## RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II (Quadrilaterals) Ex 16.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1

**Question 1.**

**Define the following terms:**

**(i) Quadrilateral**

**(ii) Convex Quadrilateral.**

**Solution:**

(i) Quadrilateral: A closed figure made of four line segments is called a quadrilateral such that:

(a) no three points of them are collinear

(b) the line segments do not intersect except at their ends points.

(ii) Convex quadrilateral: A quadrilateral is called a convex quadrilateral of the line containing any side of the quadrilateral has the remaining vertices on the same side of it. In the figure, quadrilateral ABCD is a convex quadrilateral.

**Question 2.**

**In a quadrilateral, define each of the following:**

**(i) Sides**

**(ii) Vertices**

**(iii) Angles**

**(iv) Diagonals**

**(v) Adjacent angles**

**(vi) Adjacent sides**

**(vii) Opposite sides**

**(viii) Opposite angles**

**(ix) Interior**

**(x) Exterior**

**Solution:**

(i) Sides: In a quadrilateral ABCD, form line segments AB, BC, CD and DA are called sides of the quadrilateral.

(ii) Vertices : The ends points are called the vertices of the quadrilateral. Here in the figure, A, B, C and D are its vertices.

(iii) Angles: A quadrilateral has four angles which are at their vertices. In the figure, ∠A, ∠B, ∠C and ∠D are its angles.

(iv) Diagonals: The line segment joining the opposite vertices is called diagonal. A quadrilateral has two diagonals.

(v) Adjacent Angles : The angles having a common arm (side) are called adjacent angles.

(vi) Adjacent sides : If two sides of a quadrilateral have a common end-point, these are called adjacent sides.

(vii) Opposite sides: If two sides do not have a common end-point of a quadrilateral, they are called opposite sides.

(viii) Opposite angles : The angles which are not adjacent are called opposite angles.

(ix) Interior: The region which is surrounded by the sides of the quadrilateral is called its interior.

(x) Exterior : The part of the plane made up by all points as the not enclosed by the quadrilateral, is called its exterior.

**Question 3.**

**Complete each of the following, so as to make a true statement:**

**(i) A quadrilateral has ………… sides.**

**(ii) A quadrilateral has ………… angles.**

**(iii) A quadrilateral has ……….. vertices, no three of which are …………**

**(iv) A quadrilateral has …………. diagonals.**

**(v) The number of pairs of adjacent angles of a quadrilateral is ………….**

**(vi) The number of pairs of opposite angles of a quadrilateral is ……………**

**(vii) The sum of the angles of a quadrilateral is …………**

**(viii) A diagonal of a quadrilateral is a line segment that joins two ………. vertices of the quadrilateral.**

**(ix) The sum of the angles of a quadrilateral is …………. right angles.**

**(x) The measure of each angle of a convex quadrilateral is …………. 180°.**

**(xi) In a quadrilateral the point of intersection of the diagonals lies in ………….. of the quadrilateral.**

**(xii) A point is in the interior of a convex quadrilateral, if it is in the ……….. of its two opposite angles.**

**(xiii) A quadrilateral is convex if for each side, the remaining …………. lie on the same side of the line containing the side.**

**Solution:**

(i) A quadrilateral has four sides.

(a) A quadrilateral has four angles.

(iii) A quadrilateral has four vertices, no three of which are collinear .

(iv) A quadrilateral has two diagonals.

(v) The number of pairs of adjacent angles of a quadrilateral is four .

(vi) The number of pairs of opposite angles ot a quadrilateral is two.

(vii) The sum of the angles of a quadrilateral is 360°.

(viii) A diagonal of a quadrilateral is a line segment that join two opposite vertices of the quadrilateral.

(ix) The sum of the angles of a quadrilateral is 4 right angles.

(x) The measure of each angle of a convex quadrilateral is less than 180°.

(xi) In a quadrilateral the point of intersection of the diagonals lies in interior of the quadrilateral.

(xii) A point is in the interior of a convex quadrilateral, if it is in the interior of its two opposite angles.

(xiii) A quadrilateral is convex if for each side, the remaining vertices lie on the same side of the line containing the side.

