*Q1. The angles of quadrilateral are in the ratio 3:5:9:13. Find the angles of the quadrilateral.*

**Solution:**

Let the common ratio between the angles be x.

We know that the ‘Sum of the interior angles of the quadrilateral’ =

Now,

Therefore the Angles of the quadrilateral are:

*Q2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
*

**Solution:**

**Given,**

PQ = RS

**To show,**

PQRS is a rectangle we have to prove that one of its interior angle is right angled.

**Proof,**

In

Therefore,

∠P = ∠Q (by CPCT)

also,

∠P + ∠Q = 180° (Sum of the angles on the same side of the transversal)

⇒ 2∠P = 180°

⇒ ∠P = 90°

Thus PQRS is a rectangle.

*Q3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.*

**Solution:**

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.

**Given,**

**To show,**

ABCD is parallelogram and

**Proof,**

In

Therefore,

Thus,

Similarly we can prove,

Opposites sides of a quadrilateral are equal hence

Thus,

*Q4. Show that the diagonals of a square are equal and bisect each other at right angles.*

**Solution:**

Let PQRS be a square and its diagonals PR and QS intersect each other at O.

**To show,**

**Proof,**

In

Therefore,

Thus,

Therefore, diagonals are equal.

Now,

In

Therefore, ΔAOB ≅ ΔCOD (by AAS congruence condition).

Thus,

Now,

In

Therefore,

also,

Thus,

*Q5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
*

**Solution:**

**Given,**

Let

**To prove,**

Quadrilateral

**Proof,**

In

Therefore,

Thus,

also,

⇒

Now,

In

Therefore,

Thus,

also,

also,

and

One of the interior angle is right angle.

Thus, from (i), (ii) , (iii) and (iv) the given quadrilateral ABCD is a square.

*Q6. Diagonal AC of a parallelogram ABCD bisects ∠A. Show that*

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

**Solution:**

(i)In

Therefore,

Thus,

and

Thus,

(ii)

⇒

Also,

Thus,

*Q7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.*

**Solution:**

**Let **

**Proof,**

also,

Therefore,

Similarly, we can prove that diagonal

Also, by preceding above method we can prove that diagonal

*Q8. PQRS is a rectangle in which diagonal PR bisects ∠P as well as ∠R. Show that:*

(i) PQRS is a square

(ii) diagonal QS bisects ∠Q as well as ∠S.

** Solution:**

(i)

also,

Therefore,

Thus,

(ii) In

also,

Thus,

Now,

Thus,

*Q9. In parallelogram ABCD, two points PandQ are taken on diagonal BD such that DP=BQ. Show that:*

(i)△APD≅△CQB

(ii)AP=CQ

(iii)△AQB≅△CPD

(iv)AQ=CP

(v)APCQ is a parallelogram

(i)

(ii)

(iii)

(iv)

(v)

**Solution:**

(i) In

Thus,

(ii)

(iii) In

Thus,

(iv)

(v) From (ii) and (iv), it is clear that APCQ has equal opposite sides also it has equal opposite angles. Thus,

*10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that
(i) △APB≅△CQD
(ii) AP = CQ*

**Solution:**

(i) In

AB = CD (ABCD is a parallelogram)

Thus,

(ii) AP = CQ by CPCT as

*Q11. In △ABCand△DEF, AB=DE,AB∥DE,BC=EFandBC∥EF. Vertices A, B and C are joined to vertices D, E and F respectively.*

Show that

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iii) AD∥CF and AD = CF

(iv) quadrilateral ACFD is a parallelogram

(v) AC=DF

*(vi) △ABC≅△DEF*

**Solution:**

(i)

Thus, quadrilateral ABED is a parallelogram because two opposite sides of a quadrilateral are equal and parallel to each other.

(ii) Again

Thus, quadrilateral BEFC is a parallelogram.

(iii) Since ABED and BEFC are parallelograms.

Thus,

Also,

Thus,

(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other. Thus, it is a parallelogram.

(v)

(vi)

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

Thus,

*Q12. ABCD is a trapezium in which AB∥CD and*

AD=BC .Show that

(i)∠A=∠B (i)

*(ii) ∠C=∠D*

(iii)△ABC≅△BAD

(iv) diagonal AC = diagonal BD(iii)

(iv) diagonal AC = diagonal BD

*[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]*

**Solution:**

**Construction**: Draw a line through C parallel to DA intersecting AB produced at E.

**(i)** CE = AD (Opposite sides of a parallelogram)

AD = BC (Given)

Therefor, BC = CE

**(ii)**

**(iii)** In

AB = AB (Common)

AD = BC (Given)

Thus,

**(iv) ** Diagonal AC = diagonal BD (by CPCT as