Rd Sharma Solutions for Class 8 Math Chapter 17 Understanding Shapes Iii Special Types Of Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Understanding Shapes Iii Special Types Of Quadrilaterals are extremely popular among Class 8 students for Math Understanding Shapes Iii Special Types Of Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Book of Class 8 Math Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Solutions. All Rd Sharma Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 17.10:

Question 3:

Can the following figures be parallelograms. Justify your answer.

Answer:

(i)No. This is because the opposite angles are not equal.(ii)Yes. This is because the opposite sides are equal.(ii)No, This is because the diagonals do not bisect each other.

Page No 17.10:

Question 4:

In the adjacent figure HOPE is a parallelogram. Find the angle measures x,y and z. State the geometrical truths you use to find them.

Answer:

HOP+70°=180°   (linear pair)HOP=180°-70°=110°x=HOP=110°  (opposite angles of a parallelogram are equal)EHP+HEP=180°    (sum of adjacent angles of a parallelogram is 180°)110°+40°+z=180°z=180°-150°=30°y=40° alternate angles 

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Question 5:

In the following figures GUNS and RUNS are  parallelograms. Find x and y.

Answer:

(i)Opposite sides are equal in a parallelogram. 3y-1=263y=27y=9Similarly, 3x=18x=6(ii)Diagonals bisect each other in a parallelogram. y-7=20y=27x-y=16x-27=16x=43

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Question 6:

In the following figure RISK and CLUE are parallelograms. Find the measure of x.

Answer:

In the parallelogram RISK:ISK+RKS=180°  (sum of adjacent angles of a parallelogram is 180°)ISK=180°-120°=60°Similarly, in parallelogram CLUE:CEU=CLU=70°  (opposite angles of a parallelogram are equal)In the triangle: x+ISK+CEU=180°x=180°-70°+60°x=180°-70°+60°=50°

Page No 17.10:

Question 7:

Two opposite angles of a parallelogram are (3x − 2)° and (50 − x)°. Find the measure of each angle of the parallelogram.

Answer:

Oppostie angles of a parallelogram are congurent. 3x-2°=50-x°3x°-2°=50°-x°3x°+x°=50°+2°4x°=52°x°=13°Putting the value of x in one angle:3x°-2°=39°-2°=37°Opposite angles are congurent: 50-x°=37°Let the remaining two angles be y and z.Angles y and z are congurent because they are also opposite angles. y=zThe sum of adjacent angles of a  parallelogram is equal to 180°.37°+y=180°y=180°-37°y=143°So, the anlges measure are:37°, 37°, 143° and 143°

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Question 8:

If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.

Answer:

Two adjacent angles of a parallelogram add up to 180°.Let x be the angle. x+2x3=180°5x3==180°x=72°2x3=2×72°3=108°Thus, two of the angles in the parallelogram are 108° and the other two are 72°.

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Question 9:

The measure of one angle of a parallelogram is 70°. What are the measures of the remaining angles?

Answer:

Given that one angle of the parallelogram is 70°.Since opposite angles have same value, if one is 70°, then the one directly opposite will also be 70°.So, let one angle be x°.x°+70°=180°  (the sum of adjacent angles of a parallelogram is 180° )x°=180°-70°x°=110°Thus, the remaining angles are 110°, 110° and 70°.

Page No 17.10:

Question 10:

Two adjacent angles of a parallelogram are as 1 : 2. Find the measures of all the angles of the parallelogram.

Answer:

Let the angle be A and B.The angles are in the ratio of 1:2.Measures of A and B are x° and 2x°.Then, C=A and D=B (opposite angles of a parallelogram are congruent)As we know that the sum of adjacent angles of a parallelogram is 180°. A+B=180°x°+2x°=180°3x°=180°x°=180°3=60°Thus, measure of A=60°, B=120°, C=60°  and D=120°.

Page No 17.10:

Question 11:

In a parallelogram ABCD, ∠D = 135°, determine the measure of ∠A and ∠B.

Answer:

In a parallelogram, opposite angles have the same value. D=B=135°Also, A+B+C+D=360°A+D=180° opposite angles have the same valueA=180°-135°=45°A=45°



Page No 17.11:

Question 12:

ABCD is a parallelogram in which ∠A = 70°. Compute ∠B, C and D.

