Question

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the
other two sides.

Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40
square meters​

Answer 1

Let the base of Δbe x

height=x-7

Hypotenuse =13 cm

By pythogarus theorm

x²+{x-7}²=13²

x²+x²+7²-2{x×7}=169

2x²+49-14x=169

2x²-14x-120=0

2{x²-7x-60}=0

x²-12x+5x-60=0

x{x-12}+5{x-12}=0

x=12 or -5

since negative answer we should not take in lengths so base=x=12cm

height=x-7=12-7=5cm

Given perimeter of rectangle =28m

Area=40

2{l+b}=28

l+b=28/2=14

l=14-b

Area=l×b=40

l=40/b

So, 14-b=40/b

14b-b²=40

b²-14b+40=0

b²-10b-4b+40=0

b{b-10}-4{b-10}=0

b=4

l=10

Hope it helps u...

Answer 2

1st Question :-

Answer :-

Given :-

The altitude of a right angled triangle is 7 cm less than its base . If the hypotenuse is 13 cm .

Required to find :-

• Find the other two sides ?

Theorem used :-

Pythagoras theorem

It states that ;

The sum of the squares of the two sides of the right angled triangle is equal to the squares of the hypotenuse .

This is represented as ;

( side )² + ( side )² = ( Hypotenuse )²

Solution :-

Since, it is mentioned that ;

The altitude of a right angled triangle is 7 cm less than its base . If the hypotenuse is 13 cm .

So,

Let the base of the right angled triangle be x cm

Altitude of the right angled triangle is 7 cm less than the base

So,

Let the altitude be x - 7 cm

Hypotenuse = 13 cm

Using the Pythagoras theorem ;

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

( x )² + ( x - 7 )² = ( 13 )²

Using the identity ;

• ( x - y )² = x² - 2xy + y²

x² + ( x )² - 2 ( x ) ( 7 ) + ( 7 )² = ( 13 )²

x² + x² - 14x + 49 = 169

x² + x² - 14x = 169 - 49

x² + x² - 14x = 120

2x² - 14x = 120

2x² - 14x - 120 = 0

2 ( x² - 7x - 60 ) = 0

x² - 7x - 60 = 0/2

x² - 7x - 60 = 0

x² - 12x + 5x - 60 = 0

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

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

This implies ;

=> x - 12 = 0

=> x = 12

Similarly,

=> x + 5 = 0

=> x = - 5

Since, the length can't be in negative .

Hence,

• The value of x = 12 cm

Therefore,

Base of the right angled triangle = 12 cm

Altitude of the right angled triangle = 12 - 7 = 5 cm

$\rule{400}{2}$

2nd Question :-

Answer :-

Given :-

Perimeter of the rectangle = 28 meters

Area of the rectangle = 40 m²

Required to find :-

• The dimensions of the rectangle ?

Solution :-

Given Information :-

Perimeter of the rectangle = 28 meters

Area of the rectangle = 40 m²

So,

Let's consider that ;

Length of the rectangle be ' a ' meters

Breadth of the rectangle be ' b ' meters

According to the problem ;

Perimeter of the rectangle = 2 ( length + breadth )

This implies ;

=> 2 ( a + b ) = 28

=> ( a + b ) = 28/2

=> a + b = 14

=> b = 14 - a $\tt{\bf{ Equation- 1 }}$

Consider this as equation - 1

Similarly,

Area of the rectangle = Length x Breadth

This implies ;

=> a x b = 40

Substitute the value of b from equation 1

=> a x 14 - a = 40

=> 14a - a² = 40

=> - a² + 14a = 40

=> - a² + 14a - 40 = 0

=> - 1 ( a² - 14a + 40 ) = 0

=> a² - 14a + 40 = 0 + 1

=> a² - 14a + 40 = 0

=> a² - 10a - 4a + 40

=> a ( a - 10 ) - 4 ( a - 10 )

=> ( a - 10 ) ( a - 4 )

This implies ;

a - 10 = 0

a = 10

Similarly,

a - 4 = 0

a = 4

Hence,

Length of the rectangle = 10 meters

Breadth of the rectangle = 4 meters