Find the length of the hypotenuse of an isosceles right-angled triangle whose area is 200 cm2.
Question: Find the length of the hypotenuse of an isosceles right-angled triangle whose area is $200 \mathrm{~cm}^{2} .$ Also, find its perimeter. [Given: $\sqrt{2}=1.41$ ] Solution: In a right isosceles triangle, base $=$ height $=a$ Therefore, Area of the triangle $=\frac{1}{2} \times$ base $\times$ height $=\frac{1}{2} \times a \times a=\frac{1}{2} a^{2}$ Further, given that area of isosceles right triangle = 200 cm2 $\Rightarrow \frac{1}{2} a^{2}=200$ $\Rightarrow a^{2}=400$ or, $a=\sqrt{400...
Read More →In the adjoining figure,
Question: In the adjoining figure, if AB = BC and BX = BY, then show that AX = CY Solution: We have, AB = BC (i) and BX = BY ,.,.(ii) According to Euclids axiom, if equals are subtracted from equals, the remainders are equal. So, on subtracting Eq. (ii) from Eq. (i), we get AB- BX = BC BY = AX = CY [from figure]...
Read More →Simplify:
Question: Simplify:(x2 2y2) (x+ 4y)x2y2 Solution: To simplify, we will proceed as follows: (x2 2y2) (x+ 4y)x2y2 $=\left[x^{2}(x+4 y)-2 y^{2}(x+4 y)\right] x^{2} y^{2}$ $=\left(x^{3}+4 x^{2} y-2 x y^{2}-8 y^{3}\right) x^{2} y^{2}$ $=x^{5} y^{2}+4 x^{4} y^{3}-2 x^{3} y^{4}-8 x^{2} y^{5}$ Thus, the answer is $x^{5} y^{2}+4 x^{4} y^{3}-2 x^{3} y^{4}-8 x^{2} y^{5}$...
Read More →Simplify:
Question: Simplify:x2(x+ 2y) (x 3y) Solution: To simplify, we will proceed as follows: x2(x+ 2y) (x 3y) $=\left[x^{2}(x+2 y)\right](x-3 y)$ $=\left(x^{3}+2 x^{2} y\right)(x-3 y)$ $=x^{3}(x-3 y)+2 x^{2} y(x-3 y)$ $=x^{4}-3 x^{3} y+2 x^{3} y-6 x^{2} y^{2}$ $=x^{4}-x^{3} y-6 x^{2} y^{2}$ Thus, the answer is $x^{4}-x^{3} y-6 x^{2} y^{2}$....
Read More →Look at the adjoining figure.
Question: Look at the adjoining figure. Show that length AH sum of lengths of AB + BC+CD. Thinking Process Apply the Euclids axiom, the whole is greater than the part, to show the given result. Solution: From the given figure, we have AB+ BC + CD = AD [AB, SC and CD are the parts of AD] Here, AD is also the parts of AH. According to Euclids axiom, the whole is greater than the part. i.e., AH AD. So, length AH sum of lengths of AB+ BC + CD....
Read More →It is knoyvn that x + y = 10 and that x = z.
Question: It is knoyvn that x + y = 10 and that x = z. Show that z + y = 10. Thinking Process Apply the Euclids axiom, if equals are added to equals, the wholes are equal, to show the given result. Solution: We have, x+y=10 .(i) and x = z (ii) According to Euclids axioms, if equals are added to equals, the wholes are equal. So, From Eq. (ii), x+y=z+y ..(iii) From Eqs. (i) and (iii), z+ y = 10....
Read More →Find the following product and verify the result for x = − 1, y = − 2:
Question: Find the following product and verify the result for x = 1, y = 2: $\left(\frac{1}{3} x-\frac{y^{2}}{5}\right)\left(\frac{1}{3} x+\frac{y^{2}}{5}\right)$ Solution: To multiply, we will use distributive law as follows: $\left(\frac{1}{3} x-\frac{y^{2}}{5}\right)\left(\frac{1}{3} x+\frac{y^{2}}{5}\right)$ $=\left[\frac{1}{3} x\left(\frac{1}{3} x+\frac{y^{2}}{5}\right)\right]-\left[\frac{y^{2}}{5}\left(\frac{1}{3} x+\frac{y^{2}}{5}\right)\right]$ $=\left[\frac{1}{9} x^{2}+\frac{x y^{2}}{1...