Answer:

Opposite angles of a parallelogram are equal. C=70°=A.      B=DAlso, the sum of the adjacent angles of a parallelogram is 180°. A+B=180°70°+B=180° B=110° B=110°, C=70° and D=110°

Page No 17.11:

Question 13:

The sum of two opposite angles of a parallelogram is 130°. Find all the angles of the parallelogram.

Answer:

Let the angles be A, B, C and D.It is given that the sum of two opposite angles is 130°. A+C=130°A+A=130° opposite angles of a parallelogram are sameA=65°and C=65°The sum of adjacent angles of a parallelogram is 180°.A+B=180°65°+B=180°B=180°-65°B=115° D=115° A=65°, vB=115°, C=65° and D=115°.

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Question 14:

All the angles of a quadrilateral are equal to each other. Find the measure of each. Is the quadrilateral a parallelogram? What special type of parallelogram is it?

Answer:

Let the angle be x.All the angles are equal. x+x+x+x=360°4x=360°x=90°So, each  angle is 90° and quadrilateral is a parallelogram. It is a rectangle.

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Question 15:

Two adjacent sides of a parallelogram are 4 cm and 3 cm respectively. Find its perimeter.

Answer:

We know that the opposite sides of a parallelogram are equal.Two sides are given, i.e. 4 cm and 3 cm.Therefore, the rest of the sies will also be 4 cm and 3 cm. Perimeter = Sum of all the sides of a parallelogram=4+3+4+3=14 cm

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Question 16:

The perimeter of a parallelogram is 150 cm. One of its sides is greater than the other by 25 cm. Find the length of the sides of the parallelogram.

Answer:

Opposite sides of a parallelogram are same.Let two sides of the parallelogram be x and y.Given: x=y+25Also, x+y+x+y=150  (Perimeter= Sum of all the sides of a parallelogram)y+25+y+y+25+y=1504y=150-504y=100y=1004=25 x=y+25=25+25=50 Thus, the lengths of the sides of the parallelogram are 50 cm and 25 cm.

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Question 17:

The shorter side of a parallelogram is 4.8 cm and the longer side is half as much again as the shorter side. Find the perimeter of the parallelogram.

Answer:

Given:Shorter side=4.8 cmLonger side=4.82+4.8=7.2 cmPerimeter=Sum of all the sides =4.8+4.8+7.2+7.2=24 cm

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Question 18:

Two adjacent angles of a parallelogram are (3x − 4)° and (3x + 10)°. Find the angles of the parallelogram.

Answer:

We know that the adjacent angles of a parallelogram are supplementry.Hence, 3x+10° and 3x-4° are supplementry.3x+10°+3x-4°=180°6x°+6°=180°6x°=174°x=29°First angle = 3x+10°=3×29°+10°=97°Second angle = 3x-4°=83°Thus, the angles of the parallelogram are 97°, 83°, 97° and 83°.

Page No 17.11:

Question 19:

In a parallelogram ABCD, the diagonals bisect each other at O. If ∠ABC = 30°, ∠BDC = 10° and ∠CAB = 70°. Find:
DAB, ∠ADC, ∠BCD, ∠AOD, ∠DOC, ∠BOC, ∠AOB, ∠ACD, ∠CAB, ∠ADB, ∠ACB, ∠DBC and ∠DBA.

Answer:


 ABC=30° ADC=30° opposite angle of the parallelogramand BDA=ADC-BDC=30°-10°=20°BAC=ACD=70°  (alternate angle)In  ABC:CAB+ABC+BCA=180°70°+30°+BCA=180° BCA=80°DAB=DAC+CAB=70°+80°=150°BCD=150° opposite angle of the parallelogramDCA=CAB=70°In DOC:ODC+DOC+OCD=18010°+70°+DOC=180°DOC=100°DOC+BOC=180°BOC=180°-100°BOC=80°AOD=BOC=80°   vertically opposite angles AOB=DOC=100° vertically opposite angles CAB=70° givenADB=20°DBA=BDC=10°  (alternate angle)ADB=DBC=20°  (alternate angle)

Page No 17.11:

Question 20:

Find the angles marked with a question mark shown in Fig. 17.27

Answer:

In CEB:ECB+CBE+BEC=180°   (angle sum property of a triangle)40°+90°+EBC=180° EBC=50°Also, EBC=ADC=50° opposite angle of a parallelogramIn FDC:FDC+DCF+CFD=180°50°+90°+DCF=180° DCF=40°Now, BCE+ECF+FCD+FDC=180°  (in a parallelogram, the sum of alternate angles is 180° )50°+40°+ECF+40°=180°ECF=180°-50°+40°-40°=50°

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Question 21:

The angle between the altitudes of a parallelogram, through the same vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.

