SEE Compulsory Math Question Paper With Solution
SEE 2077 (2021)
If the marked price and selling price of a bag are Rs. 5x and Rs. 4x respectively, find the discount rate.
What is the relation between the areas of parallelograms which are standing on same base and lie between the same parallel lines.
Solution:
If two parallelograms are standing on the same base and lie between same parallel lines, then their areas are equal.
Proof:
Consider we have two parallelograms ABCD and ABEF which are standing on the same base AB and in between same parallel lines CF and AB as shown in the diagram below. The question now is is there any relation between these two parallelogram.
Theorem: Parallelograms on the same base between the same parallel lines are equal in areas.
Proof: Take the figure given above. We can show the triangle ΔBCE and triangle ΔDAF are congruent. The proof is as follow
By the ASA criterion, the two triangles are now congruent and hence are equal in area. Now,
area of parallelogram ABCD = area of triangle BCE + area of quadrilateral BEDA --------- (1)
And, area of parallelogram BEFA = area of triangle ADF + area of quadrilateral BEDA --------- (2)
Clearly, from equation 1 and 2
Area of parallelogram ABCD = Area of parallelogram BEFA .
Hence Proved.
The selling price of an article including 10% VAT is Rs. 1320. Find its marked price and amount of VAT.
Solution:
Given,
Rate of VAT = 10%
S.P. with VAT = Rs. 1320
To find,
Let, the marked price be x
Here,
S.P. with VAT = MP + 10% of MP
Or, Rs. 1320 = x + 10% of x
Or, Rs. 1320 = 11x/100
∴ x = Rs. 1200
∴ Marked price (MP) of an article is Rs. 1200
and the amount of VAT = SP with VAT - MP = Rs. 1320 - Rs. 1200 = Rs. 120
Hence the VAT amount was Rs. 120.
Find the rate of interest per year if per rupee per month interest is 1 paisa.
Given:
Principal (P) = Rs. 1
Interest (I) = 1 Paisa = Rs. \(\frac1{100}\)
Time (T) = 1 Month = \(\frac1{12}\) year
To find:
Rate of interest (R) = ?
By formula,
\(R=\frac{I\times100}{P\times T}\)
= \(R=\frac{{\displaystyle\frac1{100}}\times100}{1\times{\displaystyle\frac1{12}}}\)
= \(R=\frac{12}1\)
= 12
∴ Rate of interest (R) = 12% is the required answer.
The area of isosceles triangular land whose base side length 10 meter is 60 sq meter. Find the measures of remaining sides.
Solution:
Given:
In an isosceles triangle,
Length of base (b) = 10m
Area (A) = 60 sq m
To find:
Length of each equal sides (a)
By formula,
Area of an isosceles triangle (A) = \(\frac b4\sqrt{4a^2-b^2}\)
Or, 60m^{2} = \(\frac{10}4\sqrt{4a^2-10^2}\)
Or, \(\frac{4\times60}{10}=\sqrt{4a^2-100}\)
Or, \(24=\sqrt{4a^2-100}\)
Squaring both sides we get,
\(\left(24\right)^2=\left(\sqrt{4a^2-100}\right)^2\)
Or, \(576=4a^2-100\)
Or, \(4a^2=676\)
Or, \(\frac{676}4=a^2\)
∴ a = 13 m
Hence, the measure of remaining equal sides of the triangular piece of the land is 13m.... is the required answer
If the volume and the radius of a cylinder are 1078 cubic cm and 7 cm respectively, find the height of the cylinder.
In the given solid triangular prism, if PQ ⊥ QR, PQ = 8 cm and RR' = 15 cm, find the lateral surface area of the prism
Given,
In a triangular based prism,
PQ ⊥ RQ, PQ = 8 cm, QR = 6 cm and RR' = 15 cm
To find:
Lateral surface area of the prism (LSA)
Here, In the right angled triangle PQR,
PR = \(\sqrt{PQ^2+QR^2}\)
= \(\sqrt{8^2+6^2}\)
= \(\sqrt{64+36}\)
= \(\sqrt{100}\)
= 10 cm
∴ Perimeter of the triangular base = PQ + QR + PR = 8 cm + 6 cm + 10 cm = 24 cm
Now, by formula
Lateral surface area of the prism (LSA) = perimeter of base x height
= 24 cm x 15 cm
= 360 cm^{2} is the required answer.
