CBSE Maths Test Paper 2013-Class X
General Instructions:
(i) This question paper comprises four sections – A, B, C and D. This question paper carries 40 questions. All questions are compulsory.
(ii) Section A: Q. No.1 to 11 comprises of 11 questions of one mark each.
(iii) Section B: Q. No. 12 to 20 comprises of 9 questions of two mark each.
(iv) Section C: Q. No. 21 to 30 comprises of 10 questions of three mark each.
(v) Section D: Q. No. 31 to 35 comprises of 5 questions of four mark each.
(vi) Use of Calculators are not permitted.
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Question 1 of 35
1. Question
1 point(s)Find the common difference of the Ap \(\dfrac{1}{p},\dfrac{1-p}{p},\dfrac{1-2p}{p},…. \)
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The common difference
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Question 2 of 35
2. Question
1 point(s)In Fig., Calculate the area of triangle ABC (in sq.units).
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Question 3 of 35
3. Question
1 point(s)In Fig., PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA PB, then find the length of each tangent.
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Construction: Join AC and BC.
Proof: [ latex]\because [/latex] Tangent is to the radius through the point of contact
APBC is a square.
Length of each tangent
= AC = radius = 4 cm -
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Question 4 of 35
4. Question
1 point(s)If the difference between the circumference and the radius of a circle is 37 cm, then using \(\pi=\dfrac{22}{7} \), calculate the circumference (in cm) of the circle.
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Circumference of the circle -
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Question 5 of 35
5. Question
1 point(s)Solve the following quadratic equation for \(x:4\sqrt3x^2+5x-2\sqrt3=0 \)
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Question 6 of 35
6. Question
1 point(s)Find the common difference of the A.P. \(\dfrac{1}{2b},\dfrac{1-6b}{2b},\dfrac{1-12b}{2b},…. \)
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Common difference
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Question 7 of 35
7. Question
1 point(s)A die is tossed once. Find the probability of getting an even number or a multiple of 3.
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‘an even number or multiple of 3’ are 2, 3, 4, 6,
i.e., 4 numbers
Required probability -
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Question 8 of 35
8. Question
1 point(s)Find the common difference of the A.P. \(\dfrac{1}{3q},\dfrac{1-6q}{3q},\dfrac{1-12q}{3q},…. \)
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Common difference
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Question 9 of 35
9. Question
1 point(s)A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that the drawn card is neither a jack nor an ace.
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Total number of cards = 52
Numbers of jacks = 4
Numbers of aces = 4
Card is neither a jack nor an ace = 52 – 4 – 4 = 44
Required probability -
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Question 10 of 35
10. Question
1 point(s)The sum of first \(n \) terms of an AP is \(3n^2 + 4n \). Find the 25th term of this AP.
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Question 11 of 35
11. Question
2 point(s)How many three-digit natural numbers are divisible by 7?
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“3 digits nos. ” are 100, 101, 102,….,999
3 digits nos. “divisible by 7” are:
105, 112, 119, 126,….,994As
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Question 12 of 35
12. Question
2 point(s)In Fig., a circle inscribed in triangle ABC touches its sides AB, BC and AC at points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF.
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Let
and …. [Two tangents drawn from an external point are equal
latex] AB=12\ cm[/latex]…. [Given
latex]\therefore\ x+y=12\ cm [/latex]….(i)
Similarly,…. (ii) and ….(iii)
By adding (i), (ii) and (iii)
in (ii) and (iii),
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Question 13 of 35
13. Question
2 point(s)A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that the drawn card is neither a king nor a queen.
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P(neither a king nor a queen)
(king or queen) -
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Question 14 of 35
14. Question
2 point(s)Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions \(14\ cm\times7\ cm \). Find the area of the remaining card board. [Use latex]\pi=\dfrac{22}{7} [/latex] ]
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Here
Area of the remaining card board
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Question 15 of 35
15. Question
2 point(s)For what value of \(k \), are the roots of the quadratic equation \(kx(x – 2) + 6 =0\) equal?
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Given:
Two equal roots …[Given
latex]\therefore b^2-4ac=0\\ \\ a=k,b=-2k,c=6\\ \\ \therefore(-2k)^2-4(k)(6)=0\\ \\ 4k^2-24k=0\ \Rightarrow\ 4k(k-6)=0\\ \\ 4k=0\qquad or\qquad k-6=0\\ \\ k=0\qquad or\qquad k=6 [/latex]…. [Standard form of a quadratic equation latex] ax^2+bx+c=0,a\ne0[/latex] -
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Question 16 of 35
16. Question
2 point(s)The nth term of an A.P. is given by (-4n + 15). Find the sum of first 20 terms of this A.P.
