GSEB Class 8 Maths Solutions Chapter 3 ચતુષ્કોણની સમજ Exercise 3.3

 

Question 1. સમાંતરબાજુ ચતુષ્કોણ ABCD આપેલ છે. દરેક વિધાનને તેમાં ઉપયોગ કરવામાં આવેલ વ્યાખ્યા અથવા ગુણધર્મનો ઉપયોગ કરી ને પૂરો કરો:

Answer:
(i) \( \text{AD} = \text{BC} \)
કારણ: સમાંતરબાજુ ચતુષ્કોણની સામસામેની બાજુઓની લંબાઈ સરખી હોય છે. (Reason: The lengths of opposite sides of a parallelogram are equal.)
(ii) \( \angle\text{DCB} = \angle\text{DAB} \)
કારણ: સમાંતરબાજુ ચતુષ્કોણના સામસામેના ખૂણાનાં માપ સરખાં હોય છે. (Reason: The measures of opposite angles of a parallelogram are equal.)
(iii) \( \text{OC} = \text{OA} \)
કારણ: સમાંતરબાજુ ચતુષ્કોણના વિકર્ણ પરસ્પર દુભાગે છે. (Reason: The diagonals of a parallelogram bisect each other.)
(iv) \( m\angle\text{DAB} + m\angle\text{CDA} = 180^\circ \)
કારણ: સમાંતરબાજુ ચતુષ્કોણમાં પાસપાસેના બે ખૂણાઓ પૂરક હોય છે. (Reason: In a parallelogram, two adjacent angles are supplementary.)
In simple words: For a parallelogram, opposite sides are always equal in length. Opposite angles have the same measurement. The lines that cut across from corner to corner (diagonals) cut each other exactly in half. Also, any two angles next to each other always add up to 180 degrees.

Exam Tip: Remember the fundamental properties of a parallelogram for questions involving its sides, angles, and diagonals. Clearly state the property used as a reason.

 

Question 2. નીચેના સમાંતરબાજુ ચતુષ્કોણમાં x, y અને zનાં મૂલ્ય શોધો:
(i)

Answer:
\( \text{ABCD} \) સમાંતરબાજુ ચતુષ્કોણ છે. (ABCD is a parallelogram.)
\( \therefore \angle\text{B} = \angle\text{D} \)
\( \implies y = 100^\circ \) (કારણ: સામસામેના ખૂણા) (Reason: Opposite angles)
હવે, \( y + z = 180^\circ \) (કારણ: પાસપાસેના ખૂણા પૂરક હોય છે.) (Now, \( y + z = 180^\circ \) Reason: Adjacent angles are supplementary.)
\( \therefore 100^\circ + z = 180^\circ \)
\( \therefore z = 180^\circ - 100^\circ \)
\( \therefore z = 80^\circ \)
હવે, \( x = z \) (કારણ: સામસામેના ખૂણા) (Now, \( x = z \) Reason: Opposite angles)
\( \therefore x = 80^\circ \)
આમ, \( x = 80^\circ \), \( y = 100^\circ \) અને \( z = 80^\circ \). (Thus, \( x = 80^\circ \), \( y = 100^\circ \) and \( z = 80^\circ \).)
In simple words: In a parallelogram, angles facing each other are equal. So, angle y is 100 degrees. Angles next to each other add up to 180 degrees, so angle z is 80 degrees. Also, angle x is equal to angle z, making it 80 degrees as well.

Exam Tip: For parallelograms, remember that opposite angles are equal and adjacent angles are supplementary. This helps in quickly finding unknown angles.

 

Question 2. (ii)

Answer:
આપેલ ચતુષ્કોણ સમાંતરબાજુ ચતુષ્કોણ છે. (The given quadrilateral is a parallelogram.)
સમાંતરબાજુ ચતુષ્કોણમાં પાસપાસેના ખૂણા પૂરક હોય છે. (In a parallelogram, adjacent angles are supplementary.)
\( \therefore \angle\text{A} + \angle\text{D} = 180^\circ \)
\( \therefore 50^\circ + y = 180^\circ \)
\( \implies y = 180^\circ - 50^\circ \)
\( \implies y = 130^\circ \)
સમાંતરબાજુ ચતુષ્કોણમાં સામસામેના ખૂણા સમાન હોય છે. (In a parallelogram, opposite angles are equal.)
\( \therefore x = y \)
\( \implies x = 130^\circ \)
\( \therefore z = 50^\circ \) (કારણ: સામસામેના ખૂણા) (Reason: Opposite angles)
આમ, \( x = 130^\circ \), \( y = 130^\circ \) અને \( z = 50^\circ \). (Thus, \( x = 130^\circ \), \( y = 130^\circ \) and \( z = 50^\circ \).)
In simple words: In this shape, opposite angles are equal, and angles next to each other add up to 180 degrees. So, if one angle is 50 degrees, the angle opposite it is also 50 degrees. The angles next to 50 degrees will be 130 degrees each because 50 + 130 = 180.