Find the HCF of : 6(a^{3} + 8) and 8(4a^{2} - a - 18)
Here,
First expression
= 6 ( a^{3} + 8 )
= 6 ( a^{3} + 2^{3 })
= 2 x 3 ( a + 2 ) ( a^{2} - 2a + 4 )
Second expression
= 8 ( 4a^{2} - a - 18 )
= 2 x 2 x 2 { 4a^{2} - (9 - 8)a - 18 }
= 2 x 2 x 2 { 4a^{2} - 9a + 8a - 18 )
= 2 x 2 x 2 { a(4a - 9) + 2(4a - 9)}
= 2 x 2 x 2 (4a - 9) ( a+ 2 )
∴ HCF of two given expressions is the highest common factor = 2 (a + 2)
Find the value of : \(\frac{2^{x-1}+4\times2^x}{2^{x-1}}\)
Here,
\(\frac{2^{x-1}+4\times2^x}{2^{x-1}}\)
= \(\frac{2^x\times{\displaystyle\frac12}+4\times2^x}{2^x\times{\displaystyle\frac12}}\)
= \(\frac{2^x\left({\displaystyle\frac12}+4\right)}{2^x\times{\displaystyle\frac12}}\)
= \(\frac{\displaystyle\frac{1+8}2}{\displaystyle\frac12}\)
= \(\frac92\times\frac21\)
= 9 is the required answer
Rationalize the denominator: \(\frac1{\sqrt[3]a-\sqrt[3]b}\)
Here,
\(\frac1{\sqrt[3]a-\sqrt[3]b}\)
Multiplying both numerator and denominator by \(\left(\sqrt[3]a\right)^2+\sqrt[3]a.\sqrt[3]b+\left(\sqrt[3]b\right)^2\), we get
\(\frac1{\sqrt[3]a-\sqrt[3]b}\times\frac{{\displaystyle\left(\sqrt[3]a\right)^2}{\displaystyle+}{\displaystyle\sqrt[3]a}{\displaystyle.}{\displaystyle\sqrt[3]b}{\displaystyle+}{\displaystyle{\displaystyle\left(\sqrt[3]b\right)}^2}}{\left(\sqrt[3]a\right)^2+\sqrt[3]a.\sqrt[3]b+\left(\sqrt[3]b\right)^2}\)
= \(\frac{\sqrt[3]{a^2}\times\sqrt[3]{ab}\times\sqrt[3]{b^2}}{\left(\sqrt[3]a\right)^3-\left(\sqrt[3]b\right)^3}\)
= \(\frac{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}{a-b}\)
is the required answer
If the sum of three consecutive even numbers is 36, find the numbers.
In the given figure, D is the mid-point of BC and DE⊥AC. If AC = 12 cm, and DE = 5 cm, find the area of triangle ΔABC
Solution:
Given:
D is the mid-point of BC, DE⊥AC
AC = 12 cm, DE = 5 cm
To find:
Area of triangle ΔABC
Here,
Hence the required area of triangle ΔABC is 60 cm^{2}
In the adjoining figure, O is the center of a circle. If angle ∠DAC = 25°, find the values of angle ∠BDC and angle ∠BAC.
Given:
O is the center of the circle
and ∠DAC = 25°
To find:
The values of the angles ∠BDC and ∠BAC
Here,
∴ ∠BDC = 65° and ∠BAC = 65° is the required answer
In the given figure, O is the center of circle, PG is tangent to the circle and G is the point of contact. If PG = DG, then the value of angle ∠DPG.
Given:
O is the center of circle, PG is tangent to the circle,
G is the point of contact and PG=DG
To find: The value of angle ∠DPG
Here,
∴ the value of ∠DPG is 45°, required answer.
In the given ΔXYZ, XY = 8 cm, YZ = 12 cm and the area of triangle ΔXYZ is 24 sq. cm, find the size of angle ∠XYZ
Given,
In triangle ΔXYZ, XY = 8 cm, YZ = 12 cm and area of triangle ΔXYZ is 24cm^{2}
To find: The size of angle ∠XYZ
Let,
angle ∠XYZ be θ
Now using formula
Area of triangle ΔXYZ = \(\frac12XY.YZ\sin\left(\theta\right)\)
Or, 24 = \(\frac12\cdot8\cdot12\cdot\sin\left(\theta\right)\)
Or, \(24=48\cdot\sin\left(\theta\right)\)
Or, \(\frac{24}{48}=\sin\left(\theta\right)\)
Or, \(\frac12=\sin\left(\theta\right)\)
Or, \(\sin\left(30^\circ\right)=\sin\left(\theta\right)\)
Or, 30° = θ
Hence the size of angle ∠XYZ is 30°
If the mean (\(\overline x\)) = 14 + p, the number of terms (N) = 30 and ∑fx = 600 in a continuous series. Find the value of p.