**Question 4.**

**In the figure, ABCD is a quadrilateral.**

**(i) Name a pair of adjacent sides.**

**(ii) Name a pair of opposite sides.**

**(iii) How many pairs of adjacent sides are there?**

**(iv) How many pairs of Opposite sides are there ?**

**(v) Name a pair of adjacent angles.**

**(vi) Name a pair of opposite angles.**

**(vii) How many pairs of adjacent angles are there ?**

**(viii) How many pairs of opposite angles are there ?**

**Solution:**

In the figure, ABCD is a quadrilateral

(i) Pairs of adjacent sides are AB, BC, BC, CD, CD, DA, DA, AB.

(ii) Pairs of opposite sides are AB and CD; BC and AD.

(iii) There are four pairs of adjacent sides.

(iv) There are two pairs of opposite sides.

(v) Pairs of adjacent angles are ∠A, ∠B; ∠B, ∠C; ∠C, ∠D; ∠D, ∠A.

(vi) Pairs of opposite angles are ∠A and ∠C; ∠B and ∠D.

(vii) There are four pairs of adjacent angles.

(viii) There are two pairs of opposite angles.

**Question 5.**

**The angles of a quadrilateral are 110°, 72°, 55° and x°. Find the value of x.**

**Solution:**

Sum of four angles of quadrilateral is 360°

110° + 12° + 55° + x° = 360°

⇒ 237° + x° = 360°

⇒ x° = 360° – 237° = 123°

x = 123°

**Question 6.**

**The three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.**

**Solution:**

The sum of four angles of a quadrilateral = 360°

Three angles are 110°, 50° and 40°

Let fourth angle = x

Then 110° + 50° + 40° + x° = 360°

⇒ 200° + x° = 360°

⇒ x = 360° – 200° = 160°

x = 160°

**Question 7.**

**A quadrilateral has three acute angles each measures 80°. What is the measure of fourth angle ?**

**Solution:**

Sum of four angles of a quadrilateral = 360°

Sum of three angles having each angle equal to 80° = 80° x 3 = 240°

Let fourth angle = x

Then 240° + x = 360°

⇒ x° = 360° – 240°

⇒ x° = 120°

Fourth angle = 120°

**Question 8.**

**A quadrilateral has all its four angles of the same measure. What is the measure of each ?**

**Solution:**

Let each equal angle of a quadrilateral = x

4x° = 360°

⇒ x° = \(\frac { 360 }{ 4 }\) = 90°

Each angle will be = 90°

**Question 9.**

**Two angles of a quadrilateral are of measure 65° and the other two angles are equal. What is the measure of each of these two angles ?**

**Solution:**

Measures of two angles each = 65°

Sum of these two angles = 2 x 65°= 130°

But sum of four angles of a quadrilateral = 360°

Sum of the remaining two angles = 360° – 130° = 230°

But these are equal to each other

Measure of each angle = \(\frac { 230 }{ 2 }\) = 115°

**Question 10.**

**Three angles of a quadrilateral are equal. Fourth angle is of measure 150°. What is the measure of equal angles ?**

**Solution:**

Sum of four angles of a quadrilateral = 360°

One angle = 150°

Sum of remaining three angles = 360° – 150° = 210°

But these three angles are equal

Measure of each angle = \(\frac { 210 }{ 3 }\) = 70°

**Question 11.**

**The four angles of a quadrilateral are as 3 : 5 : 7 : 9. Find the angles.**

**Solution:**

Sum of four angles of a quadrilateral = 360°

and ratio in angles = 3 : 5 : 7 : 9

Let first angles = 2x

Then second angle = 5x

third angle = 7x

and fourth angle = 9x

3x + 5x + 7x + 9x = 360°

⇒ 24x = 369°

⇒ x = \(\frac { 360 }{ 24 }\) = 15°

First angle = 3x = 3 x 15° = 45°

second angle = 5x = 5 x 15° = 75°

third angle = 7x = 7 x 15° = 105°

and fourth angle = 9x = 9 x 15° = 135°

**Question 12.**

**If the sum of the two angles of a quadrilateral is 180°, what is the sum of the remaining two angles ?**

**Solution:**

Sum of four angles of a quadrilateral = 360°

and sum of two angle out of these = 180°

Sum of other two angles will be = 360° – 180° = 180°

**Question 13.**

In the figure, find the measure of ∠MPN.