Exam Tip: Always identify the type of quadrilateral first. For a parallelogram, remember that adjacent angles sum to 180° and opposite angles are equal.

 

Question 2. (iii)

Answer:
\( \text{ABCD} \) સમાંતરબાજુ ચતુષ્કોણ છે. (ABCD is a parallelogram.)
તેમાં \( \angle\text{AMD} \) કાટખૂણો છે. (In it, \( \angle\text{AMD} \) is a right angle.)
\( \therefore m\angle\text{BMC} = 90^\circ \) અને \( m\angle\text{AMD} = 90^\circ \) (અભિકોણ) (So \( m\angle\text{BMC} = 90^\circ \) and \( m\angle\text{AMD} = 90^\circ \) (Vertically opposite angles))
\( \implies x = m\angle\text{BMC} = 90^\circ \)
\( \triangle\text{BMC} \) ના ત્રણે ખૂણાઓનાં માપનો સરવાળો \( 180^\circ \) થાય. (The sum of the measures of the three angles of \( \triangle\text{BMC} \) is \( 180^\circ \).)
\( \therefore y + 90^\circ + 30^\circ = 180^\circ \)
\( \therefore y + 120^\circ = 180^\circ \)
\( \therefore y = 180^\circ - 120^\circ \)
\( \therefore y = 60^\circ \)
\( \text{ABCD} \) માં \( \overline{\text{BC}} \parallel \overline{\text{AD}} \) અને તેમની છેદિકા \( \overline{\text{BD}} \). (In parallelogram ABCD, \( \overline{\text{BC}} \parallel \overline{\text{AD}} \) and transversal \( \overline{\text{BD}} \).)
\( \therefore y = z \) (યુગ્મકોણો) (So \( y = z \) (Alternate interior angles))
\( \therefore z = 60^\circ \) (કારણ: \( y = 60^\circ \)) (Reason: \( y = 60^\circ \))
આમ, \( x = 90^\circ \), \( y = 60^\circ \) અને \( z = 60^\circ \). (Thus, \( x = 90^\circ \), \( y = 60^\circ \) and \( z = 60^\circ \).)
In simple words: First, find x. Since vertically opposite angles are equal, x is 90 degrees. Next, use the fact that angles in a triangle add up to 180 degrees to find y. Finally, because parallel lines create equal alternate interior angles, z is the same as y.

Exam Tip: When diagonals intersect, look for vertically opposite angles. For parallel lines, remember alternate interior angles are equal. The sum of angles in a triangle is always 180°.

 

Question 2. (iv)

Answer:
\( \text{ABCD} \) સમાંતરબાજુ ચતુષ્કોણ છે. (ABCD is a parallelogram.)
\( \angle\text{D} = \angle\text{B} \) (કારણ: સામસામેના ખૂણા) (Reason: Opposite angles)
\( \therefore y = 80^\circ \)
\( m\angle\text{A} + m\angle\text{D} = 180^\circ \) (કારણ: પાસપાસેના પૂરક હોય છે.) (Adjacent angles are supplementary.)
\( \therefore x + 80^\circ = 180^\circ \)
\( \therefore x = 180^\circ - 80^\circ \)
\( \therefore x = 100^\circ \)
\( m\angle\text{A} = m\angle\text{BCD} \) (કારણ: સામસામેના ખૂણા) (Opposite angles are equal.)
\( \therefore 100^\circ = m\angle\text{BCD} \)
હવે, \( z + m\angle\text{BCD} = 180^\circ \) (કારણ: રૈખિક જોડના ખૂણા) (Now, \( z + m\angle\text{BCD} = 180^\circ \) Reason: Linear pair angles)
\( \therefore z + 100^\circ = 180^\circ \)
\( \therefore z = 180^\circ - 100^\circ \)
\( \therefore z = 80^\circ \)
આમ, \( x = 100^\circ \), \( y = 80^\circ \) અને \( z = 80^\circ \). (Thus, \( x = 100^\circ \), \( y = 80^\circ \) and \( z = 80^\circ \).)
In simple words: First, angle y is 80 degrees because it's opposite to the given 80-degree angle. Next, angles x and y are adjacent, so they add up to 180 degrees, making x 100 degrees. Finally, z is a linear pair with angle BCD (which is 100 degrees), so z is 80 degrees.

Exam Tip: Remember that opposite angles of a parallelogram are equal, and adjacent angles sum to 180°. Also, linear pairs sum to 180°.

