Question

1. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
2. 6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
3. 1 Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
4. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Answer 1

base = 12cm and altitude = 5 cm

Step-by-step explanation:

let the base = x cm

so ,the altitude be = x-7 cm

By Phythagores theorem :-

13^2 = x ^2 + (x-7)^2

169 = x^2 + x^2 +49 -14x

2x^2 -14x + 49 - 169 = 0

2x^2 - 14x - 120 =0. ( from this eq. take 2 common)

2( x^2 -7x - 60 ) =0

x^2 -7x -60 =0

By factoriesation method :-

x^2 - 12x + 5x -60 =0

x(x-12) + 5( x -12)

(x+5) ( x -12) =0

x+5=0 , x =-5

x -12=0, x= 12

hence ,

side of rectangle cannot be negative so,

base will be = 12

altitude = 12-7= 5

THANKS

Answer 2

$\large\underline{\sf{Solution-1}}$

Let assume that

Base of a right angle triangle is 'x' cm

So,

Altitude of right angle triangle is 'x - 7' cm

It is given that, Hypotenuse = 13 cm

Now,

By using Pythagoras Theorem,

⟼ (Hypotenuse)² = (Base)² + (Altitude)²

$\rm :\longmapsto\:\sf \: {13}^{2} = {x}^{2} + {(x - 7)}^{2}$

$\rm :\longmapsto\:\sf \: 169 = {x}^{2} + {x}^{2} + 49 - 14x$

$\rm :\longmapsto\:\sf \: 169 = 2{x}^{2} + 49 - 14x$

$\rm :\longmapsto\:\sf \: 2{x}^{2} - 14x + 49 - 169 = 0$

$\rm :\longmapsto\:\sf \: 2{x}^{2} - 14x - 120 = 0$

$\rm :\longmapsto\:\sf \: 2({x}^{2} - 7x - 60) = 0$

$\rm :\longmapsto\:\sf \: {x}^{2} - 7x - 60= 0$

$\rm :\longmapsto\:\sf \: {x}^{2} - 12x + 5x - 60= 0$

$\rm :\longmapsto\:\sf \: x(x - 12) + 5(x - 12) = 0$

$\rm :\longmapsto\:\sf \: (x + 5)(x - 12) = 0$

$\rm :\implies\:x = 12 \: \: \: or \: \: \: x = - 5 \: \: \{rejected \}$

Hence,

⟼ Base of right angle triangle = 12 cm

⟼ Altitude of right angle triangle = 5 cm.

$\large\underline{\sf{Solution-2}}$

Let assume that rectangular field be ABCD with longer side AB, shorter side BC and diagonal AC.

⟼ Shorter side of rectangle BC= x m

⟼ Diagonal of rectangle, AC = x + 60 m

⟼ Longer side of rectangle, AB = x + 30 m

⟼ Now, in right angle triangle ABC, right angled at B,

By using Pythagoras Theorem,

⟼ AC² = AB² + BC²

$\rm :\longmapsto\: {(x + 60)}^{2} = {x}^{2} + {(x + 30)}^{2}$

$\rm :\longmapsto\: {x}^{2} + 3600 + 120x = {x}^{2} + {x}^{2} + 900 + 60x$

$\rm :\longmapsto\: {x}^{2} - 60x - 2700 = 0$

$\rm :\longmapsto\: {x}^{2} - 90x + 30x - 2700 = 0$

$\rm :\longmapsto\:x(x - 90) + 30(x - 90) = 0$

$\rm :\longmapsto\:(x - 90)(x + 30) = 0$

$\bf\implies \:x = 90 \: \: \: or \: \: \: x = - 30 \: \{rejected \}$

Hence,

Longer side of Rectangular feild, AB = 90 + 30 = 120 m

Shorter side of Rectangular feild, BC = 60 + 30 = 90 m.

$\large\underline{\sf{Solution-3}}$

Let assume that,

⟼ Side of bigger square = x m

⟼ Side of smaller square = y m

So,

According to statement,

Difference of the perimeter = 24 m

⟼ 4x - 4y = 24

⟼ x - y = 6

⇛ x = y + 6

According to statement again,

⟼ Sum of area of two squares = 468 m²

It implies,

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 468$

On substituting the value of x, we get

$\rm :\longmapsto\: {(y + 6)}^{2} + {y}^{2} = 468$

$\rm :\longmapsto\: {y}^{2} + 36 + 12y + {y}^{2} = 468$

$\rm :\longmapsto\: 2{y}^{2} + 36 + 12y - 468 = 0$

$\rm :\longmapsto\: 2{y}^{2} + 12y - 432 = 0$

$\rm :\longmapsto\: 2({y}^{2} + 6y - 216) = 0$

$\rm :\longmapsto\: {y}^{2} + 6y - 216 = 0$

$\rm :\longmapsto\: {y}^{2} + 18y - 12y - 216 = 0$

$\rm :\longmapsto\:y(y + 18) - 12(y + 18) = 0$

$\rm :\longmapsto\:(y - 12)(y + 18) = 0$

$\bf\implies \:y = 12 \: \: \: or \: \: \: y = - 18 \: \{reject \}$

Hence,

The side of bigger square = x = y + 6 = 12 + 6 = 18 m

The side of smaller square = y = 12 m

$\large\underline{\sf{Solution-4}}$

Let assume that larger number be x

and

Let smaller number be y

So,

According to statement,

The square of smaller number is 8 times the largest number.

$\rm :\implies\: {y}^{2} = 8x - - - (1)$

Now,

Difference of the squares of 2 numbers is 180.

$\rm :\longmapsto\: {x}^{2} - {y}^{2} = 180$

$\rm :\longmapsto\: {x}^{2} - 8x = 180$

$\rm :\longmapsto\: {x}^{2} - 8x - 180 = 0$

$\rm :\longmapsto\: {x}^{2} - 18x + 10x - 180 = 0$

$\rm :\longmapsto\:x(x - 18) + 10(x - 18) = 0$

$\rm :\longmapsto\:(x - 18)(x + 10) = 0$

$\bf\implies \:x = 18 \: \: \: or \: \: \: x = - 10 \: \{reject \}$

Now, Substitute the value of x in equation (1), we get

$\rm :\longmapsto\: {y}^{2} = 8 \times 18 = 144 = {(12)}^{2}$

$\bf\implies \:y = 12$

Hence,

The larger number is 18

The smaller number is 12.