Answer:



Draw a parallelogram ABCD.Drop a perpendicular from B to the side AD, at the point E.Drop perpendicular from B to the side CD, at the point F. In the quadrilateral BEDF:EBF=60°,BED=90°BFD=90° EDF=360°-(60°+90°+90°)=120°In a parallelogram, opposite angles are congruent and adjacent angles are supplementary. In the parallelogram ABCD: B=D=120°A=C=180°-120°=60°

Page No 17.11:

Question 22:

In Fig. 17.28, ABCD and AEFG are parallelograms. If ∠C = 55°, what is the measure of ∠F?

Answer:

Both the parallelograms ABCD and AEFG are similar. C=A=55° (opposite angles of a parallelogram are equal) A=F=55°     (opposite angles of a parallelogram are equal)

Page No 17.11:

Question 23:

In Fig. 17.29, BDEF and DCEF are each a parallelogram. Is it true that BD = DC? Why or why not?

Answer:

In parallelogram BDEF BD=EF           ...(i)  (opposite sides of a parallelogram are equal)In parallelogram DCEF        CD=EF       ...(ii)     (opposite sides of a parallelogram are equal)From equations (i) and (ii)BD=CD       



Page No 17.12:

Question 24:

In Fig. 17.29, suppose it is known that DE = DF. Then, is ΔABC isosceles? Why or why not?
Fig. 17.29

Answer:

In FDE:DE=DF  FED=DFE.............(i)   (angles opposite to equal sides)In the IIgm BDEF: FBD= FED.......(ii)  (opposite angles of a parallelogram are equal)In the IIgm  DCEF:DCE=DFE......(iii)  (opposite angles of a parallelogram are equal)From equations (i), (ii) and (iii):FBD=DCEIn ABC:If FBD=DCE, then AB=AC (sides opposite to equal angles).Hence, ABC is isosceles.

Page No 17.12:

Question 25:

Diagonals of parallelogram ABCD intersect at O as shown in Fig. 17.30. XY contains O, and X, Y are points on opposite sides of the parallelogram. Give reasons for each of the following:
(i) OB = OD
(ii) ∠OBY = ∠ODX
(iii) ∠BOY = ∠DOX
(iv) ∆BOY ≅ ∆DOX
Now, state if XY is bisected at O.

Answer:

(i) Diagonals of a parallelogram bisect each other.
(ii) Alternate angles
(iii) Vertically opposite angles
(iv)
 In BOY and DOX:OB=OD  (diagonals of a parallelogram bisect each other)OBY=ODX   (alternate angles)BOY=DOX  (vertically opposite angles)


ASA congruence:
XO = YO (c.p.c.t)
So, XY is bisected at O.

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Question 26:

In Fig. 17.31, ABCD is a parallelogram, CE bisects ∠C and AF bisects ∠A. In each of the following, if the statement is true, give a reason for the same:


(i) ∠A = ∠C
(ii) FAB=12A
(iii) DCE=12C
(iv) CEB=FAB
(v) CE || AF

Answer:

(i) True, since opposite angles of a parallelogram are equal.
(ii) True, as AF is the bisector of A.
(iii) True, as CE is the bisector of C.
(iv) True
               CEB = DCE........(i)  (alternate angles)
               DCE=  FAB.........(ii)    (opposite angles of a parallelogram are equal)
            
  From equations (i) and (ii):
              CEB = FAB
                
(v) True, as corresponding angles are equal (CEB = FAB).

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Question 27:

Diagonals of a parallelogram ABCD intersect at O. AL and CM are drawn perpendiculars to BD such that L and M lie on BD. Is AL = CM? Why or why not?

Answer:


In ΔAOL and ΔCMO:AOL=COM( vertically opposite angle)....(i)ALO=CMO=90° (each right angle).....(ii)Using angle sum property: AOL+ALO+LAO=180°..........(iii)COM+CMO+OCM=180°......(iv)From equations (iii) and (iv):AOL+ALO+LAO=COM+CMO+OCMLAO=OCM  (from equations (i) and (ii) )In ΔAOL and ΔCMO:ALO=CMO (each right angle)AO=OC (diagonals of a parallelogram bisect each other)LAO=OCM   (proved above)So, ΔAOL is congruent to ΔCMO (SAS).AL=CM [cpct]

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Question 28:

Points E and F lie on diagonal AC of a parallelogram ABCD such that AE = CF. What type of quadrilateral is BFDE?