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Question 17 of 35
17. Question
2 point(s)For what value of \(k \), the roots of the quadratic equation \(kx(x – 2\sqrt5) + 10 = 0\), are equal?
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….[ latex]\because [/latex] Roots are equal
As -
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Question 18 of 35
18. Question
2 point(s)Find the value of \(x \) for which the points \((x, -1), (2,1)\) and \((4, 5)\) are collinear.
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Given: Paints
As [ latex]\because [/latex] Given 3 pts. are collinear -
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Question 19 of 35
19. Question
2 point(s)From a point P on the ground, the angle of elevation of the top of a 10m tall building is 30°. A flagstaff is fixed at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (Take \(\sqrt3=1.73 \))
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Let
be the building and be the flagstaff.
In right,
(Distance)
In right
Length of the flagstaff,
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Question 20 of 35
20. Question
2 point(s)The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.
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…[Given
latex]a+23a=2(a+9d)\quad [\because a_n=a+(n-1)d\\ \\ 23d=2a+18d-a\\ \\ 23d-18d=a\quad\Rightarrow\quad a=5d\ ….(i) /latex]
To prove with the help of
From (ii) and (iii), L.H.S = R.H.S -
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Question 21 of 35
21. Question
3 point(s)Construct a triangle with sides 5 cm, 4 cm and 6 cm. Then construct another triangle whose sides are \(\dfrac{2}{3} \) times the corresponding sides of first triangle.
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Steps of Construction:
Draw with .
Draw ray making an acute angle with .
Locate 3 equal points on.
Join.
Join.
From point draw .
is the required triangle. -
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Question 22 of 35
22. Question
3 point(s)The horizontal distance between two poles is 15 m. The angle of depression of the top of first pole as seen from the top of second pole is 30°. If the height of the second pole is 24 m, find the height of the first pole. [Use latex]\sqrt3=1.732 [/latex] ]
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Let
be the pole and be the pole
In,
Height of the first pole,
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Question 23 of 35
23. Question
3 point(s)Prove that the points (7,10), (-2,5) and (3, -4) are the vertices of an isosceles right triangle.
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Let
be the vertices of a triangle…[By converse of Pythagoras theorem
latex]\triangle ABC [/latex] is an isosceles rt.
From (i) & (ii), Points A, B, C are the vertices of an isosceles right triangle. -
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Question 24 of 35
24. Question
3 point(s)Find the ratio in which the y-axis divides the line segment joining the points (-4, -6) and (10,12). Also find the coordinates of the point of division.
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Let
Let be point on -axis
Let
Coordinates of C = Coordinates of C
Required ratio and Required Point -
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Question 25 of 35
25. Question
3 point(s)In Fig., AB and CD are two diameters of a circle with centre O, which are perpendicular to each other. QB is the diameter of the smaller circle. If QA = 7 cm, find the area of the shaded region. [Use \(\pi=\dfrac{22}{7} \) ]
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Reqd. shaded area -
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Question 26 of 35
26. Question
3 point(s)A vessel is in the form of a hemispherical 66 bowl surmounted by a hollow cylinder of same diameter. The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the total (inner) suface area of the vessel. [Use latex]\pi=\dfrac{22}{7} [/latex]
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Inner surface area of the vessel = C.S. area of Hemisphere + C.S. area of Cylinder
…[C.S. area = curved surface area
latex]=2\times\dfrac{22}{7}\times7(7+6)\\ \\ =44\times13=572\ cm^2 [/latex] -
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Question 27 of 35
27. Question
3 point(s)A wooden toy was made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the volume of wood in the toy. [Use latex]\pi=\dfrac{22}{7} [/latex] ]
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Here
Vol. of wood in the toy = Volume of cylinder – 2(Volume of hemisphere)
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Question 28 of 35
28. Question
3 point(s)In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc. [Use latex]\pi=\dfrac{22}{7} [/latex] ]
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(i) Length of the arc:
Length of the arc
(ii) Area of the sector formed by the arc:
Area of minor sector -
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Question 29 of 35
29. Question
3 point(s)Solve the following for \(x:\dfrac{1}{2a+b+2x}=\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2x} \)
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-
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Question 30 of 35
30. Question
3 point(s)Sum of the areas of two squares is \(400\ cm^2 \). If the difference of their perimeters is \(16\ cm \), find the sides of the two squares.
