve for x. I found that the number is 84.
• Nov 20th 2007, 09:37 AM
Soroban
Hello, waterqueen7!

Here's one approach . . .

Quote:

Q 4. Rajan constructed rhombus ABCD.
He then drew a straight line RABS such that RA = AB = BS.
Prove that RD and SC when produced meet at a right angle.

Code:

                              Q                               *                           *  *                         *      *                     D*-----------*C                   *α/          /β*               *  /          /  *             *    /          /    *         *α      /θ          /θ    β*       * - - - - *-----------* - - - - *       R        A          B        S

We have: . $AB = BC = CD = DA = RA = BS$

Let $\theta = \angle DAB = \angle CBS$
.Then: . $\angle DAR = 180^o - \theta$

In isosceles $\Delta DAR$, let: . $\alpha = \angle DRA = \angle RDA$
Since the sum of its angles is 180°: . $(180^o-\theta) + 2\alpha \:=\:180^o\quad\Rightarrow\quad \alpha = \frac{\theta}{2}$ .[1]

In isosceles $\Delta CBS$, let: . $\beta = \angle CSB = \angle BCS$
Since the sum of its angles is 180°: . $\theta + 2\beta \:=\:180^o\quad\Rightarrow\quad \beta \:=\:90^o - \frac{\theta}{2}$ .[2]

In $\Delta QRS\!:\;\;\angle Q + \alpha + \beta \;=\;180^o\quad\Rightarrow\quad \angle Q \;=\;180^o - \alpha - \beta$

Substitute [1] and [2]: . $\angle Q \;=\;180^o - \frac{\theta}{2} - \left(90^o - \frac{\theta}{2}\right) \;=\;90^o$

Therefore: . $\angle Q$ is a right angle.