Answer:



In the IIgm ABCD:AO=OC......(i)  (diagonals of a parallelogram bisect each other)AE=CF.......(ii)  (given)Subtracting (ii) from (i):AO-AE=OC-CFEO=OF......... (iii)In DOE and BOF:EO=OF (proved above)DO=OB  (diagonals of a parallelogram bisect each other)DOE =BOF  (vertically opposite angles)By SAS congruence: DOE BOF DE=BF (c.p.c.t)In BOE and DOF:EO=OF (proved above)DO=OB  (diagonals of a parallelogram bisect each other)DOF =BOE  (vertically opposite angles)By SAS congruence: DOE BOF DF=BE (c.p.c.t)Hence, the pair of opposite sides are equal. Thus, DEBF is a parallelogram.

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Question 29:

In a parallelogram ABCD, AB = 10 cm, AD = 6 cm. The bisector of ∠A meets DC in E, AE and BC produced meet at F. Find te length CF.

Answer:



AE is the bisector of A. DAE=BAE =x  BAE=AED=x  (alternate angles)Since opposite angles in ADE are equal, ADE is an isosceles triangle. AD=DE= 6 cm  (sides opposite to equal angles)AB=CD=10 cm CD= DE+ ECEC=CD-DEEC=10-6=4 cmDEA=CEF=x  (vertically opposite angle)EAD=EFC=x  (alternate angles)Since opposite angles in EFC are equal, EFC is an isosceles triangle. CF=CE= 4 cm (sides opposite to equal angles) CF= 4cm



Page No 17.16:

Question 1:

Which of the following statements are true for a rhombus?
(i) It has two pairs of parallel sides.
(ii) It has two pairs of equal sides.
(iii) It has only two pairs of equal sides.
(iv) Two of its angles are at right angles.
(v) Its diagonals bisect each other at right angles.
(vi) Its diagonals are equal and perpendicular.
(vii) It has all its sides of equal lengths.
(viii) It is a parallelogram.
(ix) It is a quadrilateral.
(x) It can be a square.
(xi) It is a square.

Answer:

(i) True
(ii) True
(iii) True
(iv) False
(v) True
(vi) False

Diagonals of a rhombus are perpendicular, but not equal.

(vii) True
(viii) True

It is a parallelogram because it has two pairs of parallel sides.

(ix) True

It is a quadrilateral because it has four sides.

(x) True

It can be a square if each of the angle is a right angle.

(xi) False

It is not a square because each of the angle is a right angle in a square.



Page No 17.17:

Question 2:

Fill in the blanks, in each of the following, so as to make the statement true:
(i) A rhombus is a parallelogram in which ......
(ii) A square is a rhombus in which ......
(iii)  A rhombus has all its sides of ...... length.
(iv) The diagonals of a rhombus ...... each other at ...... angles.
(v) If the diagonals of a parallelogram bisect each other at right angles, then it is a ......

Answer:

(i) A rhombus is a parallelogram in which adjacent sides are equal.
(ii) A square is a rhombus in which all angles are right angled.
(iii) A rhombus has all its sides of equal length.
(iv) The diagonals of a rhombus bisect each other at right angles.
(v) If the diagonals of a parallelogram bisect each other at right angles, then it is a rhombus.

Page No 17.17:

Question 3:

The diagonals of a parallelogram are not perpendicular. Is it a rhombus? Why or why not?

Answer:

No, it is not a rhombus. This is because diagonals of a rhombus must be perpendicular.

Page No 17.17:

Question 4:

The diagonals of a quadrilateral are perpendicular to each other. Is such a quadrilateral always a rhombus? If your answer is 'No', draw a figure to justify your answer.

Answer:

No, it is not so.
Diagonals of a rhombus are perpendicular and bisect each other. Along with this, all of its sides are equal. In the figure given below, the diagonals are perpendicular to each other, but do not bisect each other.

Page No 17.17:

Question 5:

ABCD is a rhombus. If ∠ACB = 40°, find ∠ADB.