 

Question 2. (v)

Answer:
\( \text{ABCD} \) સમાંતરબાજુ ચતુષ્કોણ છે. (ABCD is a parallelogram.)
\( \therefore m\angle\text{B} = m\angle\text{D} \) (કારણ: સામસામેના ખૂણા) (Reason: Opposite angles)
\( \therefore y = 112^\circ \)
\( m\angle\text{A} + m\angle\text{B} = 180^\circ \) (કારણ: પાસપાસેના ખૂણા પૂરક હોય છે.) (Reason: Adjacent angles are supplementary.)
\( \therefore (40^\circ + x) + 112^\circ = 180^\circ \)
\( \therefore 40^\circ + x + 112^\circ = 180^\circ \)
\( \therefore x + 152^\circ = 180^\circ \)
\( \therefore x = 180^\circ - 152^\circ \)
\( \therefore x = 28^\circ \)
હવે, \( \overline{\text{DC}} \parallel \overline{\text{AB}} \) અને \( \overleftrightarrow{\text{AC}} \) તેમની છેદિકા છે. (Now, \( \overline{\text{DC}} \parallel \overline{\text{AB}} \) and \( \overleftrightarrow{\text{AC}} \) is their transversal.)
\( \therefore z = x \) (યુગ્મકોણો) (Alternate interior angles)
\( \therefore z = 28^\circ \)
આમ, \( x = 28^\circ \), \( y = 112^\circ \) અને \( z = 28^\circ \). (Thus, \( x = 28^\circ \), \( y = 112^\circ \) and \( z = 28^\circ \).)
In simple words: First, find y by using the property that opposite angles in a parallelogram are equal. Then, use the property that adjacent angles sum to 180 degrees to find x. Finally, use the alternate interior angles property (formed by parallel lines and a transversal) to find z.

Exam Tip: Always remember the basic properties of parallelograms: opposite angles are equal, and consecutive angles are supplementary. Also, recognize alternate interior angles formed by transversals intersecting parallel sides.

 

Question 3. શું ચતુષ્કોણ ABCD, સમાંતરબાજુ ચતુષ્કોણ થઈ શકે, જો
(i) \( \angle\text{D} + \angle\text{B} = 180^\circ \)?
(ii) \( \text{AB} = \text{DC} = 8 \) સેમી, \( \text{AD} = 4 \) સેમી અને \( \text{BC} = 4.4 \) સેમી?
(iii) \( \angle\text{A} = 70^\circ \) અને \( \angle\text{C} = 65^\circ \)?

Answer:
(i) ના, અહીં \( \angle\text{D} \) અને \( \angle\text{B} \) એ પાસપાસેના બે ખૂણા નથી. તેથી ચતુષ્કોણ \( \text{ABCD} \) સમાંતરબાજુ ચતુષ્કોણ ન થઈ શકે. (No, here \( \angle\text{D} \) and \( \angle\text{B} \) are not adjacent angles. Hence, quadrilateral \( \text{ABCD} \) cannot be a parallelogram.)
(ii) ના, અહીં \( \text{AD} \neq \text{BC} \) (કારણ: \( \text{AD} = 4 \) સેમી અને \( \text{BC} = 4.4 \) સેમી). તેથી ચતુષ્કોણ \( \text{ABCD} \) સમાંતરબાજુ ચતુષ્કોણ ન થઈ શકે. (No, here \( \text{AD} \neq \text{BC} \) (Reason: \( \text{AD} = 4 \) cm and \( \text{BC} = 4.4 \) cm). Hence, quadrilateral \( \text{ABCD} \) cannot be a parallelogram.)
(iii) ના, \( m\angle\text{A} \neq m\angle\text{C} \) (કારણ: \( m\angle\text{A} = 70^\circ \) અને \( m\angle\text{C} = 65^\circ \)). તેથી, ચતુષ્કોણ \( \text{ABCD} \) સમાંતરબાજુ ચતુષ્કોણ ન થઈ શકે. (No, \( m\angle\text{A} \neq m\angle\text{C} \) (Reason: \( m\angle\text{A} = 70^\circ \) and \( m\angle\text{C} = 65^\circ \)). Hence, quadrilateral \( \text{ABCD} \) cannot be a parallelogram.)
In simple words: A quadrilateral can only be a parallelogram if its opposite sides are equal, its opposite angles are equal, and adjacent angles add up to 180 degrees. If any of these conditions are not met, like having unequal opposite sides or angles, it's not a parallelogram. Even if two opposite angles add up to 180 degrees, it doesn't automatically mean it's a parallelogram, as this could also apply to an isosceles trapezoid.

