Maharashtra Board Class 8 Maths Chapter 8 Quadrilateral Constructions and Types Set 8.2 Solutions

Practice Set 8.2 Class 8 Answers Chapter 8 Quadrilateral: Constructions And Types Maharashtra Board

Quadrilateral: Constructions And Types Class 8 Maths Chapter 8 Practice Set 8.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 8.2 Chapter 8 Solutions Answers

 

Question 1. Draw a rectangle ABCD such that I(AB) = 6.0 cm and I(BC) = 4.5 cm.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक आयत ABCD का रफ चित्र है। भुजा AB 6 सेमी और भुजा BC 4.5 सेमी है। कोण B पर एक समकोण है, जो आयत की विशेषता है।
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आयत ABCD का अंतिम निर्माण है। भुजा AB 6 सेमी, भुजा BC 4.5 सेमी है। सभी कोण 90 डिग्री के हैं। यह चित्र आयत की सभी निर्दिष्ट मापों और कोणों को दर्शाता है।In simple words: To draw a rectangle, you first draw a base, then construct 90-degree angles at its ends, and mark off the adjacent side lengths to complete the figure.

🎯 Exam Tip: Accuracy in measuring lengths and constructing 90-degree angles using a protractor or compass is crucial for precise rectangle construction.

 

Question 2. Draw a square WXYZ with side 5.2 cm.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वर्ग WXYZ का रफ चित्र है। वर्ग की प्रत्येक भुजा 5.2 सेमी है। कोण W पर एक समकोण है, जो वर्ग की विशेषता है।
ℹ️ चित्र व्याख्या (Diagram Explanation): यह वर्ग WXYZ का अंतिम निर्माण है। वर्ग की प्रत्येक भुजा 5.2 सेमी है और सभी कोण 90 डिग्री के हैं। यह चित्र वर्ग की सभी निर्दिष्ट मापों और कोणों को दर्शाता है।In simple words: To draw a square, you draw one side, then construct 90-degree angles at both ends, and mark off the same side length to complete the square.

🎯 Exam Tip: Ensure all four sides are drawn with exactly the same length and all corners are perfect right angles for accurate square construction.

 

Question 3. Draw a rhombus KLMN such that its side is 4 cm and \( m\angle K = 75^\circ \).
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समचतुर्भुज KLMN का रफ चित्र है। समचतुर्भुज की प्रत्येक भुजा 4 सेमी है और कोण K 75 डिग्री है।
ℹ️ चित्र व्याख्या (Diagram Explanation): यह समचतुर्भुज KLMN का अंतिम निर्माण है। प्रत्येक भुजा 4 सेमी है और कोण K 75 डिग्री है। इस चित्र में समचतुर्भुज की सभी निर्दिष्ट मापों और कोणों को दर्शाया गया है, जहाँ विपरीत कोण समान होते हैं।In simple words: To draw a rhombus, you start with one side and its given angle, then use a compass to mark off the equal side lengths to form the parallelogram.

🎯 Exam Tip: Accurately drawing the initial angle and then using a compass to ensure all four sides are of equal length is vital for correct rhombus construction.

 

Question 4. If diagonal of a rectangle is 26 cm and one side is 24 cm, find the other side.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक आयत ABCD को दर्शाता है। विकर्ण AC 26 सेमी है और एक भुजा BC 24 सेमी है। कोण B पर एक समकोण है, जो आयत की विशेषता है। Let ABCD be the rectangle.
I(BC) = 24cm, I(AC) = 26cm
In \( \triangle\text{ABC} \),
\( m\angle ABC = 90^\circ \quad \dots[\text{Angle of a rectangle}] \)

\( \implies [I(AC)]^2 = [I(AB)]^2 + [I(BC)]^2 \)
\( \dots[\text{Pythagoras theorem}] \)

\( \implies (26)^2 = [I(AB)]^2 + (24)^2 \)

\( \implies (26)^2 - (24)^2 = [I(AB)]^2 \)

\( \implies (26 + 24) (26 - 24) = [I(AB)]^2 \)
\( \dots\left[ a^2 - b^2 = (a + b)(a - b) \right] \)

\( \implies 50 \times 2 = [I(AB)]^2 \)

\( \implies 100 = [I(AB)]^2 \)
i.e. \( [I(AB)]^2 = 100 \)

\( \implies I(AB) = \sqrt{100} \)
\( \dots[\text{Taking square root of both sides}] \)

\( \implies I(AB) = 10 \text{ cm} \)

\( \implies \) The length of the other side is 10 cm.In simple words: For a rectangle, any diagonal forms a right-angled triangle with two adjacent sides, allowing us to use the Pythagorean theorem to find an unknown side.

🎯 Exam Tip: Remember the Pythagorean theorem (\( a^2 + b^2 = c^2 \)) is fundamental for solving problems involving diagonals and sides of rectangles and squares.