**Solution:**

In the figure, OMPN is a quadrilateral in which

∠O = 45°, ∠M = ∠N = 90° (PM ⊥ OA and PN ⊥ OB)

Let ∠MPN = x°

∠O + ∠M + ∠N + ∠MPN = 360° (Sum of angles of a quadrilateral)

⇒ 45° + 90° + 90° + x° = 360°

⇒ 225° + x° = 360°

⇒ x° = 360° – 225°

⇒x = 135°

∠MPN = 135°

**Question 14.**

**The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles ?**

**Solution:**

The sides of a quadrilateral ABCD are produced in order, forming exterior angles ∠1, ∠2, ∠3 and ∠4.

Now ∠DAB + ∠1 = 180° (Linear pair) ……(i)

Similarly,

∠ABC + ∠2 = 180°

∠BCD + ∠3 = 180°

and ∠CDA + ∠4 = 180°

Adding, we get

∠DAB + ∠1 + ∠ABC + ∠2 + ∠BCD + ∠3 + ∠CDA + ∠4 = 180° + 180° + 180° + 180° = 720°

⇒ ∠DAB + ∠ABC + ∠CDA + ∠ADC + ∠1 + ∠2 + ∠3 + ∠4 = 720°

But ∠DAB + ∠ABC + ∠CDA + ∠ADB = 360° (Sum of angles of a quadrilateral)

360° + ∠1 + ∠2 + ∠3 + ∠4 = 720°

⇒ ∠l + ∠2 + ∠3 + ∠4 = 720° – 360° = 360°

Sum of exterior angles = 360°

**Question 15.**

**In the figure, the bisectors of ∠A and ∠B meet at a point P. If ∠C = 100° and ∠D = 50°, find the measure of ∠APB.**

**Solution:**

In quadrilateral ABCD,

∠D = 50°, ∠C = 100°

PA and PB are the bisectors of ∠A and ∠B.

In quadrilateral ABCD,

∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)

⇒ ∠A + ∠B + 100° + 50° = 360°

⇒ ∠A + ∠B + 150° = 360°’

⇒ ∠A + ∠B = 360° – 150° = 210°

and \(\frac { 1 }{ 2 }\) ∠A + \(\frac { 1 }{ 2 }\) ∠B = \(\frac { 210 }{ 2 }\) = 105°

(PA and PB are bisector of ∠A and ∠B respectively)

∠PAB + ∠PBA = 105°

⇒ ∠PAB + ∠PBA + ∠APB = 180° (Sum of angles of a triangle)

⇒ 105° + ∠APB = 180°

⇒ ∠APB = 180° – 105° = 75°

∠APB = 75°

**Question 16.**

**In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measure of each angle of the quadrilateral.**

**Solution:**

Sum of angles A, B, C and D of a quadrilateral = 360°

i.e. ∠A + ∠B + ∠C + ∠D = 360°

But ∠A = ∠B = ∠C = ∠D = 1 : 2 : 4 : 5

Let ∠A = x,

Then ∠B = 2x

∠C = 4x

∠D = 5x

x + 2x + 4x + 5x = 360°

⇒ 12x = 360°

⇒ x = \(\frac { 360 }{ 12 }\) = 30°

∠A = x = 30°

∠B = 2x = 2 x 30° = 60°

∠C = 4x = 4 x 30° = 120°

∠D = 5A = 5 x 30° = 150°

**Question 17.**

**In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that ∠COD = \(\frac { 1 }{ 2 }\) (∠A + ∠B).**

**Solution:**

In quadrilateral ABCD,

**Question 18.**

**Find the number of sides of a regular polygon when each of its angles has a measures of**

**(i) 160°**

**(ii) 135°**

**(iii) 175°**

**(iv) 162°**

**(v) 150°.**

**Solution:**

In a n-sided regular polygon, each angle

**Question 19.**

**Find the number of degrees in each exterior angle of a regular pentagon.**

**Solution:**

In a pentagon or a polygon, sum of exterior angles formed by producing the sides in order, is four right angles or 360°