Exam Tip: To determine if a quadrilateral is a parallelogram, check all its properties: opposite sides are parallel and equal, opposite angles are equal, and diagonals bisect each other. If any required property is not met, it is not a parallelogram.

 

Question 4. એક એવા ચતુષ્કોણની કાચી (Rough) આકૃતિ દોરો કે જે સમાંતરબાજુ ચતુષ્કોણ ના હોય પરંતુ સામસામેના ખૂણાની એક જોડી સમાન હોય.

Answer:
અહીં ચતુષ્કોણ \( \text{ABCD} \) દોર્યો છે. (Here, quadrilateral \( \text{ABCD} \) is drawn.)
આકૃતિમાં જુઓ. (See the figure.)
\( \angle\text{B} = \angle\text{D} \).
છતાં \( \text{ABCD} \) એ સમાંતરબાજુ ચતુષ્કોણ નથી. (Still, \( \text{ABCD} \) is not a parallelogram.)
In simple words: The shape shown is a kite. It has one pair of opposite angles that are equal (angle B and angle D), but it doesn't have all the other properties of a parallelogram, like both pairs of opposite sides being parallel.

Exam Tip: A kite is a classic example of a quadrilateral with exactly one pair of opposite angles equal. Remember its defining characteristics for such questions.

 

Question 5. સમાંતરબાજુ ચતુષ્કોણમાં બે પાસપાસેના ખૂણાનાં માપનો ગુણોત્તર 3 : 2 છે, તો ચતુષ્કોણના બધા જ ખૂણાનાં માપ શોધો.

Answer:
ધારો કે સમાંતરબાજુ ચતુષ્કોણ \( \text{ABCD} \) માં \( \angle\text{A} \) અને \( \angle\text{B} \) નું માપનું પ્રમાણ \( 3 : 2 \) છે. (Suppose in parallelogram \( \text{ABCD} \), the ratio of measures of \( \angle\text{A} \) and \( \angle\text{B} \) is \( 3 : 2 \).)
જો \( \angle\text{A} = 3x \) છે, તો \( \angle\text{B} = 2x \) છે. (If \( \angle\text{A} = 3x \), then \( \angle\text{B} = 2x \).)
હવે, \( m\angle\text{A} + m\angle\text{B} = 180^\circ \) (કારણ: પાસપાસેના બે ખૂણા પૂરક હોય છે.) (Now, \( m\angle\text{A} + m\angle\text{B} = 180^\circ \) Reason: Adjacent angles are supplementary.)
\( \therefore 3x + 2x = 180^\circ \)
\( \therefore 5x = 180^\circ \)
\( \therefore x = \frac{180^\circ}{5} \)
\( \therefore x = 36^\circ \)
\( m\angle\text{A} = 3x = 3 \times 36^\circ = 108^\circ \) અને
\( m\angle\text{B} = 2x = 2 \times 36^\circ = 72^\circ \)
હવે, \( m\angle\text{A} = m\angle\text{C} \) અને \( m\angle\text{B} = m\angle\text{D} \) (કારણ: સામસામેના ખૂણા) (Now, \( m\angle\text{A} = m\angle\text{C} \) and \( m\angle\text{B} = m\angle\text{D} \) Reason: Opposite angles)
\( \therefore m\angle\text{C} = 108^\circ \) અને \( m\angle\text{D} = 72^\circ \)
આમ, ચતુષ્કોણ \( \text{ABCD} \) માં \( \angle\text{A} = 108^\circ \), \( \angle\text{B} = 72^\circ \), \( \angle\text{C} = 108^\circ \) અને \( \angle\text{D} = 72^\circ \). (Thus, in quadrilateral \( \text{ABCD} \), \( \angle\text{A} = 108^\circ \), \( \angle\text{B} = 72^\circ \), \( \angle\text{C} = 108^\circ \) and \( \angle\text{D} = 72^\circ \).)
In simple words: Since adjacent angles in a parallelogram add up to 180 degrees, and their ratio is 3:2, we can find the value of x. Then, we use x to calculate the measures of angle A and angle B. Finally, because opposite angles in a parallelogram are equal, we can find the measures of angle C and angle D.

Exam Tip: When given ratios of angles in a parallelogram, represent them with a variable (e.g., 3x, 2x). Remember that adjacent angles are supplementary, and opposite angles are equal to find all angle measures.

 

Question 6. એક સમાંતરબાજુ ચતુષ્કોણમાં પાસપાસેના ખૂણાની એક જોડના ખૂણાનાં માપ સમાન છે, તો ચતુષ્કોણના બધા જ ખૂણાનાં માપ શોધો.