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Let the side of Large square
Let the side of small square
According to the question, …(i) …. area of square = (side) ….[ latex]\because [/latex] Perimeter of square = 4 sides ….[Dividing both sides by 4
latex]x=4+y [/latex]
Putting the value of in equation (i), Side of small square and side of large square -
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Question 31 of 35
31. Question
4 point(s)If the sum of first 7 terms of an A.P. is 49 and that of first 17 terms is 289, find the sum of its first n terms.
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As
From (i) and (ii), we have -
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Question 32 of 35
32. Question
4 point(s)In Fig., l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectively. .Another tangent at C intersects the line / at D and m at E. Prove that \(\angle \)DOE = 90°.
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Proof: Let
be and be
…[Consecutive interior angles
latex]\dfrac{1}{2}\angle XDE+\dfrac{1}{2}\angle X’ED=\dfrac{1}{2}(180^o)\\ \\ \angle 1+\angle2=90^o [/latex]…. [OD is equally inclined to the tangents
In latex]\triangle DOE [/latex],
…[Angle-sum-property of a latex]\triangle [/latex]
… [Proved -
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Question 33 of 35
33. Question
4 point(s)The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, find the height of the building.
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Let
be the tower and let be the building
In right
In right
Height of the building, -
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Question 34 of 35
34. Question
4 point(s)The three vertices of a parallelogram ABCD are A(3, -4), B(-l, -3) and C(-6, 2). Find the coordinates of vertex D and find the area of ABCD.
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Let
Since diagonals of bisect each other
mid-point BD = mid-point AC
4th point,
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Question 35 of 35
35. Question
4 point(s)Water is flowing through a cylindrical pipe, of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m/s. Determine the rise in level of water in the tank in half an hour.
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Radius of tank,
Internal radius of cylindrical pipe,
Length of water flow in 1 second
Length of water flow in 30 minutes,
Volume of water in cylinder tank
= Volume of water flow from cylindrical pipe in half an hour
As Level of water in cylinder tank rises in half an hour -
[ latex]\because [/latex] Tangent is 
to the radius through the point of contact
APBC is a square.
and 
…. [Two tangents drawn from an external point are equal
…. (ii) and 
….(iii)![2(x+y+z)=30\\ \\ x+y+z=15\\ \\ ...[\because x+y=12\\ \\ z=15-12=3 /latex] Putting the value of [latex]z](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-c3f66bcf54688745d2c265b6e8f92783_l3.png "Rendered by QuickLaTeX.com")
in (ii) and (iii),
(king or queen)
![\begin{array}{l} a_{n}=-4 n+15 \\ \text { Put } n=1, \quad a_{1}=-4(1)+15=11 \\ \text { Put } n=2, \quad a_{2}=-4(2)+15=7 \\ \therefore \quad d=a_{2}-a_{1}=7-11=-4 \\ \text { As } S_{n}=\frac{n}{2}[2 a+(n-1) d], \quad n=20 \\ \therefore \quad S_{20}=\frac{20}{2}[2(11)+(20-1)(-4)] \\ \quad=10(22-76) \\ \quad=10(-54)=-540 \end{array}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-60c57a3c5426ccea444c374f8003608b_l3.png "Rendered by QuickLaTeX.com")
….[ latex]\because [/latex] Roots are equal
[ latex]\because [/latex] Given 3 pts. are collinear
be the building and 
be the flagstaff.
,
(Distance) 
…[Given
with the help of 
with 
.
making an acute angle with 
.
.
.
draw 
.
is the required triangle.
be the 
pole and 
be the 
pole
,![\tan30^o=\dfrac{DE}{AD}\\ \\ \dfrac{1}{\sqrt3}=\dfrac{DE}{15}\\ \\ DE=\dfrac{15}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}\\ \\ =\dfrac{15\sqrt3}{3}=5\sqrt3\\ \\ =5\ (1.732)=8.66\ m\ ....[\because\sqrt3=1.732 /latex] [latex]\therefore](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-81b4e05ba1da36c72ac86807350cb364_l3.png "Rendered by QuickLaTeX.com")
Height of the first pole,
be the vertices of a triangle
…[By converse of Pythagoras theorem
be point on 
-axis
![\left(\dfrac{10k-4}{k+1},\dfrac{12k-6}{k+1}\right)=(0,y)\\ \\ \dfrac{10k-4}{k+1}=0\qquad y=\dfrac{12k-6}{k+1}\\ \\ 10k-4=0\qquad y=\dfrac{12\left(\dfrac{2}{5}\right)-6}{\dfrac{2}{5}+1}\ \text{....[From (i)}\\ \\ 10k=4\qquad y=\dfrac{\dfrac{24-30}{5}}{\dfrac{2+5}{5}}=\dfrac{-6}{7}\\ \\ k=\dfrac{4}{10}=\dfrac{2}{5}\ ....(i) /latex] [latex]\therefore](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-1cbbfa0ba244a1b796e4650cb22131fc_l3.png "Rendered by QuickLaTeX.com")
Required ratio 
and Required Point 
…[C.S. area = curved surface area
…(i) ….