 

Question 5. Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.
Answer:
In rhombus ABCD,

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समचतुर्भुज ABCD को दर्शाता है। इसके विकर्ण AC और BD हैं, जो बिंदु O पर प्रतिच्छेद करते हैं। विकर्ण AC की लंबाई 16 सेमी और विकर्ण BD की लंबाई 12 सेमी है। I(AC) = 16 cm and I(BD) = 12 cm.
Let the diagonals of rhombus ABCD intersect at point O.
\( I(AO) = \frac{1}{2} I(AC) \)
\( \dots[\text{Diagonals of a rhombus bisect each other}] \)

\( \implies I(AO) = \frac{1}{2} \times 16 \)

\( \implies I(AO) = 8 \text{ cm} \)
Also, \( I(DO) = \frac{1}{2} I(BD) \)
\( \dots[\text{Diagonals of a rhombus bisect each other}] \)

\( \implies I(DO) = \frac{1}{2} \times 12 \)

\( \implies I(DO) = 6 \text{ cm} \)
In \( \triangle\text{ADO} \),
\( m\angle DOA = 90^\circ \)
\( \dots[\text{Diagonals of a rhombus are perpendicular to each other}] \)

\( \implies [I(AD)]^2 = [I(AO)]^2 + [I(DO)]^2 \)
\( \dots[\text{Pythagoras theorem}] \)
\( = (8)^2 + (6)^2 \)
\( = 64 + 36 \)

\( \implies [I(AD)]^2 = 100 \)

\( \implies I(AD) = \sqrt{100} \)
\( [\text{Taking square root of both sides}] \)

\( \implies I(AD) = 10 \text{ cm} \)

\( \implies I(AB) = I(BC) = I(CD) = I(AD) = 10 \text{ cm} \)
\( \dots[\text{Sides of a rhombus are congruent}] \)
Perimeter of rhombus ABCD
\( = I(AB) + I(BC) + I(CD) + I(AD) \)
\( = 10+10+10+10 \)
\( = 40 \text{ cm} \)

\( \implies \) The side and perimeter of the rhombus are 10 cm and 40 cm respectively.In simple words: The diagonals of a rhombus intersect at a right angle and bisect each other, forming right-angled triangles from which we can find the side length using Pythagoras theorem.

🎯 Exam Tip: Clearly label the intersection point of diagonals and remember their properties (perpendicular bisection) to form right-angled triangles for calculation of side length and perimeter.

 

Question 6. Find the length of diagonal of a square with side 8 cm.
Answer:
Let XYWZ be the square of side 8cm.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वर्ग XYWZ को दर्शाता है। वर्ग की प्रत्येक भुजा 8 सेमी है। विकर्ण XW वर्ग के दो विपरीत शीर्षों को जोड़ता है। seg XW is a diagonal.
In \( \triangle\text{XYW} \),
\( m\angle XYW = 90^\circ \)
\( [\text{Angle of a square}] \)

\( \implies [I(XW)]^2 = [I(XY)]^2 + [I(YW)]^2 \)
\( \dots[\text{Pythagoras theorem}] \)
\( = (8)^2 + (8)^2 \)
\( = 64 + 64 \)

\( \implies [I(XW)]^2 = 128 \)

\( \implies I(XW) = \sqrt{128} \)
\( \dots[\text{Taking square root of both sides}] \)
\( = \sqrt{64 \times 2} \)
\( = 8 \sqrt{2} \text{ cm} \)

\( \implies \) The length of the diagonal of the square is \( 8 \sqrt{2} \text{ cm} \).In simple words: A square's diagonal creates a right-angled triangle with two sides, so its length can be found using the Pythagorean theorem, which simplifies to side times root 2.

🎯 Exam Tip: For a square with side 's', the diagonal 'd' can be quickly calculated using the formula \( d = s\sqrt{2} \), saving time during exams.

 

Question 7. Measure of one angle of a rhombus is \( 50^\circ \), find the measures of remaining three angles.
Answer:
Let ABCD be the rhombus.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समचतुर्भुज ABCD को दर्शाता है। कोण A का माप 50 डिग्री है। इस चित्र में समचतुर्भुज के चार शीर्ष और एक ज्ञात कोण दर्शाए गए हैं। \( m\angle A = 50^\circ \)
\( m\angle C = m\angle A \)
\( \dots[\text{Opposite angles of a rhombus are congruent}] \)

\( \implies m\angle C = 50^\circ \)
Also, \( m\angle D = m\angle B \quad \dots(\text{i}) \)
\( \dots[\text{Opposite angles of a rhombus are congruent}] \)
In ABCD,
\( m\angle A + m\angle B + m\angle C + m\angle D = 360^\circ \)
\( \dots[\text{Sum of the measures of the angles of a quadrilateral is } 360^\circ] \)