Answer:
ધારો કે \( \text{ABCD} \) એવો સમાંતરબાજુ ચતુષ્કોણ છે જેના પાસપાસેના બે ખૂણા \( \angle\text{A} \) અને \( \angle\text{B} \) એકરૂપ છે. (Suppose \( \text{ABCD} \) is a parallelogram where two adjacent angles \( \angle\text{A} \) and \( \angle\text{B} \) are congruent.)
\( \therefore \angle\text{A} = \angle\text{B} \)
હવે, \( m\angle\text{A} + m\angle\text{B} = 180^\circ \) (કારણ: પાસપાસેના બે ખૂણા પૂરક હોય છે.) (Now, \( m\angle\text{A} + m\angle\text{B} = 180^\circ \) Reason: Adjacent angles are supplementary.)
\( \therefore m\angle\text{A} + m\angle\text{A} = 180^\circ \) (કારણ: \( m\angle\text{B} = m\angle\text{A} \)) (Reason: \( m\angle\text{B} = m\angle\text{A} \))
\( \therefore 2m\angle\text{A} = 180^\circ \)
\( \therefore m\angle\text{A} = \frac{180^\circ}{2} = 90^\circ \)
\( \therefore m\angle\text{B} = 90^\circ \)
હવે, \( m\angle\text{A} = m\angle\text{C} \) અને \( m\angle\text{B} = m\angle\text{D} \) (કારણ: સામસામેના ખૂણા) (Now, \( m\angle\text{A} = m\angle\text{C} \) and \( m\angle\text{B} = m\angle\text{D} \) Reason: Opposite angles)
\( \therefore m\angle\text{C} = 90^\circ \) અને \( m\angle\text{D} = 90^\circ \)
આમ, \( \angle\text{A} = 90^\circ \), \( \angle\text{B} = 90^\circ \), \( \angle\text{C} = 90^\circ \) અને \( \angle\text{D} = 90^\circ \). આ ચતુષ્કોણ લંબચોરસ છે. (Thus, \( \angle\text{A} = 90^\circ \), \( \angle\text{B} = 90^\circ \), \( \angle\text{C} = 90^\circ \) and \( \angle\text{D} = 90^\circ \). This quadrilateral is a rectangle.)
In simple words: If two angles next to each other in a parallelogram are equal, then all four angles must be 90 degrees. This means the parallelogram is a rectangle, because adjacent angles add up to 180 degrees, and if they are equal, each must be 90 degrees. Opposite angles are also equal.

Exam Tip: If a parallelogram has one pair of adjacent angles equal, it implies all its angles are 90°, making it a rectangle. This is an important derived property of parallelograms.

 

Question 7. આકૃતિમાં એક સમાંતરબાજુ ચતુષ્કોણ HOPE દર્શાવેલ છે. x, y, z ખૂણાનાં માપ શોધો. ખૂણો શોધવા કયા ગુણધર્મનો ઉપયોગ કર્યો છે તે જણાવો.

Answer:
\( \text{HOPE} \) સમાંતરબાજુ ચતુષ્કોણ છે. (HOPE is a parallelogram.)
બાજુ \( \overline{\text{HO}} \) ને \( \text{A} \) સુધી લંબાવતા, \( \angle\text{POA} \) એ \( \angle\text{HOP} \) નો બહિષ્કોણ છે. (By extending side \( \overline{\text{HO}} \) to \( \text{A} \), \( \angle\text{POA} \) is an exterior angle of \( \angle\text{HOP} \).)
\( \therefore m\angle\text{HOP} = 180^\circ - 70^\circ = 110^\circ \) (કારણ: રૈખિક જોડના ખૂણા) (Reason: Linear pair angles)
\( x = \angle\text{HEP} = \angle\text{HOP} \) (કારણ: સામસામેના ખૂણા) (Reason: Opposite angles)
\( \therefore x = 110^\circ \)
\( \overline{\text{PO}} \parallel \overline{\text{EH}} \) અને \( \overleftrightarrow{\text{PH}} \) તેમની છેદિકા છે. ( \( \overline{\text{PO}} \parallel \overline{\text{EH}} \) and \( \overleftrightarrow{\text{PH}} \) is their transversal.)
\( \therefore \angle\text{OPH} = \angle\text{EHP} \) (કારણ: યુગ્મકોણ) (Reason: Alternate interior angles)
\( \therefore y = 40^\circ \). (As per diagram, \( \angle\text{EHP} \) is y and 40° is marked there.)
હવે, \( y + z = 70^\circ \) (આપેલ શરત અથવા સંબંધ). (Now, \( y + z = 70^\circ \) (Given condition or relation).)
\( \therefore 40^\circ + z = 70^\circ \)
\( \therefore z = 70^\circ - 40^\circ \)
\( \therefore z = 30^\circ \)
આમ, \( x = 110^\circ \), \( y = 40^\circ \) અને \( z = 30^\circ \). (Thus, \( x = 110^\circ \), \( y = 40^\circ \) and \( z = 30^\circ \).)
In simple words: First, use the linear pair property to find the internal angle at O, which gives us x. Then, identify y as the alternate interior angle with angle P (if PHO is a transversal for EH and PO). Finally, use the given relation that y plus z equals 70 degrees to find z.

