Question

$verify x = \frac{3}{4} \: \: y = \frac{5}{6} \: \: z = \frac{ - 7}{8}$
explain required ​

Answer 1

GIVEN

$verify x = \frac{3}{4} \: \: y = \frac{5}{6} \: \: z = \frac{ - 7}{8}$

TO FIND:-

• required solutions

EXPLAIN:-

$x = (y + z) = (x + y) + z \\ \\ x = (y + z) = \frac{3}{4} + ( \frac{5}{6} + \frac{ - 7}{8} ) \\ \\ = \frac{3}{4} + ( \frac{20 + ( - 21)}{24} ) \\ \\ = \frac{3}{4} + ( \frac{ - 1}{24} ) \\ \\ = \frac{3}{4} + \frac{ - 1}{24} \\ \\ = \frac{18 + ( - 1)}{24} \\ \\ = \frac{17}{24} \\ \\ (x + y ) + z = ( \frac{3}{4} + \frac{5}{6} ) + \frac{ - 7}{8} \\ \\ = ( \frac{9 + 10}{12} ) + \frac{ - 7}{8} \\ \\ = \frac{19}{12} + \frac{ - 7}{8} \\ \\ = \frac{38 + ( - 21)}{24} \\ \\ = \frac{38 - 21}{24} \\ \\ = \frac{17}{24} \\ \\ x + (y + z) = (x + y) + z$

•°• verified

Answer 2

Answer:

Step-by-step explanSide of the signal board = a

Perimeter of the signal board = 3a = 180 cm

∴ a = 60 cm

Semi perimeter of the signal board (s) = 3a/2

By using Heron’s formula,

Area of the triangular signal board will be =

Ncert solutions class 9 chapter 12-1

2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?

Ncert solutions class 9 chapter 12-2

Solution:

The sides of the triangle ABC are 122 m, 22 m and 120 m respectively.

Now, the perimeter will be (122+22+120) = 264 m

Also, the semi perimeter (s) = 264/2 = 132 m

Using Heron’s formula,

Area of the triangle =

Ncert solutions class 9 chapter 12-3

=1320 m2

We know that the rent of advertising per year = ₹ 5000 per m2

∴ The rent of one wall for 3 months = Rs. (1320×5000×3)/12 = Rs. 1650000

3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Ncert solutions class 9 chapter 12-4

Solution:

It is given that the sides of the wall as 15 m, 11 m and 6 m.

So, the semi perimeter of triangular wall (s) = (15+11+6)/2 m = 16 m

Using Heron’s formula,

Area of the message =

Ncert solutions class 9 chapter 12-5

= √[16(16-15)(16-11) (16-6)] m2

= √[16×1×5×10] m2 = √800 m2

= 20√2 m2

4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42cm.

Solution:

Assume the third side of the triangle to be “x”.

Now, the three sides of the triangle are 18 cm, 10 cm, and “x” cm

It is given that the perimeter of the triangle = 42cm

So, x = 42-(18+10) cm = 14 cmation: