Question

please tell answer of any questions it's argent​

Answer 1

Solution attached :-

$\mathtt{\huge{\underline{\red{Question\:?}}}}$

The diagonal of rhombus are 12 cm & 16cm. Find the perimeter of rhombus.

$\mathtt{\huge{\underline{\green{Answer:-}}}}$

⭐ A rhombus is a parallelogram, whose diagonals intersects each other at 90°.

Given :-

The diagonal of rhombus are 12 cm & 16cm.

To find :-

The perimeter of rhombus.

Calculation :-

According to the question,

The diagonal of rhombus are 12 cm & 16cm.

So, Half of diagonal is 6cm and 8cm.

Using Pythagoras theorem,

We know that , the point O where diangonals are meet at that point there are angle formed 90°

Side of rhombus = √p² + b²

= √6² + 8²

= √36+64

=√ 100

= 10cm

So, The Side of rhombus is 10cm.

The Perimeter of rhombus= 4a

4a , where a is 10.

= 4×10

= 40cm

The Perimeter of rhombus is 40cm.

✏

________________________________

SᝪᏞᐯᗴ ᗩᑎᎩ Ꭲᗯᝪ ᔑᑌᗷᔑᎢᏆᎢᑌᎢᗴᔑ :-

$\mathtt{\huge{\underline{\pink{Question\:?}}}}$

Draw a histogram to represent the following data :-

Class interval :-10-20 ,20-30,30-40,40-50

Frequency :- 7.............,17...........,18,........10

$\mathtt{\huge{\underline{\green{Answer:-}}}}$

Refers to the attachment

$\mathtt{\huge{\underline{\red{Question\:?}}}}$

The altitude of triangle is three-fifth of the length of the corresponding base. If the altitude is decreased by, 4 CM and the corresponding base is increased by 10 CM, the area of the triangle remains same. Find the base and the altitude of the triangle?

$\mathtt{\huge{\underline{\green{Answer:-}}}}$

Given :-

• The altitude of triangle is three-fifth of the length of the corresponding base.

• If the altitude is decreased by, 4 CM and the corresponding base is increased by 10 CM, the area of the triangle remains same.

To Find :-

• The base and the altitude of the triangle.

Calculation :-

Let the base of triangle is X cm.

So, the altitude is 3/5x cm.

According to the question,

Area of the triangle = 1/2 × base × height.

= 1/2 × x × 3/5x

= 3/10x².........(1)

If the altitude is decreased by, 4 CM and the corresponding base is increased by 10 CM, the area of the triangle remains same.

• Base of triangle = (x - 4) cm

• Altitude of triangle = 3/5x + 4

So, the Area of triangle :- 1/2 × (x - 4) × 3/5x + 4

= (x - 4) / 2 × 3x + 20 /5

= x × 3x + x × 20 - 4×3x - 4×20 / 2

= 3x² + 20x - 12x - 100 / 2×5

= 3x² + 8x - 100 / 10........(2)

Equating 1 & 2

➡ 3x²/10 = 3x² + 8x - 100 / 10

➡ 3x² = 3x² + 8x - 100

➡ 8x = 100

➡ x = 100/8

➡ x = 12.5

Base = x = 12.5.

Altitude= 3x²/10 = 3×12.5/10 = 12 cm.

_____________________________________

Answer 2

Question:

The diagonal of a rhombus are 12cm and 16cm. find the perimeter of the rhombus.

Solution:

[See figure 1]

Diagonal 1 = AC = 16cm

Diagonal 2 = BD = 12cm

[We know that the diagonal of a rhombus bisect each other at 90°]

So,

We get,

AO = CO = 8cm each. [Both are half of AC, i.e diagonal 1]

BO = OD = 6cm each. [Both are half of AD, i.e diagonal 2]

We get 4 right-angled triangles - ΔAOD, ΔDOC, ΔCOB, and ΔAOB.

We can take one triangle to get the hypotenuse side.