Answer:


In a rhombus, the diagonals  are perpendicular. BPC=90°From  BPC, the sum of angles is 180°. CBP+BPC+PBC=180°CBP=180°-BPC-PBCCBP=180°-40°-90°=50° ADB=CBP=50°  (alternate angle)

Page No 17.17:

Question 6:

If the diagonals of a rhombus are 12 cm and 16cm, find the length of each side.

Answer:


All sides of a rhombus are equal in length.The diagonals intersect at 90° and the sides of the rhombus form right triangles.One leg of these right triangles is equal to 8 cm and the other is equal to 6 cm.The sides of the triangle form the hypotenuse of these right triangles.So, we get:(82+62)cm2=(64+36)cm2=100 cm2The hypotneuse is the square root of 100cm2. This makes the hypotneuse equal to 10.Thus, the side of the rhombus is equal to 10 cm.

Page No 17.17:

Question 7:

Construct a rhombus whose diagonals are of length 10 cm and 6 cm.

Answer:


1. Draw AC equal to 10 cm.
2. Draw XY, the right bisector of AC, meeting it at O.
3. With O as centre and radius equal to half of the length of the other diagonal,
    i.e. 3 cm, cut OB = OD = 3 cm.
4. Join AB, AD and CB, CD.

Page No 17.17:

Question 8:

Draw a rhombus, having each side of length 3.5 cm and one of the angles as 40°.

Answer:



1. Draw a line segment AB of 3.5 cm.
2. Draw BAX equal to 40°.
3. With A as centre and the radius equal to AB, cut AD at 3.5 cm.
4. With D as centre, cut an arc of radius 3.5 cm.
5. With B as centre, cut an arc of radius 3.5 cm. This arc cuts the arc of step 4 at C.
6. Join DC and BC.

Page No 17.17:

Question 9:

One side of a rhombus is of length 4 cm and the length of an altitude is 3.2 cm. Draw the rhombus.

Answer:


1. Draw a line segment AB of 4 cm.
2. Draw a perpendicular XY on AB, which intersects AB at P.
3. With P as centre, cut PE at 3.2 cm.
4. Draw a line WZ that passes through E. This line should be parallel to AB.
5. With A as centre, draw an arc of radius 4 cm that cuts WZ at D.
6. With D as centre and radius 4 cm, cut line DZ. Label it as point C.
4. Join AD and CB.

Page No 17.17:

Question 10:

Draw a rhombus ABCD, if AB = 6 cm and AC = 5 cm.

Answer:



1. Draw a line segment AC of 5 cm.
2. With A as centre, draw an arc of radius 6 cm on each side of AC.
3. With C as centre, draw an arc of radius 6 cm on each side of AC. These arcs intersect the arcs of step 2 at B and D.
4. Join AB, AD, CD and CB.

Page No 17.17:

Question 11:

ABCD is a rhombus and its diagonals intersect at O.
(i) Is ∆BOC ≅ ∆DOC? State the congruence condition used?
(ii) Also state, if ∠BCO = ∠DCO.

Answer:


(i) Yes

In BCO andDCO:OC=OC (common)BC=DC (all sides of a rhombus are equal)BO=OD (diagonals of a rhomus bisect each other)By SSS congruence:BCO DCO

(ii) Yes
By c.p.c.t:
BCO=DCO

Page No 17.17:

Question 12:

Show that each diagonal of a rhombus bisects the angle through which it passes.

Answer:


In AED and DEC: AE=EC  (diagonals bisect each other)AD=DC (sides are equal)DE=DE (common)By SSS congruence: AED CEDADE=CDE (c.p.c.t)Similarly, we can prove AEB and BEC, BEC and DEC, AED and AEB are congruent to each other.Hence, diagonal of a rhombus bisects the angle through which it passes.

Page No 17.17:

Question 13:

ABCD is a rhombus whose diagonals intersect at O. If  AB = 10 cm, diagonal BD = 16 cm, find the length of diagonal AC.

Answer:


We know that the diagonals of a rhombus bisect each other at right angles. BO=12BD=(12×16) cm=8cmAB=10 cm andAOB=90°From right OAB: AB2=AO2+BO2AO2=(AB2BO2)AO2=(10)2-(8)2cm2AO2=(100-64) cm2=36cm2AO=36 cm=6cm AC=2×AO=(2×6) cm=12 cm

Page No 17.17:

Question 14:

The diagonals of a quadrilateral are of lengths 6 cm and 8 cm. If the diagonals bisect each other at right angles, what is the length of each side of the quadrilateral?