Exam Tip: For complex diagrams, identify parallel lines and transversals to find alternate interior or corresponding angles. Always use linear pairs for exterior angles, and apply the properties of parallelograms consistently.

 

Question 8. નીચેની આકૃતિ GUNS અને RUNS સમાંતરબાજુ ચતુષ્કોણ છે. x અને y શોધો. (લંબાઈ સેમીમાં છે.)
(i)

Answer:
\( \text{GUNS} \) એ સમાંતરબાજુ ચતુષ્કોણ છે. (GUNS is a parallelogram.)
\( \therefore \text{GS} = \text{NU} \) અને \( \text{SN} = \text{GU} \) (કારણ: સામસામેની બાજુઓ) (Reason: Opposite sides)
\( \therefore 3x = 18 \) અને \( 26 = 3y - 1 \)
હવે, \( 3x = 18 \)
\( \therefore x = \frac{18}{3} \)
\( \therefore x = 6 \)
તથા \( 26 = 3y - 1 \)
\( \therefore 3y = 26 + 1 \)
\( \therefore 3y = 27 \)
\( \therefore y = \frac{27}{3} \)
\( \therefore y = 9 \)
આમ, \( x = 6 \) સેમી અને \( y = 9 \) સેમી. (Thus, \( x = 6 \) cm and \( y = 9 \) cm.)
In simple words: Since GUNS is a parallelogram, its opposite sides are equal in length. Set the expressions for opposite sides equal to each other to form two simple equations, and then solve for x and y.

Exam Tip: For parallelograms, remember that opposite sides are always equal in length. This allows you to set up algebraic equations to find unknown side lengths or variables.

 

Question 8. (ii)

Answer:
\( \text{RUNS} \) એ સમાંતરબાજુ ચતુષ્કોણ છે. (RUNS is a parallelogram.)
\( \text{RUNS} \) ના વિકર્ણો પરસ્પર દુભાગે છે. (The diagonals of RUNS bisect each other.)
\( \therefore y + 7 = 20 \)
\( \therefore y = 20 - 7 \)
\( y = 13 \)
અને \( x + y = 16 \) માં \( y = 13 \) મૂકતાં, (And substituting \( y = 13 \) in \( x + y = 16 \),)
\( x + 13 = 16 \)
\( \therefore x = 16 - 13 \)
\( \therefore x = 3 \)
આમ, \( x = 3 \) સેમી અને \( y = 13 \) સેમી. (Thus, \( x = 3 \) cm and \( y = 13 \) cm.)
In simple words: In a parallelogram, the diagonals cut each other in half. This means the two parts of each diagonal are equal in length. Set up equations using these equal segments to find the values of y and then x.

Exam Tip: A key property of parallelograms is that their diagonals bisect each other. This means the intersection point divides each diagonal into two equal segments, which is useful for solving problems involving segment lengths.

 

Question 9. ઉપરની આકૃતિમાં RISK અને CLUE સમાંતરબાજુ ચતુષ્કોણ છે, તો x શોધો.

Answer:
અહીં \( \text{RISK} \) સમાંતરબાજુ ચતુષ્કોણ છે. (Here \( \text{RISK} \) is a parallelogram.)
\( m\angle\text{R} + m\angle\text{K} = 180^\circ \) (કારણ: પાસપાસેના બે ખૂણા પૂરકકોણ) (Reason: Adjacent angles are supplementary.)
\( \therefore m\angle\text{R} + 120^\circ = 180^\circ \)
\( \therefore m\angle\text{R} = 180^\circ - 120^\circ \)
\( \therefore m\angle\text{R} = 60^\circ \)
હવે \( \angle\text{R} \) અને \( \angle\text{S} \) એ સમાંતરબાજુ \( \text{RISK} \) ના સામસામેના ખૂણા છે. (Now \( \angle\text{R} \) and \( \angle\text{S} \) are opposite angles of parallelogram \( \text{RISK} \).)
\( \therefore m\angle\text{S} = 60^\circ \)
હવે \( \text{CLUE} \) એ સમાંતરબાજુ ચતુષ્કોણ છે. (Now \( \text{CLUE} \) is a parallelogram.)
\( m\angle\text{E} = m\angle\text{L} = 70^\circ \) (કારણ: સામસામેના ખૂણા) (Reason: Opposite angles)
હવે, \( \triangle\text{ESI} \) ના ત્રણે ખૂણાનાં માપનો સરવાળો \( 180^\circ \) થાય છે. (Now, the sum of the measures of the three angles of \( \triangle\text{ESI} \) is \( 180^\circ \).)
\( \therefore m\angle\text{E} + m\angle\text{S} + x = 180^\circ \)
\( \therefore 70^\circ + 60^\circ + x = 180^\circ \)
\( \therefore 130^\circ + x = 180^\circ \)
\( \therefore x = 180^\circ - 130^\circ \)
\( \therefore x = 50^\circ \)
આમ, \( x = 50^\circ \). (Thus, \( x = 50^\circ \).)
In simple words: First, use the properties of parallelograms to find the measure of angle S in RISK (adjacent to K) and angle E in CLUE (opposite to L). Then, recognize that these two angles, along with x, form a triangle. Since the angles in a triangle add up to 180 degrees, you can easily calculate x.