area of square = (side)
….[ latex]\because [/latex] Perimeter of square = 4 sides
….[Dividing both sides by 4
in equation (i),![\begin{aligned} &\begin{aligned} &(4+y)^{2}+y^{2}=400 \\ \Rightarrow & y^{2}+8 y+16+y^{2}-400=0 \\ \Rightarrow & 2 y^{2}+8 y-384=0 \end{aligned}\\ &\Rightarrow y^{2}+4 y-192=0 \text { ...[Dividing both sides by 2 }\\ &\Rightarrow y^{2}+16 y-12 y-192=0\\ &\begin{array}{l} \Rightarrow y(y+16)-12(y+16)=0 \\ \Rightarrow(y-12)(y+16)=0 \end{array}\\ &\begin{array}{lll} \Rightarrow y-12=0 & \text { or } & y+16=0 \\ \Rightarrow y=12 & \text { or } & y=-16 \end{array} \end{aligned}/latex][Neglecting negative value][latex]\therefore](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-1d6a3d416f843523a8443721396f0fe9_l3.png "Rendered by QuickLaTeX.com")
Side of small square 
and side of large square 
![S_n=\dfrac{n}{2}[2a+(n-1)d]\\ \\ S_7=49\qquad S_{17}=289\\ \\ \Rightarrow\dfrac{7}{2}(2a+6d)=49\qquad \dfrac{17}{2}(2a+16d)=289\\ \\ \dfrac{1}{2}(2a+6d)=\dfrac{49}{7}\qquad\dfrac{1}{2}(2a+16d)=\dfrac{289}{17}\\ \\ a+3d=7\ …(i)\qquad a+8d=17\ ....(ii)](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-93a6a248272b11194c81834964d9c9a0_l3.png "Rendered by QuickLaTeX.com")
![a+3d=7\\\_a+8d=\_17\\ \\ -5d=-10\qquad\therefore d=2\\ \\ a+3(2)=7\ ....\text{[From (i)}\\ \\ a+6=7\qquad\therefore a=7-6=1 /latex][latex] \begin{aligned}\begin{array}{l} \text { As } S_{n}=\dfrac{n}{2}[2 a+(n-1) d] \\ \therefore \quad S_{n}=\dfrac{n}{2}[2(1)+(n-1) \cdot 2] \\ \quad=\dfrac{n}{2}[2+2 n-2] \\ \quad=\dfrac{n}{2}[2 n]=n^{2} \end{array} \end{aligned}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-c269410587429ac37ba06ebfc691caec_l3.png "Rendered by QuickLaTeX.com")
be 
and 
be 
…[Consecutive interior angles
…[Angle-sum-property of a latex]\triangle [/latex]
… [Proved
be the tower and let 
be the building
bisect each other
![\begin{array}{l} \operatorname{ar}(\Delta)=\dfrac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \\ \therefore \operatorname{ar}(\Delta \mathrm{ABC})=\dfrac{1}{2}[3(-3-2)+(-1)(2+4) +(-6)(-4+3)] \\ =\dfrac{1}{2}[-15-6+6] \mathrm{s}=\dfrac{-15}{2} \\ =\dfrac{15}{2} \mathrm{sq} . \text { units } \ldots[\because \text { ar of } \Delta \text { cannot be -ve } \end{array} /latex] Since diagram of a parallelogram divides it into 2 equal areas [latex]\begin{aligned} \therefore \quad \operatorname{ar}(\|^{ \mathrm{gm}} \mathrm{ABCD}) &=2(\operatorname{ar} \Delta \mathrm{ABC})=2\left(\dfrac{15}{2}\right) \\ &=15\ \mathrm{sq} . \text { units } \end{aligned}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-2205ef3039f45ed98c396091b17900fd_l3.png "Rendered by QuickLaTeX.com")