Answer:


Let the given quadrilateral be ABCD in which diagonals AC is equal to 6 cm and BD is equal to 8 cm.Also, it is given that the diagonals bisect each other at right angle, at point O. AO=OC=12AC=3 cmAlso, OB=OD=12BD=4 cmIn right AOB:AB2=OA2+OB2AB2=(9+16) cm2AB2=25 cm2AB=5 cmThus, the length of each side of the quadrilateral is 5 cm.



Page No 17.22:

Question 1:

Which of the following statements are true for a rectangle?
(i) It has two pairs of equal sides.
(ii) It has all its sides of equal length.
(iii) Its diagonals are equal.
(iv) Its diagonals bisect each other.
(v) Its diagonals are perpendicular.
(vi) Its diagonals are perpendicular and bisect each other.
(vii) Its diagonals are equal and bisect each other.
(viii) Its diagonals are equal and perpendicular, and bisect each other.
(ix) All rectangles are squares.
(x) All rhombuses are parallelograms.
(xi) All squares are rhombuses and also rectangles.
(xii) All squares are not parallelograms.

Answer:

(i) True
(ii) False
(iii) True
(iv) True
(v) False
(vi) False
The diagonals are not perpendicular to each other.
(vii) True
(viii) False
The diagonals are not perpendicular to each other.
(ix) False
All sides are not equal.
(x) True
(xi) True
(xii) False
All squares are parallelogram.

Page No 17.22:

Question 2:

Which of the following statements are true for a square?
(i) It is a rectangle.
(ii) It has all its sides of equal length.
(iii) Its diagonals bisect each other at right angle.
(iv) Its diagonals are equal to its sides.

Answer:

(i) True
(ii) True
(iii) True
(iv) False
This is because the hypotenuse in any right angle triangle is always greater than its two sides.



Page No 17.23:

Question 3:

Fill in the blanks in each of the following, so as to make the statement true:
(i) A rectangle is a parallelogram in which .....
(ii) A square is a rhombus in which .....
(iii) A square is a rectangle in which .....

Answer:

(i) A rectangle is a parallelogram in which each angle is a right angle.

(ii) A square is a rhombus in which each angle is a right angle.

(iii) A square is a rectangle in which the adjacent sides are equal.

Page No 17.23:

Question 4:

A window frame has one diagonal longer than the other. Is the window frame a rectangle? Why or why not?

Answer:

No, since diagonals of a rectangle are equal.

Page No 17.23:

Question 5:

In a rectangle ABCD, prove that ∆ACB ≅ ∆CAD.

Answer:


In ∆ACB and ∆CAD:
AB = CD (rectangle property)

AD = BC (rectangle property)

AC ( common side )

Hence, by SSS criterion, it is proved that ACBCAD.

Page No 17.23:

Question 6:

The sides of a rectangle are in the ratio 2 : 3, and its perimeter is 20 cm. Draw the rectangle.

Answer:



Let the side be x cm and y cm.So, we have: 2x+y=20 Sides are in the ratio 2:3. y=3x2Putting the value of y:2x+3x2=20 2x+3x2=105x=20x=4 y=3×42=6Thus, sides of the rectangle will be 4 cm and 6 cm.ABCD is the rectangle having sides 4 cm and 6 cm.

Page No 17.23:

Question 7:

The sides of a rectangle are in the ratio 4 : 5. Find its sides if the perimeter is 90 cm.

Answer:

Let the side be x cm and y cm.So, we have: 2x+y=90 Sides are in the ratio 4:5.  y=5x4Putting the value of y:2x+5x4=90 4x+5x4=459x=180x=20 y=5×204=25Thus, the sides of the rectangle will be 20 cm and 25 cm.

Page No 17.23:

Question 8:

Find the length of the diagonal of a rectangle whose sides are 12 cm and 5 cm.

Answer:


Using Pythagoras theorem:AD2+DC2=AC252+122=AC225+14=AC2169=AC2AC=169=13 cmThus, length of the diagonal is 13 cm.

Page No 17.23:

Question 9:

Draw a rectangle whose one side measures 8 cm and the length of each of whose diagonals is 10 cm.