Exam Tip: For problems with multiple overlapping parallelograms, break down the figure into individual parallelograms and apply their angle properties (adjacent angles supplementary, opposite angles equal) to find specific angles. Then look for triangles formed by the intersecting lines to find the unknown angle.

 

Question 10. નીચેની આકૃતિ સમલંબ ચતુષ્કોણ કેવી રીતે છે, તે સમજાવો. કઈ બે બાજુ પરસ્પર સમાંતર છે?

Answer:
અહીં \( \square\text{LMNK} \) માં \( m\angle\text{L} + m\angle\text{M} = 80^\circ + 100^\circ = 180^\circ \). (Here in \( \square\text{LMNK} \), \( m\angle\text{L} + m\angle\text{M} = 80^\circ + 100^\circ = 180^\circ \).)
એટલે કે \( \angle\text{L} \) અને \( \angle\text{M} \) એ પૂરકકોણો છે. (This means \( \angle\text{L} \) and \( \angle\text{M} \) are supplementary angles.)
પક્ષ \( \overline{\text{NM}} \parallel \overline{\text{KL}} \) અને \( \overleftrightarrow{\text{ML}} \) છેદવાથી બનતા છેદિકાની એક જ બાજુના અંતઃકોણો છે. (Sides \( \overline{\text{NM}} \parallel \overline{\text{KL}} \) and \( \overleftrightarrow{\text{ML}} \) forms consecutive interior angles on the same side of the transversal.)
\( \therefore \overline{\text{NM}} \parallel \overline{\text{KL}} \)
\( \square\text{NMNK} \) માં એક જ જોડીની બાજુઓ \( \overline{\text{NM}} \parallel \overline{\text{KL}} \) છે. તેથી આ ચતુષ્કોણ સમલંબ ચતુષ્કોણ છે. (In \( \square\text{NMNK} \), only one pair of sides \( \overline{\text{NM}} \parallel \overline{\text{KL}} \) is parallel. Hence this quadrilateral is a trapezoid.)
સમાંતર બાજુઓ \( \overline{\text{NM}} \) અને \( \overline{\text{KL}} \) છે. (The parallel sides are \( \overline{\text{NM}} \) and \( \overline{\text{KL}} \).)
In simple words: This figure is a trapezoid because the sum of the angles on one side (angle L and angle M) is 180 degrees. This means that the top side (NM) and the bottom side (KL) are parallel to each other. A trapezoid is a quadrilateral with exactly one pair of parallel sides.

Exam Tip: A quadrilateral is a trapezoid if and only if it has exactly one pair of parallel sides. You can check for parallel lines by seeing if consecutive interior angles sum to 180° when a transversal intersects them.

 

Question 11. આપેલી આકૃતિમાં જો \( \overline{\text{AB}} \parallel \overline{\text{DC}} \) હોય, તો \( m\angle\text{C} \) શોધો.

Answer:
\( \square\text{ABCD} \) માં \( \overline{\text{AB}} \parallel \overline{\text{DC}} \). (In \( \square\text{ABCD} \), \( \overline{\text{AB}} \parallel \overline{\text{DC}} \).)
\( \text{ABCD} \) એ સમલંબ ચતુષ્કોણ છે. (ABCD is a trapezoid.)
અહીં \( \overline{\text{AB}} \parallel \overline{\text{DC}} \) ની છેદિકા \( \overleftrightarrow{\text{BC}} \). (Here \( \overleftrightarrow{\text{BC}} \) is the transversal of \( \overline{\text{AB}} \parallel \overline{\text{DC}} \).)
\( \therefore m\angle\text{B} + m\angle\text{C} = 180^\circ \) (કારણ: છેદિકાની એક જ બાજુના અંતઃકોણો પૂરકકોણ હોય છે.) (Reason: Interior angles on the same side of the transversal are supplementary.)
\( \therefore 120^\circ + m\angle\text{C} = 180^\circ \)
\( \therefore m\angle\text{C} = 180^\circ - 120^\circ = 60^\circ \)
આમ, \( m\angle\text{C} = 60^\circ \). (Thus, \( m\angle\text{C} = 60^\circ \).)
In simple words: Since the top and bottom sides are parallel, and the side BC cuts across them, the angles next to each other on that transversal (angle B and angle C) must add up to 180 degrees. If angle B is 120 degrees, then angle C must be 60 degrees.

