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Ask QuestionPosted by Santosh Terli 4 years ago
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Santosh Terli 4 years ago
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Gaurav Seth 4 years ago
Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l.
Answer:
To construct: A line parallel to given line when perpendicular line is also given.
Steps of construction:
(a) Draw a line l and take a point P on it.
(b) At point P, draw a perpendicular line n.
(c) Take PX = 4 cm on line n.
(d) At point X, again draw a perpendicular line m.
It is the required construction.
Posted by Bhavana Jc 4 years ago
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∆∆∆©¶€¥€¥{°Π°}¥€¥€©¶∆∆∆ . 4 years ago
Posted by Likhith Manohar Vadapalli 4 years ago
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Gaurav Seth 4 years ago
Construction : Join AC
In 
DAC,
S and R are mid points of DA and DC respectively.
By mid point theorem
SR || AC and SR = 
...(1)
Similarly in BAC,
PQ || AC and PQ = ...(2)
From (1) and (2),
SR || PQ and SR = PQ
Hence PQRS is a parallelogram.
<div class="qsnDetails" style="margin:0px; padding:0px; text-align:start; -webkit-text-stroke-width:0px"> </div>Posted by Likhith Manohar Vadapalli 4 years ago
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Prashant Rai 4 years ago
Posted by Sankalp Badgaiyan 4 years ago
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Posted by Anchal Yadav 4 years ago
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Gaurav Seth 4 years ago
Let 100x = 0.323232.............. (i)
10000x = 32.323232............... (ii)
(ii) - (i) = 32.323232.... - 0.323232...
10000x - 100x = 32
9900x = 32
x = 32/9900
Making 32/9900 in simplest form
HCF of 32,9900 = 4
divide them with 4
32/9900 = 8/2475
Posted by Tanisha Dahiya 4 years ago
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Gaurav Seth 4 years ago
Variable
In Maths, a variable is an alphabet or term that represents an unknown number. The variables are specially used in the case of algebraic expression or algebra. The variable is a quantity that can be changed or which is not fixed according to the mathematical operation performed. Usually, in algebra, we express an unknown number using the term ‘x’ and ‘y’. But this is not particular, and we can use any alphabet.
Linear Equations
An equation is said to be linear equation in two variables if it is written in the form of ax + by + c=0, where a, b & c are real numbers and the coefficients of x and y, i.e a and b respectively, are not equal to zero.
For example, 10x+4y = 3 and -x+5y = 2 are linear equations in two variables.
Posted by Abhishek Singh 4 years ago
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Yogita Ingle 4 years ago
Ratio of sides
12: 17:25
let them be
12x,17x, 25x respectively
perimeter of a triangle = sum of all sides
540 = 12x,17x, 25x
540 = 54x
x = 10
all sides measure
12x = 12×10 = 120
17x = 17× 10 = 170
25x= 25 × 10 = 250
it's semipetimeter = 540/2 = 270
using heron's formula area of the triangle =√ (s)(s-a)(s-b)(s-c)}
where s is the semipetimeter and a,b,c
area the sides of the triangle.
√ {( 270)(270-120)(270-170)(270-250)}
= 9000cm^2.
Posted by Abhishek Singh 4 years ago
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Yogita Ingle 4 years ago
MidPoint Theorem Statement
The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”
Construction- Extend the line segment DE and produce it to F such that, EF=DE.
In the triangle, ADE, and also the triangle CFE
EC= AE —– (given)
∠CEF = ∠AED {vertically opposite angles}
EF = DE { by construction}
hence,
△ CFE ≅ △ ADE {by SAS}
Therefore,
∠CFE = ∠ADE {by c.p.c.t.}
∠FCE= ∠DAE {by c.p.c.t.}
and CF = AD {by c.p.c.t.}
The angles, ∠CFE and ∠ADE are the alternate interior angles. Assume CF and AB as two lines which are intersected by the transversal DF.