So,

ΔAOD

∠AOD = 90°, base = OD, perpendicular = AO, hypotenuse = AD

By Pythagoras theorem,

(Hypotenuse)² = (base)² + (perpendicular)²

⇒ (AD)² = (6)² + (8)²

⇒ (AD)² = 36 + 64

⇒ (AD)² = 100

⇒ AD = 10 cm  [Taking square roo on both sides.]

So, one side of the rhombus = 10cm,

We know that all sides of a rhombus if equal so, each side of the rhombus = 10cm

Hence

The perimeter of the rhombus = sum of all side

= 10+10+10+10

= 40cm

- - - - -

Question 4.

(2) The altitude of a triangle is three-fifth of the length of the corresponding base. If the altitude is decreased by 4cm and the corresponding base increased by 10cm, the area of the triangle remains the same. Find the base and altitude of the triangle.

SOLUTION:

Let the base of the triangle be y cm.

Give - The altitude is three-fifth of the corresponding base.

It means the altitude is $\frac{3}{5}\ of\ y$ = $\frac{3y}{5}$

The area of the triangle

= $\boxed{\bf{\frac{1}{2} \times base \times altitude}}$

So, the area will be

= $\frac{1}{2} \times y \times \frac{3y}{5}$

= $\frac{y}{2} \times \frac{3y}{5}$

= $\frac{3y^{2}}{10}\ cm^{2}$

Now,

The base of the new triangle = (y+10) cm

And

The altitude of a new triangle = $\frac{3y}{5}-4$ = $\frac{3y-20}{5}$

So, we will get a new area of the new triangle,

= $\frac{1}{2} \times (y+10) \times (\frac{3y-20}{5})$ cm sq.

= $(\frac{y}{2}+\frac{10}{2}) \times (\frac{3y-20}{5})$

= $(\frac{y+10}{2}) \times (\frac{3y-20}{5})$

= $\frac{(y+10)(3y-20)}{10}$

Now,

According to the question,

The area will remain the same.

So,

$\implies \frac{3y^{2}}{10}= \frac{(y+10)(3y-20)}{10}$

$\implies \frac{3y^{2}}{10}= \frac{3y^{2}-20y+30y-200}{10}$

$\implies \frac{3y^{2}}{10}= \frac{3y^{2}+10y-200}{10}$

$\implies 10(3y^{2}) = 10(3y^{2}+10y-200)$

$\implies 30y^{2} =30y^{2}+100y-2000$

$\implies -100y = 30y^{2}-30y^{2}-2000$

$\implies y = \frac{-20\not0\not0}{-1\not0\not0}$

$\implies y = 20$

Hence,

The base = y = 20cm

And the altitude = $\frac{3y}{5} = \frac{3 \times 20}{5} = \frac{60}{5} = \boxed{\bf 12cm}$

Hence,

The Base of the triangle = 20cm

And the altitude of the triangle = 12cm.

- - - - -

Question.4.

(1) Draw the histogram to represent the following data.

Class interval - (10-20) ; (20-30) ; (30-40) ; (40-50)

Frequency      -    7      ;      17     ;      18     ;     10

The solution is in the attached images(Figure 2).

- - - - -

Question 4

Find the value of x, y, and z in the following figure. ABCD is a parallelogram.

SOLUTION:

See figure 3.

Given ABCD is a parallelogram. And BD is diagonal.

∠BAD = 120°

Some properties of parallelogram-

• The diagonal bisects the angle equally.
• The adjacent angles sum is equal to 180°.
• The opposite angles are equal.
• The opposite sides are equal.

Now,

∠BAD = ∠z = 120° (Opposite angles are equal)

Now,

∠(x+y) + 120° = 180°  (Adjacent angles sum is 180°)

⇒ ∠(x+y) = 180° - 120°

⇒ ∠(x+y) = 60°

Now,

∠x = ∠y (diagonal bisect ∠ABC equally)

⇒ ∠x + ∠y = 60°

⇒ ∠x + ∠x = 60°

[as ∠x = ∠y, putting ∠x instead of y]

⇒ 2∠x = 60°

⇒ ∠x = 60° ÷ 2

⇒ ∠x = 30°

hence,

∠x = 30°,

∠y = 30°

∠z = 120°