Find the probability of getting a number divisible by 3 or a square number when a fair dice is thrown.
Solution:
Let, the event of numbers divisible by 3 be A
and the event of square number be B.
Then,
S = {1, 2, 3, 4, 5, 6}
A = {3, 6}
B = {1, 4}
∴ n(S) = 6, n(A) = 2 and n(B) = 2
To find: n(A∪B)
Here,
A and B are mutually exclusive events. Then by formula
P(A∪B) = p(A) + p(B)
= \(\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}\)
= \(\frac26+\frac26\)
= \(\frac23\)
∴ The probability of getting a number divisible by 3 or a square number is 2/3.
From a bag containing 10 red and 6 black balls of the same shape and size, two balls are taken randomly in succession without replacement. Show the probabilities of all outcomes in a tree diagram.
In a survey of a group of 180 students, 50 students say to like cricket game only, 30 students say to like basketball game only and 50 students say do not like both games.
Let, the set of students who like cricket be C and the set of students who like basketball be B.
Then, n(U) = 180, n_{o}(C) = 50, n_{o}(B) = 30 and \(n\left(\overline{C\cup B}\right)\) = 50
To find: n(C):n(B) = ?
Let, n(C∩B) = x.
Now, representing above information in Venn-diagram we get,
From above Venn-diagram
n_{o}(C) + n_{o}(B) + n(C∩B) + \(n\left(\overline{C\cup B}\right)\) = n(U)
Or, 50 + 30 + x + 50 = 180
Or, 130 + x = 180
Or, x = 180 - 130
Or, x = 50
∴ n(C∩B) = x = 50
Here, n(C) = n_{o}(C) + n(C∩B) = 50 + 50 = 100
And, n(B) = n_{o}(B) + n(C∩B) = 30 + 50 = 80
So, n(C):n(B) = 100:80 = \(\frac54\) = 5:4
Hence, the ratio of the students who like cricket game and basketball game is 5:4
A tourist bought a Nepali cap with 20% discount and 13% value added tax. When returning to his country, the vat amount Rs. 83.2 returned back to him at the airport. What was the marked price of the cap? Find it.
Solution:
Given:
Rate of discount = 20%
Rate of VAT ( value added tax ) = 13%
Amount of VAT = Rs. 83.20
To find: Marked price (M.P)
Let, marked price of cap be Rs. x
By formula,
Selling price (SP) = MP - Discount % of MP
= x - 20% of x
= \(x-\frac{20}{100}\times x\)
= \(x-\frac x5\)
= \(\frac{5x-x}5\)
= \(\frac{4x}5\)
Again, by formula
Amount of VAT = VAT % of SP
Or, \(83.20=13\%\times\frac{4x}5\)
Or, \(83.20=\frac{\displaystyle13}{100}\times\frac{4x}5\)
Or, \(83.20=\frac{13x}{125}\)
Or, \(x=\frac{83.20\times125}{13}\)
Or, x = Rs. 800
Hence the marked price of the cap is Rs. 800 is the required answer
The ratio of the sides of a triangular shape of land is 5:6:7. If its perimeter is 1800 m, what is the area of the land? Find it.
Solution:
Given:
Perimeter of triangular piece of land (P) = 1800m
Ratio of the length of triangular sides (a:b:c) = 5:6:7
To find: Area of the triangular piece of land (A)
Let, the common unit for the length of sides be x.