Read More →Find the area of a right-angled triangle, the radius of whose circumcircle measure 8 cm
Question: Find the area of a right-angled triangle, the radius of whose circumcircle measure 8 cm and the altitude drawn to the hypotenuse measures 6 cm. Solution: Given: Radius = 8 cmHeight = 6 cmArea=?In a right-angled triangle, the centre of the circumcircle is the mid-point of the hypotenuse. Hypotenuse $=2 \times$ (radius of circumcircle) for a right triangle $=2 \times 8$ $=16 \mathrm{~cm}$ So, hypotenuse $=16 \mathrm{~cm}$ Now, base = 16 cm and height = 6 cm Area of the triangle $=\frac{...
Read More →Two salesmen make equal sales during the month of August.
Question: Two salesmen make equal sales during the month of August. In September, each salesman doubles his sale of the month of August. Compare their sales in September. Solution: Let the equal sale of two salesmen in August be $x$. In September each salesman doubles his sale of August. Thus, sale of first salesman is $2 x$ and sale of second salesman is $2 x$. According to Euclid's axioms, things which are double of the same things are equal to one another. So, in September their sales are aga...
Read More →Attempt to prove Euclid's fifth
Question: Attempt to prove Euclid's fifth postulate using the other postulates and axioms led to the discovery of several other geometries. Solution: True All attempts to prove the fifth postulate as a theorem led to a great achievement in the creation of several other geometries. These geometries are quite different from Euclidean geometry and called non-Euclidean geometry....
Read More →Two distinct intersecting
Question: Two distinct intersecting lines cannot be parallel to the same line. Solution: True Since, it is an equivalent version of Euclid's fifth postulate....
Read More →"For every line L and for every point P
Question: "For every line $L$ and for every point $P$ not lying on a given line $L$, there exists a unique line $m$ passing through $P$ and parallel to $L^{n}$ is known as Playfair's axiom. Solution: True Since, it is an equivalent version of Euclid's fifth postulate and it is known as Playfair's axiom....
Read More →Find the following product and verify the result for x = − 1, y = − 2:
Question: Find the following product and verify the result for x = 1, y = 2:(x2y1) (32x2y) Solution: To multiply, we will use distributive law as follows: $\left(x^{2} y-1\right)\left(3-2 x^{2} y\right)$ $=x^{2} y\left(3-2 x^{2} y\right)-1 \times\left(3-2 x^{2} y\right)$ $=3 x^{2} y-2 x^{4} y^{2}-3+2 x^{2} y$ $=5 x^{2} y-2 x^{4} y^{2}-3$ $\therefore\left(x^{2} y-1\right)\left(3-2 x^{2} y\right)=5 x^{2} y-2 x^{4} y^{2}-3$ Now, we put $x=-1$ and $y=-2$ on both sides to verify the result. $\operato...
Read More →The statements that are
Question: The statements that are proved are called axioms. Solution: False Because the statements that are proved are called theorems....
Read More →The hypotenuse of a right-angled triangle measures 65 cm and its base is 60 cm.
Question: The hypotenuse of a right-angled triangle measures 65 cm and its base is 60 cm. Find the length of perpendicular and the area of the triangle. Solution: Hypotenuse = 65 cmBase = 60 cmIn a right-angled triangle, $(\text { Hypotenuse })^{2}=(\text { B ase })^{2}+(\text { P erpendicular })^{2}$ $\Rightarrow(65)^{2}=(60)^{2}+(\text { perpendicular })^{2}$ $\Rightarrow(65)^{2}-(60)^{2}=(\text { perpendicular })^{2}$ $\Rightarrow(\text { perpendicular })^{2}=(65-60)(65+60)$ $\Rightarrow(\te...
Read More →If a quantity B is a part of another
Question: If a quantity $B$ is a part of another quantity $A$, then $A$ can be written as the sum of $B$ and some third quantity C. Solution: True Since, it is one of the Euclid's axioms....
Read More →The things which are double of the
Question: The things which are double of the same thing are equal to one another. Solution: True Since, it is one of the Euclid's axioms....
Read More →The edges of a surface are curves.
Question: The edges of a surface are curves. Solution: False Because the edges of surfaces are lines....
