Question

Please solve this all ........

urgent

100 points

write answer will be mark as brainalest​

Answer 1

1. scalene triangle

2. equilateral triangle

3. hypotenuse² = side²+side²

4. 180°

5. false

questions not clear

Answer 2

Section A -

( a ) In a scalene triangles , no two sides are of equal length .

( b ) In an equilateral traingle , all sides are equal .

( c ) For any right angled triangle , the sum of the square of the base and it's corresponding perpendicular side is equal to the square of the hypotenuse.

This can be mathematically stated as -

( Base )² + ( Perpendicular )² = ( Hypotenuse ) ²

( d ) The sum of the angles in a triangle is 180°

( e ) True. The difference between the length of any two sides of a triangle is always less than the third side. This is an application for triangle inequalities.

Section B -

Q1 - The angles of a triangle are ( 3x - 10 )°, ( 2x + 25 )°, and ( x + 15 )° . Find the angles of the triangle.

Solution -

Here , the three given angles are ( 3x - 10 )°, ( 2x + 25 )°, and ( x + 15 )° .

Now , we know that the sum of the angles in a triangle is 180° .

So , ( 3x - 10 )° + ( 2x + 25 )° + ( x + 15 ) ° = 180°

=> 6x + 30° = 180°

=> 6x = 150°

=> x = 25°

Angle 1 => 3x - 10 => 65°

Angle 2 => 2x + 25 => 75°

Angle 3 => x + 15° = 40°

Thus , the three angles of that triangle are 65°, 75° and 40° respectively

Question 2 => Find AC

Solution

Here , on observing the above figure carefully , we can see that , ∆ ABT and ∆ TCD are right angled triangles .

So , we can use Pythagoras theorem .

Pythagoras Theorem -

For any right angled triangle , the sum of the square of the base and it's corresponding perpendicular side is equal to the square of the hypotenuse.

This can be mathematically stated as -

( Base )² + ( Perpendicular )² = ( Hypotenuse ) ²

In ∆ ABT ,

[ AT ] ² + 4² = 5

=> [ AT ]² = 9

=> [ AT ] = 3 cm .

In ∆ TCD ,

[ CT ] ² + 5² = 13²

=> [ CT ]² = 144

=> [ CT ] = 12 cm .

AC = AT + TC

=> 12 cm + 3 cm

=> 15 cm .

Question 3 -

Part 1

According to the exterior angle property , the exterior angle of a triangle is equal to the sum of its interior opposite angles .

So ,

x = 58° + 79°

=> x = 137°

Part 2 -

All the angles in this triangle are equal . So, this is an equilateral triangle.

In an equilateral triangle , each angle is 60°

Question 4 - ABCD is a quadrilateral. Then prove that -

AB + BC + CD + DA < 2 ( AC + BD )

Solution -

In a right triangle , the hypotenuse is the largest side.

So.

In ∆ ABC

=> AB < AC

=> BC < AC

Adding -

=> ( AB + BC ) < 2AC

In ∆ ABD

=> AD < BD

=> CD < BD

Adding -

=> ( AD + CD ) < 2AC

Adding -

AB + BC + CD + DA < 2 ( AC + BD )

Hence Proved

Question 5 -

A tree is broken at a height of 5 m from the ground and the top touches the ground at a distance of 12 m from the base of the tree .

Find the original height of the tree.

Solution -

See the attachment 1 .

∆ABC is an right angled triangle . So , the Pythagoras theorem is applicable here.

Pythagoras Theorem -

For any right angled triangle , the sum of the square of the base and it's corresponding perpendicular side is equal to the square of the hypotenuse.

This can be mathematically stated as -

( Base )² + ( Perpendicular )² = ( Hypotenuse ) ²

Here ,

[ AB ] ² + [ BC ]² = [ AC ] ²

=> { AC } ² = { 12 }² + { 5 }²

=> { AC }² = { 13 }²

=> AC = 13 m .

Original length of the tree

=> AC + AB

=> 13m + 5m

=> 18 m .

Thus , the original height of the tree is 18 m .

Question 6 -

Find the perimeter of the rectangle , whose length is 40 cm and the diagonal is 41 cm.

Solution -

See the second attachment.

Here , the diagonal is 41 cm and one of the sides is 40 cm .

The triangle formed Is a right angled ∆

So. the Pythagoras theorem is applicable here.

Pythagoras Theorem -

For any right angled triangle , the sum of the square of the base and it's corresponding perpendicular side is equal to the square of the hypotenuse.

This can be mathematically stated as -

( Base )² + ( Perpendicular )² = ( Hypotenuse ) ²

Here ,

[ L ] ² + [ 40 ] ² = [ 41 ] ²

=> [ L ]² + 1600 = 1681

=> [ l ] ² = 9

=> Length of the rectangle = 9 cm .

The breadth is given as 40 cm .

Perimeter

=> 2 [ L + B ]

=> 2 × 49

=> 98 cm .

Thus , the required perimeter of the rectangle is 98 cm .

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