Each exterior angle = \(\frac { 360 }{ 5 }\) = 72°

**Question 20.**

**The measure of angles of a hexagon are x°, (x – 5)° (x – 5)°, (2x – 5)°, (2x – 5)°, (2x + 20)°. Find the value of x.**

**Solution:**

We know that the sum of interior angels of a hexagon = 720° (180° x 4)

⇒ x + x – 5 + x – 5 + 2x – 5 + 2x – 5 + 2x + 20 = 720°

⇒ 9x – 20 + 20 = 720

⇒ 9x = 720

⇒ x = \(\frac { 720 }{ 9 }\) = 80°

x = 80°

**Question 21.**

**In a convex hexagon, prove that the sum of all interior angles is equal to twice the sum of its exterior angles formed by producing the sides in the same order.**

**Solution:**

In a convex hexagon ABCDEF, its sides AB, BG, CD, DE, EF and FA are produced in order forming exterior angles a, b, c, d, e, f

∠a + ∠b + ∠c + ∠d + ∠e + ∠f = 4 right angles (By definition)

By joining AC, AD, and AE, 4 triangles ABC, ACD, ADE and AEF are formed

In ∆ABC,

∠1 + ∠2 + ∠3 = 180° = 2 right angle (Sum of angles of a triangle) …… (i)

Similarly,

In ∆ACD,

∠4 +∠5 + ∠6 = 180° = 2 right angles

In ∆ADE,

∠1 + ∠8 + ∠9 = 2 right angles …(iii)

In ∆AEF,

∠10 + ∠11 + ∠12 = 2 right angles …(iv)

Joining (i), (ii), (iii) and (iv)

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 + ∠9 + ∠10 + ∠11 + ∠12 = 8 right angles

⇒ ∠2 + ∠3 + ∠5 + ∠6 + ∠8 + ∠9 + ∠11 + ∠12 + ∠1 + ∠4 + ∠7 + ∠10 = 8 right angles

⇒ ∠B + ∠C + ∠D + ∠E +∠F + ∠A = 8 right angles

⇒ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 2 (∠a + ∠b + ∠c + ∠d + ∠e + ∠f)

Sum of all interior angles = 2(the sum of exterior angles)

Hence proved.

**Question 22.**

**The sum of the interior angles of a polygon is three times the sum of its exterior angles. Determine the number of sides of the polygon.**

**Solution:**

Let number of sides of a regular polygon = n

Each interior angle = \(\frac { 2n – 4 }{ n }\) right angles

Sum of all interior angles = \(\frac { 2n – 4 }{ n }\) x n

right angles = (2n – 4) right angles

But sum of exterior angles = 4 right angles

According to the condition,

(2n – 4) = 3 x 4 (in right angles)

⇒ 2n – 4 = 12

⇒ 2n = 12 + 4 = 16

⇒ n = 8

Number of sides of the polygon = 8

**Question 23.**

**Determine the number of sides of a polygon whose exterior and interior angles are in the ratio 1 : 5.**

**Solution:**

Ratio in exterior angle and interior angles of a regular polygon = 1 : 5

But sum of interior and exterior angles = 180° (Linear pair)

By cross multiplication:

6n – 12 = 5n

⇒ 6n – 5n = 12

⇒ n = 12

Number of sides of polygon is 12

**Question 24.**

**PQRSTU is a regular hexagon. Determine each angle of ∆PQT.**

**Solution:**

In regular hexagon, PQRSTU, diagonals PT and QT are joined.

In ∆PUT, PU = UT

∠UPT = ∠UTP

But ∠UPT + ∠UTP = 180° – ∠U = 180° – 120° = 60°

∠UPT = ∠UTP = 30°

∠TPQ = 120° – 30° = 90° (QT is diagonal which bisect ∠Q and ∠T)

∠PQT = \(\frac { 120 }{ 2 }\) = 60°

Now in ∆PQT,

∠TPQ + ∠PQT + ∠PTQ = 180° (Sum of angles of a triangle)

⇒ 90° + 60° + ∠PTQ = 180°

⇒ 150° + ∠PTQ = 180°

⇒ ∠PTQ = 180° – 150° = 30°

Hence in ∆PQT,

∠P = 90°, ∠Q = 60° and ∠T = 30°

Hope given RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 are helpful to complete your math homework.

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