Answer

Verified

450.6k+ views

Hint: Consider the two statements p and q. Prove that they both are true. if p and q are true then $p\to q$ also becomes true. If both are true $p\leftrightarrow q$ also becomes true i.e. p implies q and q implies p.

Complete step-by-step answer:

p: Every square is a rectangle.

(i) A rectangle is a quadrilateral with all 4 angles $90{}^\circ $.

From this definition, you can prove that the opposite sides are parallel and of the same lengths. A rectangle can be tall & thin, short & fat or all the sides can have the same length.

(ii) A square is a quadrilateral with all 4 angles right angles and all 4 sides of same length.

So a square is a special kind of rectangle, it is one where all the sides have the same length. Thus every square is a rectangle because it is a quadrilateral with all 4 angles right angles. However not every rectangle is a square, to be a square its sides must have the same length.

$\therefore $ p: Every square is a rectangle is true.

q: Every rhombus is a kite.

When all the angles are also $90{}^\circ $ the kite will be square. The sum of its side, opposite angles are equal. Therefore the rhombus is symmetrical about each of its diagonals. So a rhombus can become a kite. The difference between a rhombus and kite is that a kite doesn’t always have 4 equal sides or 2 pairs of parallel sides like rhombus.

$\therefore $ Every rhombus is a kite, but every kite is not a rhombus.

$\therefore $ q is true.

So we found out that both p and q are true.

$p\to q$ is T, it means that p implies q.

Also we know that $T\leftrightarrow T$ is true.

$\therefore $ $p\leftrightarrow q$ is true, it means that p implies q and q implies p.

Hence the truth values of $p\to q$ and $p\leftrightarrow q$ is T, T.

Note: If it was given,

p: Every rectangle is square.

q: Every kite is rhombus.

Then $p\to q$ would have been F.

As both the statements are wrong.

So, $p\leftrightarrow q$ is false.

So, the value of $p\to q$ and $p\leftrightarrow q$ becomes F, F or from the table of logical implication here p and q are T.

$\therefore $ $p\to q$ is T and $q\to p$is T i.e. case 1 is true.

Complete step-by-step answer:

p: Every square is a rectangle.

(i) A rectangle is a quadrilateral with all 4 angles $90{}^\circ $.

From this definition, you can prove that the opposite sides are parallel and of the same lengths. A rectangle can be tall & thin, short & fat or all the sides can have the same length.

(ii) A square is a quadrilateral with all 4 angles right angles and all 4 sides of same length.

So a square is a special kind of rectangle, it is one where all the sides have the same length. Thus every square is a rectangle because it is a quadrilateral with all 4 angles right angles. However not every rectangle is a square, to be a square its sides must have the same length.

$\therefore $ p: Every square is a rectangle is true.

q: Every rhombus is a kite.

When all the angles are also $90{}^\circ $ the kite will be square. The sum of its side, opposite angles are equal. Therefore the rhombus is symmetrical about each of its diagonals. So a rhombus can become a kite. The difference between a rhombus and kite is that a kite doesn’t always have 4 equal sides or 2 pairs of parallel sides like rhombus.

$\therefore $ Every rhombus is a kite, but every kite is not a rhombus.

$\therefore $ q is true.

So we found out that both p and q are true.

$p\to q$ is T, it means that p implies q.

Also we know that $T\leftrightarrow T$ is true.

$\therefore $ $p\leftrightarrow q$ is true, it means that p implies q and q implies p.

Hence the truth values of $p\to q$ and $p\leftrightarrow q$ is T, T.

p | q | $p\to q$ | $q\to p$ |

T | T | T | T |

T | F | F | T |

F | T | T | F |

F | F | T | T |

Note: If it was given,

p: Every rectangle is square.

q: Every kite is rhombus.

Then $p\to q$ would have been F.

As both the statements are wrong.

So, $p\leftrightarrow q$ is false.

So, the value of $p\to q$ and $p\leftrightarrow q$ becomes F, F or from the table of logical implication here p and q are T.

$\therefore $ $p\to q$ is T and $q\to p$is T i.e. case 1 is true.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Guru Purnima speech in English in 100 words class 7 english CBSE

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Select the word that is correctly spelled a Twelveth class 10 english CBSE