The, a = 5x, b = 6x and c = 7x
We know that,
Perimeter of a triangle (P) = a + b + c
Or, 1800 m = 5x + 6x + 7x
Or, 18x = 1800 m
Or, x = 100 m
∴ a = 5x = 5*100 m = 500 m
b = 6x = 6*100 m = 600 m
c = 7x = 7*100 m = 700 m
Again,
Semi-peremeter (s) = half of the perimeter = \(\frac P2\)
= \(\frac{100}2\)
= 900 m
Now, by formula,
Area of triangle (A) = \(\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\)
= \(\sqrt{900\left(900-500\right)\left(900-600\right)\left(900-700\right)}\)
= \(\sqrt{21600000000m^4}\)
= \(\sqrt{\left(146969.38m^2\right)^2}\)
= 146969.38 m^{2}
Hence the area of triangular piece of land is 146969.38 m^{2} is the required answer
Find the HCF of a^{3} + 1 +2a^{2} + 2a, a^{3} - 1 and a^{4} + a^{2} + 1
Solution:
Here, first expression is
a^{3} + 1 +2a^{2} + 2a
= a^{3} + 1^{3} + 2a(a+1)
= (a+1)(a^{2} - a +1) + 2a(a+1)
= (a+1) (a^{2} - a +1 + 2a)
= (a+1) (a^{2} + a +1)
Second expression,
= a^{3} - 1
= a^{3} - 1^{3}
= (a-1)(a^{2} + a +1)
Third expression,
a4 + a2 + 1
= (a^{2})^{2} + (1)^{2} + a^{2}
= (a^{2} + 1)^{2} - 2a^{2} + a^{2}
= (a^{2} + 1)^{2} - a^{2}
= (a^{2} + a +1) (a^{2} - a +1)
Hence highest common factor (HCF) from all three expressions is (a^{2} + a +1)
Solve: \(\frac{6x-49}{7+\sqrt{6x}}=6-\frac{\sqrt{6x}{\displaystyle-}{\displaystyle11}}3\)
Here,
\(\frac{6x-49}{7+\sqrt{6x}}=6-\frac{\sqrt{6x}{\displaystyle-}{\displaystyle11}}3\)
Or, \(\frac{\left(\sqrt{6x}\right)^2-\left(7\right)^2}{\sqrt{6x}+7}=\frac{\displaystyle18-\left(\sqrt{6x}-11\right)}3\)
Or, \(\frac{\left(\sqrt{6x}+7\right)\left(\sqrt{6x}-7\right)}{\sqrt{6x}+7}=\frac{\displaystyle18-\left(\sqrt{6x}-11\right)}3\)
Or, \(3\left(\sqrt{6x}-7\right)=29-\left(\sqrt{6x}-11\right)\)
Or, \(3\sqrt{6x}-21=29-\sqrt{6x}\)
Or, \(3\sqrt{6x}+\sqrt{6x}=29+11\)
Or, \(4\sqrt{6x}=50\)
Or, \(\sqrt{6x}=\frac{\displaystyle50}4\)
Or, \(\sqrt{6x}=\frac{25}2\)
Squaring both sides we get,
\(\left(\sqrt{6x}\right)^2=\left(\frac{25}2\right)^2\)
Or, \(6x=\frac{625}4\)
Or, \(x=\frac{625}{24}\)
Or, \(x=26\frac1{24}\)
is the required answer.
Prove that the opposite angles of cyclic quadrilateral are supplementary.
Solution:
Given:
O is the center of the circle and ABCD is a cyclic quadrilateral.
To prove:
i. ∠ABC + ∠ADC = 180°
ii. ∠BAD + ∠BCD = 180°
Construction: Radii OA and OC are drawn
Proof:
Statement | Reasons | |
1. | ∠ABC = \(\frac12\) ∠AOC | ∠ABC is a circumference angle and ∠AOC is central angle standing on same arc ADC. |
2. | ∠ADC = \(\frac12\) reflex ∠AOC | Similar as reason 1. |
3. | ∠ABC + ∠ADC = \(\frac12\angle AOC+\frac12reflex\angle AOC\) | From statement 1. and 2. by addition axiom. |
4. | ∠AOC + reflex ∠AOC = 360° | Sum of angles at the center of the circle |
5. | ∠ABC + ∠ADC = \(\frac12\) 360° = 180° | From statements 3 and 4 by substitution axiom. |
6. | ∠ABC + ∠ADC + ∠BAD + ∠BCD = 360° | Sum of the angles of a quadrilateral. |
7. | 180° + ∠BAD + ∠BCD = 360° Or, ∠BAD + ∠BCD = 360° - 180° = 180° |
From statements 5 and 6 by substitution axiom |
8. | ∠ABC + ∠ADC = 180° and, ∠BAD + ∠BCD = 180° |
From statement 5 and 7 |
hence proved
Construct a triangle ΔABC in which AB = 5.4 cm, BC = 7 cm and angle ∠BAC = 60°. Also construct a parallelogram equal in area to the triangle ΔABC and having a side of length 7.2 cm
Given,
In ΔABC,
AB = 5.4 cm,
BC = 7 cm,
and ∠ABC = 60°
To construct: ΔABC of given data and a parallelogram equal in area to the triangle ΔABC and having a side of length 7.2 cm
hence ΔABC of given data and parallelogram CDEF equal in area to the triangle ΔABC and having side DE = CF = 7.2 cm is constructed.