Read More →Solve this
Question: $2 y-3 z=0$ $x+3 y=-4$ $3 x+4 y=3$ Solution: These equations can be written as $0 x+2 y-3 z=0$ $x+3 y+0 z=-4$ $3 x+4 y+0 z=3$ $D=\left|\begin{array}{ccc}0 2 -3 \\ 1 3 0 \\ 3 4 0\end{array}\right|$ $=0(0-0)-2(0-0)-3(4-9)$ $=15$ $D_{1}=\left|\begin{array}{ccc}0 2 -3 \\ -4 3 0 \\ 3 4 0\end{array}\right|$ $=0(0-0)-2(0-0)-3(-16-9)$ $=75$ $D_{2}=\left|\begin{array}{ccc}0 0 -3 \\ 1 -4 0 \\ 3 3 0\end{array}\right|$ $=0(0-0)-0(0-0)-3(3+12)$ $=-45$ $D_{3}=\left|\begin{array}{ccc}0 2 0 \\ 1 3 -4 ...
Read More →The boundaries of the solids are curves.
Question: The boundaries of the solids are curves. Solution: False Because the boundaries of the solids are surfaces....
Read More →Solve this
Question: $2 y-3 z=0$ $x+3 y=-4$ $3 x+4 y=3$ Solution: These equations can be written as $0 x+2 y-3 z=0$ $x+3 y+0 z=-4$ $3 x+4 y+0 z=3$ $D=\left|\begin{array}{ccc}0 2 -3 \\ 1 3 0 \\ 3 4 0\end{array}\right|$ $=0(0-0)-2(0-0)-3(4-9)$ $=15$ $D_{1}=\left|\begin{array}{ccc}0 2 -3 \\ -4 3 0 \\ 3 4 0\end{array}\right|$ $=0(0-0)-2(0-0)-3(-16-9)$ $=75$ $D_{2}=\left|\begin{array}{ccc}0 0 -3 \\ 1 -4 0 \\ 3 3 0\end{array}\right|$ $=0(0-0)-0(0-0)-3(3+12)$ $=-45$ $D_{3}=\left|\begin{array}{ccc}0 2 0 \\ 1 3 -4 ...
Read More →Euclidean geometry is valid
Question: Euclidean geometry is valid only for curved surfaces. Solution: False Because Euclidean geometry is valid only for the figures in the plane but on the curved surfaces, it fails....
Read More →Solve this
Question: $2 y-3 z=0$ $x+3 y=-4$ $3 x+4 y=3$ Solution: These equations can be written as $0 x+2 y-3 z=0$ $x+3 y+0 z=-4$ $3 x+4 y+0 z=3$ $D=\left|\begin{array}{ccc}0 2 -3 \\ 1 3 0 \\ 3 4 0\end{array}\right|$ $=0(0-0)-2(0-0)-3(4-9)$ $=15$ $D_{1}=\left|\begin{array}{ccc}0 2 -3 \\ -4 3 0 \\ 3 4 0\end{array}\right|$ $=0(0-0)-2(0-0)-3(-16-9)$ $=75$ $D_{2}=\left|\begin{array}{ccc}0 0 -3 \\ 1 -4 0 \\ 3 3 0\end{array}\right|$ $=0(0-0)-0(0-0)-3(3+12)$ $=-45$ $D_{3}=\left|\begin{array}{ccc}0 2 0 \\ 1 3 -4 ...
Read More →The base of a right-angled triangle measures 48 cm and its hypotenuse measures 50 cm.
Question: The base of a right-angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle. Solution: Base = 48 cmHypotenuse =50 cmFirst we will find the height of the triangle; let the height be 'p'. $\Rightarrow(\text { Hypotenuse })^{2}=(\text { base })^{2}+p^{2}$ $\Rightarrow 50^{2}=48^{2}+p^{2}$ $\Rightarrow p^{2}=50^{2}-48^{2}$ $\Rightarrow p^{2}=(50-48)(50+48)$ $\Rightarrow p^{2}=2 \times 98$ $\Rightarrow p^{2}=196$ $\Rightarrow p=14 \mathrm{~cm}$ Area o...
Read More →‘Lines are parallel, if they do not intersect’
Question: Lines are parallel,if they do not intersect is stated in the form of (a)an axiom (b)a definition (c)a postulate (d)a proof Solution: (b) 'Lines are parallel, if they do not intersect' is the definition of parallel lines....
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