Exam Tip: For trapezoids, consecutive interior angles between parallel sides are supplementary. You can find unknown angles by setting their sum to 180°. Use this property to find unknown angles.

 

Question 12. આપેલી આકૃતિમાં જો \( \overline{\text{SP}} \parallel \overline{\text{RQ}} \) હોય, તો \( \angle\text{P} \) અને \( \angle\text{S} \) નું માપ શોધો. (જો તમે \( m\angle\text{R} \) શોધતા હોય, તો શું, \( m\angle\text{P} \) શોધવાની અન્ય પદ્ધતિઓ હશે?)

Answer:
અહીં \( \text{PQRS} \) માં \( \overline{\text{SP}} \parallel \overline{\text{RQ}} \) છે. (Here in \( \text{PQRS} \), \( \overline{\text{SP}} \parallel \overline{\text{RQ}} \).)
\( \therefore \text{PQRS} \) સમલંબ ચતુષ્કોણ છે. (Therefore, \( \text{PQRS} \) is a trapezoid.)
\( \overline{\text{SP}} \parallel \overline{\text{RQ}} \) ની છેદિકા \( \overleftrightarrow{\text{PQ}} \). ( \( \overleftrightarrow{\text{PQ}} \) is the transversal of \( \overline{\text{SP}} \parallel \overline{\text{RQ}} \).)
\( m\angle\text{P} + m\angle\text{Q} = 180^\circ \) (કારણ: છેદિકાની એક જ બાજુના અંતઃકોણો) (Reason: Interior angles on the same side of the transversal)
\( \therefore m\angle\text{P} + 130^\circ = 180^\circ \)
\( \therefore m\angle\text{P} = 180^\circ - 130^\circ \)
\( \therefore m\angle\text{P} = 50^\circ \)
હવે \( \text{PQRS} \) માં \( \angle\text{R} \) કાટખૂણો છે. (Now in \( \text{PQRS} \), \( \angle\text{R} \) is a right angle.)
\( \overline{\text{SP}} \parallel \overline{\text{RQ}} \) અને \( \overleftrightarrow{\text{RS}} \) છેદિકા છે. ( \( \overline{\text{SP}} \parallel \overline{\text{RQ}} \) and \( \overleftrightarrow{\text{RS}} \) is a transversal.)
\( \therefore m\angle\text{S} + m\angle\text{R} = 180^\circ \)
\( \therefore m\angle\text{S} + 90^\circ = 180^\circ \)
\( \therefore m\angle\text{S} = 180^\circ - 90^\circ \)
\( \therefore m\angle\text{S} = 90^\circ \)
હા, ચતુષ્કોણના બધા ખૂણાઓનાં માપનો સરવાળો \( 360^\circ \) થાય છે, તે પરથી પણ \( \angle\text{P} \) અને \( \angle\text{R} \) શોધી શકીએ. (Yes, we can also find \( \angle\text{P} \) and \( \angle\text{R} \) from the fact that the sum of all angles in a quadrilateral is \( 360^\circ \).)
\( \therefore m\angle\text{P} + m\angle\text{Q} + m\angle\text{R} + m\angle\text{S} = 360^\circ \)
\( \therefore m\angle\text{P} + 130^\circ + 90^\circ + 90^\circ = 360^\circ \)
\( \therefore m\angle\text{P} + 310^\circ = 360^\circ \)
\( \therefore m\angle\text{P} = 360^\circ - 310^\circ \)
\( \therefore m\angle\text{P} = 50^\circ \).
In simple words: First, find angle P by using the property that angles between parallel lines and a transversal add up to 180 degrees. Then, do the same to find angle S. Yes, you can also find angle P by using the fact that all four angles in any quadrilateral always add up to 360 degrees.

Exam Tip: For trapezoids, consecutive interior angles between parallel sides are supplementary. You can find unknown angles by setting their sum to 180°. Alternatively, the sum of all interior angles of any quadrilateral is 360°, which can be used as a cross-check or a direct method if other angles are known.