Verify experimentally that the circumference angles of circle standing on the same arc are equal.
Statement: The circumference angles of a circle standing on the same arc are equal.
Construction figures:
Two circles of different radii with center 'O' are drawn. In each circle, circumference angle ∠ABC and ∠ADC are drawn which are standing on the same arc AC
To verify :- ∠ABC = ∠ADC
Method of verification:
By using a protractor, ∠ABC and ∠ADC are measured and the result are tabulated below.
Figure | ∠ABC | ∠ADC | Results |
i. | 65º | 65º | ∠ABC = ∠ADC |
ii. | 56º | 56º | ∠ABC = ∠ADC |
Conculsion:- hence, the circumference angles of a circle standing on the same arc are equal.
Hence verified
From the roof of a house, the angle of depression of the top of a tree 20 ft high was found to be 30°. If the distance between the house and the tree is 10√3 ft, find the height of the house.
Solution:
Let, AB be the height of the house CD be the height of a tree, GC = BD be the distance between the house and the tree, and ∠FAC be the angle of depression of the top of the tree from the house as shown in the figure below.
Then, according to the question
GB = CD = 20 ft, GC = BD = 10√3 ft, ∠FAC = 30°, AB = ?
here, in right angled triangle ΔAGC,
\(\tan\left(30^\circ\right)=\frac{AG}{GC}\)
Or, \(\frac1{\sqrt3}=\frac{AG}{10\sqrt3ft}\)
Or, \(\sqrt3AG=10\sqrt3ft\)
Or, \(AG=\frac{10\sqrt3}{\sqrt3}ft\)
Or, AG = 10 ft.
Now, AB = AG + GB = 10 ft + 20 ft = 30 ft
hence, the height of the house (AB) = 30 feet is the required answer.
If the median of the following data is 19, find the value of p
Age in Years | 6 - 12 | 12 - 18 | 18 - 24 | 24 - 30 | 30 - 36 | 36 - 42 |
No. of Students | 4 | 10 | p | 4 | 3 | 3 |
Solution:
Preparing the cumulative frequency table to calculate median from the given data, we get
Age in Years X |
No. of students f |
Cumulative frequency cf |
6 - 12 | 4 | 4 |
12 - 18 | 10 | 14 |
18 - 24 | p | 14+p |
24 - 30 | 4 | 18+p |
30 - 36 | 3 | 21+p |
36 - 42 | 3 | 24+p |
N = 24 + p |
Here, median (Md) = 19 lies in (18 - 24)^{th} class
∴ Median class = 18 - 24
Median (Md) = \(L+\frac{{\displaystyle\frac N2}-cf}f\times i\)
Where,
L = 18, \(\frac N2=\frac{24+p}2\), cf = 14, f = p, i = 6
∴ \(19=18+\frac{\left({\displaystyle\frac{24+p}2}-14\right)}p\times6\)
Or, \(19-18=\left(\frac{24+p-28}2\times\frac1p\right)\times6\)
Or, \(1=\frac{3\left(p-4\right)}4\)
Or, p = 3p - 12
Or, 2p = 12
Or, p = 6
Hence the required value of p is 6
A sum of Rs. 150000 amounts to Rs. 262500 at a certain rate of interest in 5 years. Find the sum of money that amounts to Rs. 198375 at the same rate of compound interest in 2 years.
Solution:
Here,
In the first case,
Principal (P) = Rs. 150000
Time (T) = 5 years
Simple amount (A) = Rs. 252600
Rate of interest (R) = ?
We know,
Simple Interest (I) = Simple Amount (A) - Principle (P)
= Rs. 252600 -Rs. 150000
= Rs. 112500
Now, using formula for simple rate (R)
R = \(\frac{I\times100}{P\times T}\)
= \(\frac{112500\times100}{150000\times5}\)
= \(\frac{1125}{75}\)
= 15
∴ Rate of simple interest (R) = 15% p.a. (per annum)
Again, according to the second condition,
Rate of interest (R) = 15% p.a,
Time (T) = 2 years,
Compound amount (CA) = Rs. 198375,
Principal (P) = ?
By formula,
\(CA=p\left(1+\frac R{100}\right)^T\)
Or, \(198375=p\left(1+\frac{15}{100}\right)^2\)
Or, \(198375=p\left(1+0.15\right)^2\)
Or, \(198375=p\left(1.15\right)^2\)
Or, \(198375=1.3225P\)
Or, P = 150000
Hence principal (P) is Rs 150000 is the required answer.