Answer:

(i) Draw a side AB, equal to 8 cm.
(ii) With A as the centre, draw an arc of length 10 cm.
(iii) Draw ABX = 90°, which intersects the arc at C.
(iv)Draw BAY = 90°.
(v) With C as the centre, draw an arc of length 8 cm.
(vi) Join CD.
Thus, ABCD is the required rectangle.

Page No 17.23:

Question 10:

Draw a square whose each side measures 4.8 cm.

Answer:


 (i) Draw side AB=4.8 cm.(ii) From A, make an angle of 90° and cut it at 4.8 cm and mark it point D.(iii) From B, make an angle of 90° and cut it at 4.8 cm and mark it point C.(iv) Join C and D.Thus, ABCD is the required square.

Page No 17.23:

Question 11:

Identify all the quadrilaterals that have:
(i) Four sides of equal length
(ii) Four right angles

Answer:

(i) If all four sides are equal, then it can be either a square or a rhombus.

(ii) All four right angles, make it either a rectangle or a square.

Page No 17.23:

Question 12:

Explain how a square is
(i) a quadrilateral?
(ii) a parallelogram?
(iii) a rhombus?
(iv) a rectangle?

Answer:

(i) Since a square has four sides, it is a quadrilateral.

(ii) Since the opposite sides are parallel and equal, it is a parallelogram.

(iii) Since the diagonals bisect each other and all the sides are equal, it is a rhombus.

(iv) Since the opposite sides are equal and all the angles are right angles, it is a rectangle.

Page No 17.23:

Question 13:

Name the quadrilaterals whose diagonals:
(i) bisect each other
(ii) are perpendicular bisector of each other
(iii) are equal.

Answer:

(i) Rhombus, parallelogram, rectangle and square
(ii) Rhombus and square
(iii) Rectangle and square

Page No 17.23:

Question 14:

ABC is a right-angled trianle and O is the mid-point of the side opposite to the right angle. Explain why O is equidistant from A, B and C.

Answer:

(i) Construct a triangle ABC right angle at B.(ii) Suppose O is the mid point of AC.(iii)Complete the rectangle ABCD having AC as its diagonal.Since diagonals of a rectangle are equal and they bisect each other, O is the midpoint of both AC and BD. OA=OB=OC

Page No 17.23:

Question 15:

A mason has made a concrete slab. He needs it to be rectangular. In what different ways can he make sure that it is rectangular?

Answer:

(i) By measuring each angle - Each angle of a rectangle is 90°.

(ii) By measuring the length of the diagonals - Diagonals of a rectangle are equal.

(iii) By measuring the sides of rectangle - Each pair of opposite sides are equal.



Page No 17.9:

Question 1:

Given below is a parallelogram ABCD. Complete each statement along with the definition or property used.
(i) AD =
(ii) ∠DCB =
(iii) OC =
(iv) ∠DAB + ∠CDA =

Answer:

The correct figure is

(i)AD=BC (opposite sides of a parallelogram are equal)(ii)DCB=BAD opposite angles are equal(iii)OC=OA diagonals of a prallelogram bisect each other(iv)DAB+CDA=180° the sum of two adjacent angles of a parallelogram is 1800

Page No 17.9:

Question 2:

The following figures are parallelograms. Find the degree values of the unknowns x, y, z.

Answer:

(i)Opposite angles of a parallelogram are same. x=z and y=100°Also, y+z=180°   (sum of adjacent angles of a quadrilateral is 180°)z+100°=180°x=180°-100°x=80° x=80°, y=100° and z=80°(ii)Opposite angles of a parallelogram are same. x=y and RQP=100°PSR+SRQ=180°y+50°=180°x=180°-50°x=130° x=130°, y=130°  Since y and z are alternate angles, z=130°.(iii)Sum of all angles in a triangle is 180°.30°+90°+z=180°z=60°Opposite angles are equal in parallelogram. y=z=60°and x=30°  (alternate angles)(iv)x=90°   (vertically opposite angle)Sum of all angles in a triangle is 180°. y+90°+30°=180°y=180°-(90°+30°)=60°y=z=60°  (alternate angles)(v)Opposite angles are equal in a parallelogram. y=80°y+x=180°     x=180°-100°=80°z=y=80°   (alternate angles)(vi)y=112°  (opposite angles are equal in a parallelogram)In UTW :x+y+40°=180° (angle sum property of a triangle)x=180°-(112°-40°)=28°Bottom left vertex=180°-112°=68° z=x=28° (alternate angles)



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