\( \implies 50^\circ + m\angle B + 50^\circ + m\angle D = 360^\circ \)

\( \implies m\angle B + m\angle D + 100^\circ = 360^\circ \)

\( \implies m\angle B + m\angle D = 360^\circ - 100^\circ \)

\( \implies m\angle B + m\angle B = 260^\circ \quad \dots[\text{From (i)}] \)

\( \implies 2m\angle B = 260^\circ \)

\( \implies m\angle B = \frac{260}{2} \)

\( \implies m\angle B = 130^\circ \)

\( \implies m\angle D = m\angle B = 130^\circ \quad \dots[\text{From (i)}] \)

\( \implies \) The measures of the remaining angles of the rhombus are \( 130^\circ, 50^\circ \) and \( 130^\circ \).In simple words: In a rhombus, opposite angles are equal, and the sum of adjacent angles is 180 degrees, allowing you to find all angles from just one given angle.

🎯 Exam Tip: Remember that adjacent angles in a rhombus are supplementary (sum to \( 180^\circ \)), and opposite angles are congruent (equal), which simplifies finding unknown angles.

 

Maharashtra Board Class 8 Maths Chapter 8 Quadrilateral: Constructions And Types Practice Set 8.2 Intext Questions And Activities

 

Question 1. Construct a rectangle PQRS by taking two convenient adjacent sides. Name the point of intersection of diagonals as T. Using divider and ruler, measure the following lengths.
(i) lengths of opposite sides, seg QR and seg PS.
(ii) lengths of seg PQ and seg SR.
(iii) lengths of diagonals PR and QS.
(iv) lengths of seg PT and seg TR, which are parts of the diagonal PR.
(v) lengths of seg QT and seg TS, which are parts of the diagonal QS.
Observe the measures. Discuss about the measures obtained by your classmates.
Answer:
Solution:
Draw a rectangle PQRS such that, I(PQ) = 3 cm and I(QR) = 4 cm.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक आयत PQRS का रफ चित्र है। भुजा PQ 3 सेमी और भुजा QR 4 सेमी है। विकर्ण बिंदु T पर प्रतिच्छेद करते हैं। Steps of construction:
(i) As shown in the rough figure, draw seg QR of length 4 cm.
(ii) Placing the centre of the protractor at point Q, draw ray QW making an angle of \( 90^\circ \) with seg QR.
(iii) By taking a distance of 3 cm on the compass and placing it at point Q, draw an arc on ray QW. Name the point as P.
(iv) Draw ray PV and ray RU making an angle of \( 90^\circ \) with seg PQ and seg QR respectively.
(v) Name the point of intersection of ray PV and ray RU as S.
PQRS is the required rectangle.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आयत PQRS का अंतिम निर्माण है। भुजा PQ 3 सेमी, भुजा QR 4 सेमी है। विकर्ण PR और QS बिंदु T पर प्रतिच्छेद करते हैं, जिनकी मापों का अवलोकन किया गया है। From the figure,
(i) I(QR) = I(PS) = 4 cm
(ii) I(PQ) = I(SR) = 3 cm
(iii) I(PR) = I(QS) = 5 cm
(iv) I(PT) = I(TR) = 2.5 cm
(v) I(QT) = I(TS) = 2.5 cm
From the above measures, we can say that for any rectangle,
(i) Opposite sides are congruent.
(ii) Diagonals are congruent.
(iii) Diagonals bisect each other.In simple words: By constructing a rectangle and measuring its sides and diagonals, we can observe that opposite sides are equal, diagonals are equal, and diagonals bisect each other.

🎯 Exam Tip: Precision in construction and measurement is vital to verify the fundamental geometric properties of rectangles, such as congruent opposite sides and diagonals, and diagonal bisection.

 

Question 2. Draw a square by taking convenient length of side. Name the point of intersection of its diagonals as E. Using the apparatus in a compass box, measure the following lengths.
(i) lengths of diagonal AC and diagonal BD.
(ii) lengths of two parts of each diagonal made by point E.
(iii) all the angles made at the point E.
(iv) parts of each angle of the square made by each diagonal, (e.g. \( \angle\text{ADB} \) and \( \angle\text{CDB} \)).
Observe the measures. Also observe the measures obtained by your classmates and discuss about them.
Answer:
Solution:
Draw a square ABCD such that its side is 5cm

