Now if the length of the rectangle is reduced by 5 units and breadth is increased by 2 units then new length and breadth will be (a−5) units and (b+2) units.

Then new area will be (a−5)(b+2).

Then according to the problem,

(a−5)(b+2)−ab=−80

or, 2a−5b=−70.......(1).

Now if length of the rectangle is increased by 10 units and breadth is decreased by 5 units then new length and breadth will be (a+10) units and (b−5) units.

Then new area will be (a+10)(b−5).

Then according to the problem,

(a+10)(b−5)−ab=50

or, 10b−5a=100

or, 2b−a=20

or, 4b−2a=40......(2).

Now adding (1) and (2) we get

−b=−30

or, b=30.

Putting the value of b in (1) we get, a=40.

Now a+b=40+30=70.

3 0

3 years ago

Kerstin has a square tile. The perimeter of the tile is 32 inches. What is the length of the tile?

Troyanec [42]

If it is a square all sides are equal, so the perimeter is equal to 4 times the length of one side...

4s=32

s=8in

4 0

3 years ago

Read 2 more answers

Find side lengths C and H

Nuetrik [128]

Answer:

$\displaystyle c=\frac{2\sqrt{3}}{3}\\\\\displaystyle h=7\sqrt{2}\\$

Step-by-step explanation:

Right Triangles

A right triangle can be identified by the fact it has an internal angle of 90°. In a right triangle, the trigonometric ratios stand.

Let's consider the triangle to the left. We need to calculate side c, one of the legs of the triangle. We can use the angle adjacent to it (60°) or the angle opposite to it (30°) with the appropriate trigonometric ratio.

We'll use the adjacent angle, and

$\displaystyle tan60^o=\frac{2}{c}$

Solving for c

$\displaystyle c=\frac{2}{tan60^o}=\frac{2}{\sqrt{3}}$

Rationalizing

$\displaystyle c=\frac{2}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$

Now for the triangle to the right. The side h is the hypotenuse. Again, any of the two angles can be used (though they are equal, for it's an isosceles triangle). For any of them it is true that

$\displaystyle sin45^o=\frac{7}{h}$

Solving for h

$\displaystyle h=\frac{7}{sin45^o}=\frac{7}{\frac{\sqrt{2}}{2}}=\frac{14}{\sqrt{2}}$

Rationalizing

$\displaystyle h=\frac{14}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{14\sqrt{2}}{2}=7\sqrt{2}$

3 0

4 years ago