A water tank is formed with the combination of cylinder and hemisphere. The total height of the tank is 14 m and base area is 38.5 sq.m. If the tank is filled with water at the rate of 24 paisa per liter, what is the total cost to fill the tank with water. Find it.
Given:
Total height of the tank = 14 m
Area of the circular base = 38.5 m^{2}
To find: Per liter cost of water required to fill the tank (T)
By formula,
Area of the circular base = πr^{2}
Or, 38.5 = \(\frac{22}7\times r^2\)
Or, \(r^2=\frac{269.5}{22}\)
Or, \(r^2=\left(3.5m\right)^2\)
Or, r = 3.5 m
Hence the radius of cylindrical base is 3.5 m
Here, height of the cylindrical part = total height - radius of the hemispherical part = 14 m - 3.5 m = 10.5 m
Now, by formula,
Volume of the cylindrical part (V_{1}) = Area of the base x height
= 38.5 x 10.5
= 404.25 m^{3}
and volume of the hemispherical part
(V_{2}) = \(\frac23\mathrm{πr}^3\)
= \(\frac23\times\frac{22}7\times\left(3.5m\right)^3\)
= \(\frac{44\times42.875m^3}{21}\)
= 89.8333 m^{3}
So, total volume of the tank (V) = V_{1} + V_{2} = 404.25 m^{3} + 89.8333 m^{3} = 494.0833 m^{3}
We know that, 1m^{3} = 1000 liters
So, 494.0833 m^{3} = 494083.3 liters
∴ Total cost of water required to fill the tank (T) = per liter cost x quantity of water = 24 paisa x 494083.3 liters = 24/100 x 494083.3 liters = Rs. 118580 ans.
Find the value of x in : \(\sqrt{\frac x{1-x}}+\sqrt{\frac{1-x}x}=\frac{13}6\)
Here,
\(\sqrt{\frac x{1-x}}+\sqrt{\frac{1-x}x}=\frac{13}6\)
Or, \(\frac{\sqrt x}{\sqrt{1-x}}+\sqrt{\frac{1-x}x}=\frac{13}6\)
Or, \(\frac{\left(\sqrt x^2\right)+\left(\sqrt{1-x}\right)^2}{\sqrt{1-x\;}\sqrt x}=\frac{13}6\)
Or,\(\frac{\displaystyle x+\left(1-x\right)}{\sqrt{x-x^2}}=\frac{13}6\)
Or,\(\frac{\displaystyle x+1-x}{\sqrt{x-x^2}}=\frac{13}6\)
Or,\(\frac{\displaystyle1}{\sqrt{x-x^2}}=\frac{13}6\)
Squaring both sides, we get
\(\left(\frac{\displaystyle1}{\sqrt{x-x^2}}\right)^2=\left(\frac{13}6\right)^2\)
Or,\(\frac1{x-x^2}=\frac{169}{36}\)
Or,36=169(x-x^{2})
Or,36=169x-169x^{2
}Or,169^{2} - 169x+36=0
Or,169^{2} - (117 - 52)x + 36 = 0
Or,169^{2} - 117x - 52x + 36 = 0
Or,13x(13x-9)-4(13x-9) = 0
Or,(13x-9) (13x-4) = 0
Either, 13x-9=0
Or,13x=9
Or,\(x=\frac9{13}\)
Or, 13x - 4 = 0
Or, 13x=4
Or, \(x=\frac4{13}\)
In the figure, M is the mid point of AE. Prove that the area of triangle ΔABE is equal to the area of parallelogram ABCD.
Solution:
Given:
M is the mid point of AE and ABCD is a parallelogram
To prove: Area of triangle ΔABE = area of parallelogram ABCD
Construction: BM is joined.
Proof:
Statements | Reasons | |
1. | area of triangle ΔAMB = 1/2 area of parallelogram ABCD Or, Area of parallelogram ABCD = 2 x area of triangle ΔAMB |
Both are standing on the same base and between the same parallel lines. |
2. | area of triangle ΔAMB = 1/2 area of triangle ΔABE Or, area of triangle ΔABE = 2 x area of triangle ΔAMB |
Median BM bisects triangle ΔABE |
3. | area of triangle ΔABE = area of parallelogram ABCE | from statements 1 and 2, being double of the same area triangle ΔAMB |
4. | ∴ Area of triangle ΔABE = area of parallelogram ABCD | from statement 3 |
Hence proved
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