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वर्ग ABCD का रफ चित्र है। वर्ग की प्रत्येक भुजा 5 सेमी है। विकर्ण बिंदु E पर प्रतिच्छेद करते हैं। Steps of construction:
(i) As shown in the rough figure, draw seg BC of length 5 cm.
(ii) Placing the centre of the protractor at point B, draw ray BP making an angle of \( 90^\circ \) with seg BC.
(iii) By taking a distance of 5 cm on the compass and placing it at point B, draw an arc on ray BP. Name the point as A.
(iv) Placing the centre of the protractor at point C, draw ray CQ making an angle of \( 90^\circ \) with seg BC.
(v) By taking a distance of 5 cm on the compass and placing it at point C, draw an arc on ray CQ. Name the point as D.
(vi) Draw seg AD.
ABCD is the required square.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह वर्ग ABCD का अंतिम निर्माण है। वर्ग की प्रत्येक भुजा 5 सेमी है। विकर्ण AC और BD बिंदु E पर प्रतिच्छेद करते हैं। From the figure,
(i) I(AC) = I(BD) \( \approx \) 7cm
(ii) I(AE) = I(EC) \( \approx \) 3.5cm,
I(BE) = I(ED) \( \approx \) 3.5cm
(iii) \( m\angle AED = m\angle BEC = m\angle CED = m\angle BEA = 90^\circ \)
(iv) Angles made by diagonal AC:
\( m\angle BAC = m\angle DAC = 45^\circ \)
\( m\angle BCA = m\angle DCA = 45^\circ \)
Angles made by diagonal BD:
\( m\angle ABD = m\angle CBD = 45^\circ \)
\( m\angle ADB = m\angle CDB = 45^\circ \)
From the above measures, we can say that for any square,
(i) Diagonals are congruent.
(ii) Diagonals bisect each other.
(iii) Diagonals are perpendicular to each other.
(iv) Diagonals bisect the opposite angles.In simple words: Constructing and measuring a square reveals that its diagonals are equal, bisect each other perpendicularly, and also bisect the square's vertex angles.

🎯 Exam Tip: Accurately drawing all 90-degree angles and equal sides ensures precise observation of a square's diagonal properties, including perpendicular bisection and angle bisection.

 

Question 3. Draw a rhombus EFGH by taking convenient length of side and convenient measure of an angle.
Draw its diagonals and name their point of Intersection as M.
(i) Measure the opposite angles of the quadrilateral and angles at the point M.
(ii) Measure the two parts of every angle made by the diagonal.
(iii) Measure the lengths of both diagonals. Measure the two parts of diagonals made by point M.
Observe the measures. Also observe the measures obtained by your classmates and discuss about them.
Answer:
Solution:
Draw a rhombus EFGH such that its side is 5 cm and \( m\angle F = 60^\circ \).

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समचतुर्भुज EFGH का रफ चित्र है। इसकी प्रत्येक भुजा 5 सेमी है और कोण F 60 डिग्री है। विकर्ण बिंदु M पर प्रतिच्छेद करते हैं। Steps of construction:
(i) As shown in the rough figure, draw seg FG of length 5 cm.
(ii) Placing the centre of the protractor at point F, draw ray FX making an angle \( 60^\circ \) with seg FG.
(iii) By taking a distance of 5 cm on the compass and placing it at point F, draw an arc on ray FX. Name the point as E.
(iv) By taking a distance of 5 cm on the compass and placing it at point E and point G, draw arcs. Name the point of intersection of arcs as H. EFGH is the required rhombus.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह समचतुर्भुज EFGH का अंतिम निर्माण है। प्रत्येक भुजा 5 सेमी है और कोण F 60 डिग्री है। विकर्ण बिंदु M पर प्रतिच्छेद करते हैं। From the figure,
(i) Opposite angles:
\( m\angle EFG = m\angle GHE = 60^\circ \),
\( m\angle FEH = m\angle HGF = 120^\circ \)
Angles at the point M:
\( m\angle EMF = m\angle FMG = m\angle GMH = m\angle HME = 90^\circ \)
(ii) Angles made by diagonal FH:
\( m\angle EFH = m\angle GFH = 30^\circ \)
\( m\angle EHF = m\angle GHF = 30^\circ \)
Angles made by diagonal EG:
\( m\angle FEG = m\angle HEG = 60^\circ \)
\( m\angle FGE = m\angle HGE = 60^\circ \)
(iii) I(FH) \( \approx \) 8.6 cm
I(EG) = 5 cm
I(FM) = I(HM) \( \approx \) 4.3 cm
I(EM) = I(GM) \( \approx \) 2.5 cm
From the above measures, we can say that for any rhombus,
(i) Opposite angles are congruent.
(ii) Diagonals bisect the opposite angles.
(iii) Diagonals bisect each other and they are perpendicular to each other.In simple words: Constructing and measuring a rhombus demonstrates that opposite angles are equal, diagonals bisect each other perpendicularly, and the diagonals also bisect the rhombus's angles.

🎯 Exam Tip: Key observations for a rhombus include perpendicular diagonals, opposite angles being equal, and diagonals bisecting vertex angles, all of which should be verified during construction and measurement.