In a similar way, ∠FCE and ∠DAE are alternate interior angles. Assume CF and AB are the two lines which are intersected by the transversal AC.
Therefore, CF ∥ AB
So, CF ∥ BD
and CF = BD {since BD = AD, it is proved that CF = AD}
Thus, BDFC forms a parallelogram.
By the use of properties of a parallelogram, we can write
BC ∥ DF
and BC = DF
BC ∥ DE
and DE = (1/2 * BC).
Hence, the midpoint theorem is Proved.
Posted by Kumar Rohan 4 years ago
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Posted by Kumar Rohan 4 years ago
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Gaurav Seth 3 years, 11 months ago
Diameter = 14 cm, radius = 14/2 = 7 cm
Or
Given : Diameter is 14 cm.
To find : The volume of a sphere ?
Solution :
The volume of a sphere is
Diameter is 14 cm.
Radius is
Substitute the value,
Therefore, the volume of a sphere is 1427.33 cm³.
Posted by K. Vignesh Krishnamurthy 4 years ago
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N Ayana 4 years ago
Posted by Tanisha Jain 4 years ago
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Gaurav Seth 4 years ago
Let the angle be x
Complementary angles : A pair of angles whose sum is 90° are called complementary angles.
Complement of x = 90-x
We are given that an angle is 10 degree more than one third of its complement .
So,
Complement = 90-x = 90-60 = 30°
Hence The greater angle is 60°
Posted by Mohit Solanki 4 years ago
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Yogita Ingle 4 years ago
□ ABCD is a parallelogram
consider Δ ACD and Δ ABD
AC = BD .... (given)
AB = DC .... (opposite sides of parallelogram)
AD = AD .... (common side)
∴Δ ACD ≅Δ ABD (sss test of congruence)
∠ BAD = ∠ CDA .... (cpct)
∠BAD+∠CDA=180∘. [Adjacent angles of parallelogram are supplementary]
so ∠ BAD and ∠ CDA are right angles as they are congruent and supplementary.
Therefor, □ ABCD is a rectangle since a
parallelogram with one right interior angle is a rectangle.
Posted by Krupal Sonkusare 4 years ago
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Gaurav Seth 3 years, 11 months ago
It is given that
∠B = 35°
∠C = 65°
AP is the bisector of ∠CAB
We have to arrange AP, BPand CP in descending order.
In ΔACP we have
∠ACP = 65°
∠CAP = 40°(As AP is the bisector of ∠CAB
So AP > CP (Sides in front or greater angle will be greater) ........(1)
In ΔABP we have
∠BAP = 40°(As AP is the bisector of ∠CAB)
Since,
∠BAP >∠ABP
So BP > AP ..........(2)
Hence
From (1) & (2) we have
BP > AP > CP
Posted by Rishika Vijapure 4 years ago
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Aayushi Nagar 4 years ago
Posted by Anshi Sheth 4 years ago
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Pluto Physicist 4 years ago
Kumar Rohan 4 years ago
Aayushi Nagar 4 years ago
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Kumar Rohan 4 years ago
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Posted by Gulshan Sinku 4 years ago
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Meghna Thapar 4 years ago
A circle inscribed inside a triangle is called the incenter, and has a center called the incenter. A circled drawn outside a triangle is called a circumcircle, and it's center is called the circumcenter. Drag around the vertices of the triangle to see where the centers lie.
All the vertices of a triangle are equidistant from the circumcenter. In an acute-angled triangle, circumcenter lies inside the triangle. In an obtuse-angled triangle, it lies outside of the triangle. Circumcenter lies at the midpoint of the hypotenuse side of a right-angled triangle.
My Guide ... 4 years ago
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My Guide ... 4 years ago
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Gaurav Seth 4 years ago
Given:
Now the first step will be that we will split the middle term therefore,
=
=
=%2B%5Csqrt%7B2%7D(x%2B2%5Csqrt%7B2%7D))
=(x%2B2%5Csqrt%7B2%7D%20))
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